Will give brainliest please help. Worth lot of points

Are they similar or why are they not

Answer:

110°

Step-by-step explanation:

Angles don't change if both are same shape (gradient)

Answer:

Step-by-step explanation:

Calculate and compare:

Answer:

3

……………………………………………………………………

Answer:

The answer is 3

Step-by-step explanation:

Answer:

answer is y=-2x

Step-by-step explanation:

-2x because every time  the dot decreases

Answer:

  65.2 g

Step-by-step explanation:

1 ounce is approximately 28.35 grams. So, 2.3 ounces is ...

  (2.3 oz) × (28.35 g/oz) ≈ 65.2 g

The caterpillar weighs approximately 65.2 grams.

_____

Additional comment

The exact conversion is 1 oz = 28.349523125 g. For a 3-significant figure answer, we only need a conversion factor accurate to 4 significant figures.

Answer:

DE = BE = x

x = 29² - 21²

= √400

= 20

x = 20

y = 20² + 99²

= √10201

= 101

y = 101

perimeter of kite ABCD : 29 + 29 + 101 + 101

= 260

Check the picture below.

and if you check your Unit Circle, you'd notice that those angles have an specific value.

[tex]\cfrac{15}{sin(60^o)}~~ = ~~\cfrac{x}{sin(45^o)}\implies \cfrac{15}{~~ \frac{\sqrt{3}}{2}~~}~~ = ~~\cfrac{x}{~~ \frac{\sqrt{2}}{2}~~}\implies \cfrac{15}{1}\cdot \cfrac{2}{\sqrt{3}}~~ = ~~\cfrac{x}{1}\cdot \cfrac{2}{\sqrt{2}}[/tex]

[tex]\cfrac{30}{\sqrt{3}}=\cfrac{2x}{\sqrt{2}}\implies \cfrac{30\sqrt{2}}{\sqrt{3}}=2x\implies \cfrac{30\sqrt{2}}{2\sqrt{3}}=x\implies \cfrac{15\sqrt{2}}{\sqrt{3}}=x \\\\[-0.35em] ~\dotfill\\\\ \stackrel{\textit{rationalizing the denominator}}{\cfrac{15\sqrt{2}}{\sqrt{3}}\cdot \cfrac{\sqrt{3}}{\sqrt{3}}\implies \cfrac{15\sqrt{6}}{3}}\implies \boxed{5\sqrt{6}=x}[/tex]

Step-by-step explanation:

Multiply:

[tex] \frac{ - 5}{7} \times \frac{ - 21}{55} = \frac{105}{385} [/tex]

Divide each side by 35:

[tex] \frac{105 \div 35}{385 \div 35} = \frac{3}{11} [/tex]

Your simplified product is 3/11.

Answer:

1,870

Step-by-step explanation:

Answer:

-5 is your answer.

Step-by-step explanation:

[tex]=y+8>3\\\\=y+8-8>3-8\\\\[/tex] [Subtracting 8 in both sides]

[tex]=y>-5[/tex]

The distance between the flagpole and the building is the number of feet between them

The building is 162 feet from the flagpole

The given parameters are:

The flagpole's shadow is 50% longer than the flagpole.

So, the length (l) of the flagpole's shadow is:

[tex]l = 40 * (1 + 50\%)[/tex]

[tex]l = 60[/tex]

The length of the building's shadow (d) is then calculated as:

[tex]60 : 40 =d : 324[/tex]

Express as fraction

[tex]\frac{60}{40}= \frac{d}{324}[/tex]

[tex]1.50 = \frac{d}{324}[/tex]

Solve for d

[tex]d = 1.50 * 324[/tex]

[tex]d = 486[/tex]

The distance (x) of the building from the flagpole is then calculated as:

[tex]x = 486 - 324[/tex]

[tex]x = 162[/tex]

Hence, the building is 162 feet from the flagpole

Read more about distance and bearing at:

https://brainly.com/question/11744248

Answer:1 cup of chocolate chips for every 1and 3/4 cup of flour

so divide 4 and 2/3 x 1 and 3/4

turn them both into fractions

4 2/3 = 14/3

1 3/4 = 7/4

14/3 / 7/4 = 14/3 x 4/7 = 56/21 = 2 and 2/3 cups of chocolate chips are needed

Step-by-step explanation:

Answer:

y = 3

Step-by-step explanation:

3^2y - 1 = (9)(27)^y - 2

3^2y-1 = 243^y - 2

log(243)/log(3) = 5

2y - 1 = 5(y - 2)

2y - 1 = 5y - 10

-5y      -5y

-3y - 1 = -10

     + 1    +1

-3y = -9

-3      -3

y = 3

Answer:

c= gx/f -x

Step-by-step explanation:

You're Welcome haha

Answer:

c = 0.63(s)

Step-by-step explanation:

brainliest please<3

Answer:

43.16cm

Step-by-step explanation:

(1.2 x 3.6)/2=2.16

3.6 x 1=3.6

2.2 x 1=2.2

8 x 2=16

8 x 2.4=19.2

Answer:

83.25 cm

Step-by-step explanation:

1/4(x) cm

x= time

1/4(333)

or 333/4

her height changes

83.25 cm

Answer:

True because when you substitute the x coordinate to find the value of y =4 and the coordinate is 4

Answer:

True

Step-by-step explanation:

[tex]\left \{ {{-2x-3y=-10} \atop {-3x+y=7}} \right.[/tex]

Rearrange like terms to the same side of the equation:

[tex]\left \{ {{-2x-3y=-10} \atop {y=7+3x}} \right.[/tex]

Substitute into one of the equations:

[tex]-2x-3(7+3x)=-10[/tex]

Apply the Distributive Property:

[tex]-2x-21-9x=-10[/tex]

Combine like terms:

[tex]-11x-21=-10[/tex]

Rearrange variables to the left side of the equation:

[tex]-11x=-10+21[/tex]

Calculate the sum or difference:

[tex]-11x=11[/tex]

Divide both sides of the equation by the coefficient of variable:

[tex]x=-\frac{11}{11}[/tex]

Cross out the common factor:

[tex]x=-1[/tex]

Substitute into one of the equations:

[tex]y=7+3[/tex] × [tex](-1)[/tex]

Calculate:

[tex]y=4[/tex]

The solution of the system is:

[tex]\left \{ {{x=-1} \atop {y=4}} \right.[/tex]

Express solutions in ordered pairs:

[tex](-1,4)[/tex]

Determine whether [tex](-1,4)[/tex] and [tex](-1,4)[/tex] are equivalent:

True

Answer:

300

Step-by-step explanation:

I think it will be 300 because

People High-Fives

2. 1

3. 3

4. 6

5. 10

6. 15

7. 21

8. 28

9. 36

10. 45.

There fore 10 people's high five will be 45 keep repeating and finding out the high fives I got 300 but I'm not sure if it's correct

The required number of high fives is 300.

Given that,
There are 25 students and one teacher in the class. After an exam, everyone high-fives everyone else to celebrate how well they did. How many high-fives were there is to be determined.

In arithmetic, combination and permutation are two different ways of grouping elements of a set into subsets. In a combination, the components of the subset can be recorded in any order. In a permutation, the components of the subset are listed in a distinctive order.

Here,
There are 25 students
n = 25
For each high five, there would be at least 2 students,
r = 2

The number of high five is given by combination,
= 25 C 2
= 25 * 24 / 2
= 600 / 2
= 300

Thus, the required number of high fives is 300.

Learn more about permutations and combinations here: https://brainly.com/question/2295036
#SPJ2

Answer:

g(f(2)) = -9

Step-by-step explanation:

You work from the inside out...

To find g(f(2)), you must first find f(2).  f(2) = 3(2) - 9 = 6 - 9 = -3.

Now find g(-3).  g(-3) = -(-3)^2 = -9.

Answer:

63/8

Step-by-step explanation:

(7x8) + 7 = 63
keep the same dinominator

Answer: it would be 63/8 as an improper fraction

Step-by-step explanation:

Answer:

...

Step-by-step explanation:

WHat THE hAIL I DONT EVEN KNO WHAT UR ASKING MY BOY

Answer:

2 1/4

Step-by-step explanation:

how am i supposed to explain this (wait until someone else answers aswell im not that smart)

Answer:

20 sales.

Step-by-step explanation:

p(x) = x² + 18x – 160

Here p(x) represents the net pay.

Given that p(x), net pay is $600

x² + 18x – 160 = 600

x² + 18x – 160 - 600 = 0

x² + 18x – 760 = 0

x² +38x - 20x - 760 = 0

x(x+38) - 20(x+38) = 0

(x-20)(x+38) = 0

x = -38, 20

x = 20

Therefore, she makes 20 sales in order to make $600.

Answer: 20 sales.

Proof of validity is shown below.

Answer:

the answer for 2 is given answer for 1 is  trans prop of answer for 3 is knl and answer for 4 is  nkl

Step-by-step explanation:

Answer:

pretty sure its d :)

Step-by-step explanation:

Answer: D

Step-by-step explanation:

To solve for the inverse of an equation, the first step is to rewrite f(x) as y.

This should leave you with the equation  y=[tex]\frac{3x}{5}[/tex]+3

Then switch y and x, which should leave you with x=[tex]\frac{3y}{5}[/tex]+3

Lastly, solve for y.

This should give you the final answer of D or y=[tex]\frac{5(x-3)}{3}[/tex]

The picture below gives a step-by-step picture.

Solution:

To solve a mixed fraction multiplication problem, you must:

After reviewing how to multiply an improper fraction, let's solve the multiplication problem.

The solution to the multiplication problem is 7/11.

[tex] \bf \nrightarrow \: 1 \frac{1}{6} \times 6 \frac{6}{11} [/tex]

Convert the mixed fraction to improper fraction by Multiplying the denominator with the integer and adding the numerator to the product of denominator and integer

[tex] \sf \underbrace{1 \frac{1}{6} = \frac{7}{6} }[/tex]

Now, Multiply the fractions

[tex] \bf \: \frac{7}{6} \times \frac{6}{11} [/tex]

Multiply the numerators and denominators

[tex] \bf \nrightarrow \: \frac{7 \times \cancel 6}{ \cancel6 \times 11} [/tex]

[tex] \boxed{ \sf \: ans \multimap \: \frac{7}{11} }[/tex]