can you help me solve this equation? 20 POINTS

Can You Help Me Solve This Equation? 20 POINTS

Answers

Answer 1

Answer:

x = 63

Step-by-step explanation:

In [tex]\odot \:O, [/tex] ML and MN are tangents from external points M at points L and N respectively. OL and ON are radii of the circle.

[tex]\implies ML\perp OL,\:\&\: MN \perp ON[/tex] (By tangent theorem)

[tex]\implies m\angle MLO = m\angle MNO = 90\degree[/tex]

[tex]m\angle LON = 117\degree[/tex] (Given)

In quadrilateral LMNO, by interior angle sum theorem, we have:

[tex] m\angle LON+m\angle MOL +m\angle MON +m\angle LMN= 360\degree[/tex]

[tex] \implies 117\degree+90\degree +90\degree +x\degree= 360\degree[/tex]

[tex] \implies 297\degree+x\degree= 360\degree[/tex]

[tex] \implies x\degree= 360\degree-297\degree[/tex]

[tex] \implies x\degree= 63\degree[/tex]

[tex] \implies\red{\boxed{\bold{ x= 63}}}[/tex]
Answer 2

Answer:

63°

Step-by-step explanation:

The central angle is 117°The angles at the tangent are 90° each (property of tangents)

Applying the Angle Sum of Quadrilaterals,

117° + 2(90°) + x° = 360°180° + x° = 243°x° = 63°

Related Questions

Please help. I dont understand a thing. lol

Answers

Answer:

39 in²

Step-by-step explanation:

Area of triangle = 1/2 * b * h, where,

b = Base of Triangle

h = Height of triangle

Area of 4 triangles = 4 * 1/2 * 3 in * 5in

Area of 4 triangles = 2 * 3 in * 5 in

Area of 4 triangles = 30 in²

Area of square = a², where,

a = Each side of square.

Area of square = (3 in) ²

Area of square = 9 in²

Total surface area of pyramid = 30 inch² + 9 inch²

Total surface area of pyramid = 39 in²

Answer:

39 in^2 (Please mark me brainliest)

Step-by-step explanation:

3*3+3*2*5

=39

HELP I DONT GET THIS AND ITS DUE IN AN HOUR! PLEASE SHOW YOUR WORK IN PART B BECAUSE I DONT KNOW WHAT TO DO! I’LL GIVE BRAINLIEST AND 20 POINTS!!

Last year’s winner set a school record by completing 37 sit-ups. Amelia wants to know how many weeks she will need to train to beat this record.

Part A

Why is Amelia’s goal of breaking the school record best described by an inequality? Explain.

Part B

Write the inequality and graph the solution. How many weeks of training
will it take for Amelia to beat the record? Show your work.

Answers

An inequality is used to show two values that are different and to show that the two values are unequal.

What is an Inequality?

This refers to the mathematical relation that is used to show the comparison between two numbers that are unequal.

We can note that because there is a previous record, for a person to beat the record, he would have to perform better than the record and be faster than the previous record holder.

Therefore, if a record of Amelia was given, this would be used to find the relation between her record and the current record of 37 sit-ups and how many weeks of training would be needed to break the record.

Hence, we can see that the question is incomplete so I gave you a general overview to help you get a better understanding.

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Which graph represents the function 3x – 2y = 6?

Answers

Answer:

graph 19 because since it is that means also like the thingy equals that number si then like when the thing is at the highest point it cab maybe actually it will equal that number

Which of the following has the same value as the expression?

(-3+5)+2

Answers

Answer:

Answer(s):

1x4

2+2

2x2

Step-by-step explanation:

Hope this helps!

8.
A circular plate has a diameter of 11 inches. An ant walking along the edge makes 1.5 revolutions in 4
minutes. Approximately how many inches does the ant travel in one minute? Round your answer to the
nearest tenth of an inch.

Answers

Answer:

[tex]12.95 \: inches[/tex]

Step-by-step explanation:

We were told that the diameter of the circular plate is 11 inches. In order to find the distance the ants travel to make one revolution we have to first find the circumference of the circular plate. Circumference is:

[tex]2\pi \: r[/tex]

The radius would be half of the diameter (11). Therefore the radius is 5.5.

[tex]r = 5.5 \: \: \: \: \: \pi = 3.14 \\ cir = 2(3.14) \: (5.5) \\ cir = 6.28 \times 5.5 \\ cir = 34.54[/tex]

Since we found the circumference of the circular plate that means that we know how much one revolution made by the ants would be (34.54). Since the ant made 1.5 revolutions we have to multiply 34.54 by 1.5.

[tex]34.54 \times 1.5 = 51.81[/tex]

So we now know that the ant walked 51.81 inches in 4 minutes, now we have to find the distance he would walk in 1 minute.

Let x be the unknown value*

[tex] 51.81 \:inches = 4 \:minutes \\ x \:inches = 1 \: minutes\\ \\ 51.81=4x \\ \\\frac{51.81}{4} = x \\ \\ 12.95 = x[/tex]

x² + y² = 8
x - y = 0
Select all of the following that are solutions to the system shown.
O (2, 2)
O (2,-2)
O (-2,2)
0 (-2,-2)

Answers

Answer:

[tex](2,2)~ \text{and}~ (-2,-2)[/tex]

Step by step explanation:

[tex]x^2 +y^2 =8~~~~~~...(i)\\\\x-y=0~~~~~~~~~...(ii)\\\\\text{From (ii):}\\\\x-y = 0 \implies x = y\\\\\text{Substitute x = y in eq (i):}\\\\~~~~~~y^2 +y^2 =8\\\\\implies 2y^2 = 8\\ \\\implies y^2 = \dfrac 82\\\\\implies y^2 = 4\\\\\implies y = \pm\sqrt 4\\\\\implies y = \pm 2\\\\\text{Substitute y = x in equation (i):}\\ \\~~~~~~x^2 +x^2 = 8\\ \\\implies 2x^2 = 8\\\\\implies x^2 = \dfrac 82\\\\\implies x^2 = 4\\\\\implies x = \pm\sqrt 4\\\\\implies x = \pm2\\\\[/tex]

[tex]\text{Hence,}~ (x,y) = (\pm2, \pm 2)[/tex]

Write the expression in repeated multiplication form. Then write the expression as a power.
(6^4)^2
The expression in repeated multiplication form is
The expression as a power is

Answers

Answer:

Step-by-step explanation:

a. (6)^(4*2) or (6^4)(6^4)(6^4)(6^4)

b. 6^8

The radius of circle a is 5 cm. the radius if circle b is 5 cm greater than the radius of circle a. the radius of circle c is 3 cm greater than the radius of circle b. the radius of circle d is 1 cm less than the radius of circle c. what is the area of each circle? how many times greater than the area of circle a is the area of circle d? the area of circle a is ___ cm2.

Answers

Answer:

Theres a lot going on in this question but i tried my best to answer it all, correctly. Sorry if something is wrong or i misread the question. check explanation for answer.

Step-by-step explanation:

I'm not sure to use pi or 3.14 so i just did both.

Circle A's radius = 5cm

Circle B's radius = 10cm

Circle C's radius = 13cm

Circle D's = 12cm

(circle A)

π = 3.14

A = πr²

A = 3.14(5²)

A = 3.14(25)

A = 78.5cm²

π = π

A = πr²

A = π(5²)

A = π(25)

A = 78.54cm²

(circle B)

π = 3.14

A = πr²

A = 3.14(10²)

A = 3.14(100)

A = 314cm²

π = π

A = πr²

A = π(10²)

A = π(100)

A = 314.16cm²

(circle C)

π = 3.14

A = πr²

A = 3.14(13²)

A = 3.14(169)

A = 530.7cm²

π = π

A = π(13²)

A = π(169)

A = 530.9cm²

(circle D)

π = 3.14

A = πr²

A = 3.14(12²)

A = 3.14(144)

A = 452.2cm²

π = π

A = πr²

A = π(12²)

A = π(144)

A = 452.4cm²

452.2 / 78.5 =5.8

circle d is 5( almost 6) times greater than circle A

Aldo is a software salesperson. His base salary is $2500 and he makes $80 more for every copy of English is Fun he sells. His total pay, p (in dollars), after selling c copies is given by the following.

p=80c+2500

Answer the following questions.

(a)
If Aldo's total pay is $5140, how many copies did he sell?
? copies
(b)
What is Aldo's total pay if he sells 23 copies?
$

Answers

Answer:

A: He sold 33 copies

B: $4340

Step-by-step explanation:

A: p=80(33) + 2500

   p=2640+ 2500

   p=5140

B: p = 80(23) + 2500

   p=1840+ 2500

   p=4340

Line segment EF begins at (-1,5) and ends at (-1, -3). The segment is translated to the right 4 units and down
2 units to form line segment E'F.
Enter the length, in units, of line segment E'F.

Answers

The coordinates for the line segment E'F' are (3, 3) and (3, -5) and the length of the segment E'F' is 8 units.

What is geometric transformation?

It is defined as the change in coordinates and the shape of the geometrical body. It is also referred to as a two-dimensional transformation. In the geometric transformation, changes in the geometry can be possible by rotation, translation, reflection, and glide translation.

Line segment EF begins at (-1, 5) and ends at (-1, -3). The segment is translated right 4 units and down 2 units to form line segment E'F'. Then the coordinate of E' and F' will be:

E' = (-1 + 4, 5 – 2) = (3, 3)

F' = (-1 + 4, -3 - 2) = (3, -5)

[tex]\rm d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}[/tex]

The length of the E'F' = √(0 + 64) = 8 units

Thus, the coordinates for the line segment E'F' are (3, 3) and (3, -5) and the length of the segment E'F' is 8 units.

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I need help!! Thanks

Answers

[tex]\text{Given that,}~ f(x) = 3x+5~ \text{and}~ g(x)=\sqrt{x^2 +3}\\\\(g \circ f)(-1)\\\\=g(f(-1))\\\\=g(3(-1) +5)\\\\=g(-3+5)\\\\=g(2)\\\\=\sqrt{2^2+3}\\\\=\sqrt{4+3}\\\\=\sqrt{7}[/tex]

what is the gfc of 8 and 24

Answers

the greatest common factor is 8

use Demoiure's Theorem to find (-2-2√3i)³

Answers

Let z = -2 - 2√3 i.

Find the modulus of z :

|z| = √((-2)² + (-2√3)²) = √16 = 4

Find the argument of z (note that z lies in the third quadrant):

arg(z) = arctan(2√3/2) - π = arctan(√3) - π = -2π/3

Then in polar form, we have

[tex]z = -2 - 2\sqrt 3\, i = 4 \exp\left(-i\dfrac{2\pi}3\right)[/tex]

By DeMoivre's theorem, the cube of z is

[tex]z^3 = \left(4 \exp\left(-i\dfrac{2\pi}3\right)\right)^3 = 4^3 \exp(-i 2\pi) = \boxed{64}[/tex]

Select all the correct answers.

If the measure of angle is 0 is 7pi/4, which statements are true?
The measure of the reference angle is 45°.
The measure of the reference angle is 30°.
tan(0) = -1
The measure of the reference angle is 60°.
cos(0) = -√2/2
sin(0) = -√2/2

Answers

The statement(s) that are true is that the measure of the reference angle is 45°.

What is the conversion of radians to degrees?

The conversion of an angle in radian to a degree can be determined by the multiplication of the given angle in radian by 180/π.

From the given information:

[tex]\mathbf{\theta = \dfrac{7 \pi}{4}}[/tex]

To degrees, we have:

[tex]\mathbf{\theta = \dfrac{7 \pi}{4} \times \dfrac{180}{\pi}}[/tex]

[tex]\mathbf{\theta =315^0}[/tex]

A reference angle is an angle that measures the remaining distance (called the terminal distance) to the x-axis.

315° is in the 4th quadrant, i.e.

= 360° - 315°

= 45°

Therefore, the statement(s) that is true is:

The measure of the reference angle is 45°.

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Answer:

sin(0)=-√2/2

tan(0) = -1

The measure of the reference angle is 45°

Step-by-step explanation:

I took the test

pls help odd numbers only

Answers

Answer:

The question asks, "find the slope".

9) -4/15

10) -1/14

11) -3/4

12) -1/4

13) 11/17

14) -18/13

15) 0

16) -30

17) -5

19) -1/5

21) 1/4

23) -1

25) -2/3

A piece of ribbon is 63 inches long. it is cut into pieces that are each 7 inches long. how many pieces of ribbon are there?

Answers

63/7 = 9

Answer is 9 ribbons

Find the zeroes of the polynomial 100x^2 - 81

Answers

Answer:

x = ±0.9

Step-by-step explanation:

The algebraic identity to be used here :

a² - b² = (a + b)(a - b)

We know that :

100x² = (10x)²81 = 9²

Therefore,

100x² - 81 (10x + 9)(10x - 9)

The zeros are :

10x + 9 = 0 ⇒ 10x = -9 ⇒ x = -9/10 ⇒ x = -0.910x - 9 = 0 ⇒ 10x = 9 ⇒ x = 9/10 ⇒ x = 0.9

Answer:

[tex]x=-\dfrac{9}{10} \quad x=\dfrac{9}{10}[/tex]

Step-by-step explanation:

[tex]\begin{aligned}100x^2-81 & =(10^2)x^2-9^2\\ & = (10x)^2-9^2\end{aligned}[/tex]

[tex]\textsf{Difference of Two Squares Formula}: \quad a^2-b^2=\left(a+b\right)\left(a-b\right)[/tex]

[tex]\implies a=10x \: \textsf{ and } \: b=9[/tex]

[tex]\implies 100x^2-81=(10x+9)(10x-9)[/tex]

[tex]\begin{aligned}\textsf{To find the zeros}: \quad 100x^2-81 & = 0\\\\\implies (10x+9)(10x-9) & =0\\\implies 10x+9 & =0 \implies x=-\dfrac{9}{10}\\\implies 10x-9 & =0 \implies x=\dfrac{9}{10}\end{aligned}[/tex]

If alpha and beta are zeroes of polynomial fx = 4x2-5x-1. Pls answer full questiob

Answers

...................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................I like 15 yr old girls.

There are three sets. A = {1,2,3}, B = {4,5} and C = {6,7,8}

What is the probability of A∩B∩C

Answers

The three sets are independent sets and the value of the probability of A∩B∩C is: 0

How to determine the probability?

The sets are given as:

A = {1,2,3}, B = {4,5} and C = {6,7,8}

The total number of elements in the three sets is:

Total = 8

The number of elements of A∩B∩C is 0, because the sets do not have a common element.

So, the probability of A∩B∩C is:

P = 0/8

Evaluate

P = 0

Hence, the probability of A∩B∩C is: 0

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A truck's engine needs 1 gallons of oil for an oil change. How many quarts of oil are needed? (Hint: 1 quart
gallon)

Answers

Answer:

4 quarts

Step-by-step explanation:

1 gallon is equal to 4 quarts

Help?????????????????

Answers

Answer:

please make the image less blurry

Step-by-step explanation:

Can someone help me with this

Answers

Ans57 , 58 , 59

Step-by-step explanation:

A water tank has a radius 5 feet and height 11 feet. How much water can it hold?

Answers

Answer:

I think it is 275πcm^3 or 863.94cm^3

Step-by-step explanation:

please correct me if I'm wrong

Can someone please help me

Answers

Answer:

1. a=3

2. a=3

3. a=9

Step-by-step explanation:

1. 9/a+9-4=12-56/8+3

We move all terms to the left:

9/a+9-4-(12-56/8+3)=0

Domain of the equation: a=0

a∈R

We add all the numbers together, and all the variables

9/a+9-4-8=0

We add all the numbers together, and all the variables

9/a-3=0

We multiply all the terms by the denominator

-3*a+9=0

We add all the numbers together, and all the variables

-3a+9=0

We move all terms containing a to the left, all other terms to the right

-3a=-9

a=-9/-3

a=+3

6. Which statement is justified by the
transitive property of equality?
F. (x+y)+ z = x + (y+z)
G. If 7x = 14, then 14 = 7x
H. If -3x = y, and y = 8z, then -3x = 8z
J. x(y-z) = xy - xz

Answers

Answer:

h

Step-by-step explanation:

The equation of the trend line is y = -0.36x + 12.6 use the equation to of the trend line to predict the wind chill for a wind speed of 23 mi/h round your answer to the nearest degree. The wind chill at 23 mi/h​

Answers

Answer:

4°C

Explanation:

Equation: y = -0.36x + 12.6

To find wind chill at 23 mi/hInsert x = 23

Simplify

y = -0.36(23)+12.6

y = 4.32°C

y = 4°C       (rounded to nearest degree)

PLEASE HELP ILL MARK BRAINEST!!!!

Answers

Answer:

1, 6, 11, and 16 in that order

Step-by-step explanation:

every time the x value increases by 1, the f(x) value increased by 5

find arc JK

BRAINLIEST,POINTS,THANKS AND RATING GIVEN!!!

Answers

Answer:

164°

Step-by-step explanation:

By the property of intersecting chords, we have:

[tex]m\angle JNK =\frac{1}{2}[m\widehat {JK}+m\widehat{LM}][/tex]

[tex]\implies 2*125\degree =(7x+31)\degree+86\degree[/tex]

[tex]\implies 250\degree =(7x+31)\degree+86\degree[/tex]

[tex]\implies 250\degree -86\degree=(7x+31)\degree[/tex]

[tex]\implies 164\degree=(7x+31)\degree[/tex]

[tex]\implies 164=7x+31[/tex]

[tex]\implies 164-31=7x[/tex]

[tex]\implies 133=7x[/tex]

[tex]\implies x = \frac{133}{7}[/tex]

[tex]\implies x = 19[/tex]

[tex]m\widehat{JK}=(7x+31)\degree [/tex]

[tex]\implies m\widehat{JK}=(7*19+31)\degree [/tex]

[tex]\implies \huge{\orange{m\widehat{JK}=164\degree}} [/tex]

Its a math problem Please help me !!! For the data sets below, which of the following statements is FALSE?

Answers

Answer:

fist bubble i am pretty sure

Step-by-step explanation:

if u want to u can answer question 16 and 17 but if u don’t just answer 16

Answers

Answer:

the answer to 16 is C i know cause i had the same thing and the answer for 17 is A.

Step-by-step explanation:

Hope thi helps

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