2 intersecting lines are shown. A line with points T, R, W intersects a line with points S, R, V at point R. Angle T R V is 95 degrees.
mAngleSRW =
°.

Answer:

Step-by-step explanation:

Answer: 50 + 0.25 is less than 90

Step-by-step explanation:

[tex]\angle R \cong \angle STV[/tex]: Alternate interior angles theorem

[tex]\angle TSV \cong \angle U[/tex]: Alternate interior angles theorem

[tex]\angle R \cong \angle TSV[/tex]: Transitive property of congruence

[tex]\angle R \cong \angle U[/tex]: Transitive property of congruence

Answer:

Below

Step-by-step explanation:

a^2 + b^2 = c^2       subtract  a^2 from both sides

b^2 =  c^2 - a^2    

   take the positive sqrt of both sides ( since you can't have a negative length)

b = sqrt (c^2 - a^2)    

Answer:

180-125 = x

x= 55 ( allied angle)

Hence, the slope which is perpendicular to the line to [tex]y = -x - 18[/tex] and passes through [tex](-6, 11)[/tex] is [tex]y=x+17[/tex].

What is the equation?

The definition of an equation is a mathematical statement that shows that two mathematical expressions are equal.

Here given that,

The line in slope-intercept form that is perpendicular to [tex]y = -x - 18[/tex] and passes through [tex](-6, 11)[/tex].

As we know the slope of line is

[tex]y=mx+c[/tex]

where [tex]m[/tex] is the slope and [tex]c[/tex] is the y intercept.

Given points are [tex](-6,11)[/tex].

If we want to describe the perpendicular line to the original line then the slope of line is

[tex]m_2=-\frac{1}{m_1}[/tex]

In our given slope of line

[tex]y = -x -18[/tex]

where [tex]m_1=-1[/tex]

So,

[tex]m_2=-\frac{1}{m_1}\\\\m_2=-\frac{1}{-1}\\\\m_2=1[/tex]

Now for the slope of second line we use the point slope from the construct equation then,

[tex]y-y_1=m_2(x-x_1)\\\\y-11=1(x-(-6))\\\\y-11=x+6\\\\y=x+6+11\\\\y=x+17[/tex]

Hence, the slope which is perpendicular to the line to [tex]y = -x - 18[/tex] and passes through [tex](-6, 11)[/tex] is [tex]y=x+17[/tex].

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Answer: As Devansh Sehta has calculated, there are 82 terms and  82=7×11+5.  When each term of  S  is divided by 7, the remainders, in order are 6, 3, 0, 4, 1, 5, 2, and these repeat.

Step-by-step explanation:

For any arithmetic sequence, record the remainders when successive terms are divided by n. Then the sum of every set of n consecutive remainders is divisible by 7. In this case, the first 77 terms add to a multiple of 7 as do the 5 remaining terms.

It can be seen that the radius and circumferance are not proportional.

        y = mx

        m = y/x

        [m] is called constant of proportionality.

Given is a graph as shown in the figure.

For the proportional relationship -

(25 - 18)/(4 - 3) should be equal to (38 - 25)/(6 - 4)

7/1 = 13/2

7/1 = 6.5

7 ≠ 6.5

Therefore, it can be seen that the radius and circumferance are not proportional.

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Answer:

<AFC

Step-by-step explanation:

Vertical angles are pairs of opposite angles that are created by two intersecting lines. As an example, the other pair of vertical angles are <CFD and <AFE.

Answer:

m=-6

Step-by-step explanation:

use slope formula to solve

m= rise/run=y2-y1/x2-x1

Using the ti-83 plus/ti-84 plus calculator, the 25th percentile of x is equal to 47.25 with a mean of 48 and a standard deviation of 12, respectively.

A dataset's mean is calculated by dividing the sum of all values by the total number of values. The term "average" is frequently used to describe it because it is the most widely used central tendency metric.

calculation

a) µ =     48    

σ =     12    

n=    115    

X =    45    

Z =(X - µ )/(σ/√n)  =     (45-48)/(12/√115)=    -2.68

P(X<45) = P(Z<-2.68) =         0.0037

b) µ =     48.00    

σ =     12.00    

n=    115.00    

P(X ≤ x) =    0.2500    

z value at  0.25=    -0.674        (excel formula =NORMSINV(0.25))

z=(x-µ)/(σ/√n)        

X=z * σ/√n +µ= -0.674*12/√115+48=        47.25

Using the ti-83 plus/ti-84 plus calculator, the 25th percentile of x is equal to 47.25 with a mean of 48 and a standard deviation of 12, respectively.

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Answer: It is an obtuse triangle.

Answer: No

Step-by-step explanation:

To know if a triangle is a right triangle, we can use Pythagorean Theorem to test.

Pythagorean Theorem: [tex]a^2+b^2=c^2[/tex]

[tex]4^2+6^2=8^2[/tex]                [exponents]

[tex]16+36=64[/tex]                [add]

[tex]52\neq 64[/tex]

Therefore, this is not a right triangle.

The given statements are:

We can then use the ASA congruence theorem to prove the triangles are congruent, aka they are mirror copies of one another.

Each "A" refers to facts (1) and (3) shown above. Fact (2) is the "S" of ASA.

ASA = angle side angle

If the equation x² + 16x + c = 0 has only one real solution, the value of c is equal to 64.

The quadratic formula includes the discriminant, which comes after the square root. The discriminant of a quadratic equation is important because it indicates the quantity and kind of solutions.  

The formula for the Discriminant of a Quadratic Equation ax²+bx +c = 0 is given by

Here we have

x² + 16x + c = 0 equation has only one real solution

Compare given equation with ax²+ bx + c = 0

=> a = 1, and b = 16  

As we know when an equation has only one real solution

then its Discriminant will be equal to zero

=> b² - 4ac = 0

By the above values,

=> 16² - 4(1)c = 0  

=> 256 - 4c = 0

=> 4c = 256

=> c = 256/4

=> c = 64

Therefore,

If the equation x² + 16x + c = 0 has only one real solution, the value of c is equal to 64.

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The speed of boat is 14 miles per hour and the speed of current 35 mile per hour

Speed:

The general formula to calculate the speed is written as

Speed = Distance/Time

Given,

A boat travel 42 mile uptream in 3 hour looking for a lot diver. It then travel downstream 70 mile in 2 hour.

Here we need to find the speed of boat and the speed of current.

While we looking into the given question, we have identified the following values,

Distance = 42 mile

time = 3 hour

Then the speed is calculated as,

=> speed = 42/3

=> speed = 14 miles per hour

Similarly, for the current is

distance = 70 miles

time = 2 hour

Then the speed is calculated as,

=> speed = 70/2 = 35 miles per hour.

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The monthly base fee of the situation is $0

From the question, we have the following parameters that can be used in our computation:

Charges per minute = $0.50 per minute

Number of minutes = 200 minutes

Total charges = $100

The above parameters mean that

Total charges = Monthly Base fee + Charges per minute x Number of minute

Substitute the known values in the above equation, so, we have the following representation

Monthly base fee + 0.50 * 200 = 100

This gives

Monthly base fee + 100 = 100

Subtract 100 from both sides

Monthly base fee = 0

Hence, the monthly base fee is $0

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On average, it is going to take DJ 7.27 hours to get to the beach.

Average refers to the quotient when a numerator is divided by a denominator.

In this situation, the average refers to the time, which is the quotient of the distance and speed.

The total distance going to the beach from his family = 800 km

The average speed = 110 km/hr

The time to cover the distance of 800 km at 110 km/hr = 7.27 hours (800/110).

Thus, DJ will take 7.27 hours to cover 800 km at 110 km/hr.

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Point ( 0 , -27 ) on the line and  ( -9 , 0 ) are not on the line

Given :

Madison is reeling in her string at a steady rate the linear equation 3y - 9x = -81 can be used to find y the number of feet of kite string she still needs to reel in after x seconds .

Points ( 0 , -27 )

when x = 0

3y - 9 * 0 = -81

3y - 0 = -81

3y = -81

y = -81 / 3

y = -27

so point ( 0 , -27 ) on the line .

when y = 0

3 * 0 - 9x = -81

0 - 9x = -81

-9x = -81

x = 81/9

x = 9

so ( -9 , 0 ) are not on the line

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The domain the function is In interval notation [7,  ∞). The range of the function is in the interval  (-1, ∞).

A mathematical phrase, rule, or law that establishes the link between an independent variable and a dependent variable (the dependent variable). A "vertical line test" can be used when examining a graph to determine whether a relation is a function. The relation is not a function if any point on the graph can be used to draw a vertical line that crosses the relation twice.

Here,

y=√(x-7)-1

x-7≥0

x≥7

at x=7

y=√(7-7)-1

y=-1

The function's domain is represented by the interval notation [7, ∞). The function's range falls within the range [-1, ∞).

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It is important to state the restrictions before simplifying rational expressions because the simplified expression may be defined for restrictions of the original.

What is rational expressions?

The ratio of two polynomials is a rational expression. F can be expressed in the form p/q, where p and q are polynomials, if f is a rational expression.

Fractional terms are seen in rational expressions. We include constraints because with some x values, the equation may become undefined. Undefined inquiries are hypothetical or ideal. The seven expressions for undefined words are  (∞-∞),∞^∞, N/0, 0⁰, 1^∞, ∞/∞ and 0×∞ respectively. N/0 is the most typical constraint on rational expressions. Any number divided by 0 is therefore undefined. For instance, if you substitute 0 for x in the case of the function f(x) = 6/x2, the outcome is 6/0, which is undefined. If you graphed this function, you would see a break at x=0.

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Answer:

Step-by-step explanation:

Given f(x) = √(3+x) and g(x) = √(3-x), you want the domains of various combinations of f and g.

The domain of each of these functions is the interval on which it is defined. The domain of the square root function is all argument values greater than or equal to zero.

The argument of the square root will be non-negative when ...

  3 +x ≥ 0

  x ≥ -3 . . . . . . subtract 3

  -3 ≤ x . . . . . . written using left-pointing arrow

The argument of the square root will be non-negative when ...

  3 -x ≥ 0

  3 ≥ x . . . . . add x

  x ≤ 3 . . . . . written using left-pointing arrow

Any combination of functions f and g will only be defined on the domain where both f and g are defined. That is, the domain will be the intersection of the domains of f and g:

  (-3 ≤ x) ∩ (x ≤ 3) = -3 ≤ x ≤ 3

The domain of f+g is the intersection of the domains of f and g.

  [-3, 3]

The domain of f-g is the intersection of the domains of f and g.

  [-3, 3]

The domain of f ·g is the intersection of the domains of f and g.

  [-3, 3]

The domain of f/g is the intersection of the domains of f and g, excluding the value of x that makes g(x) = 0. That value is x=3.

  [-3, 3)

__

Additional comment

The graphs of the combinations of functions are attached. You notice the graph of f/g tends to infinity as x → 3. It is undefined at x=3.

The equation of the straight line is 2y + 21x = 10. The correct option is (D).

To represent a straight line on a graph consider two points namely x and y intercepts of the line. To find x-intercept put y = 0 and for y-intercept put x = 0. Then draw a line passing through these two points.

The given points are (-4, 47) and (2, -16) respectively.

The equation of the straight line passing through them can be written as,

(y - 47)/(x - (-4)) = (-16 - 47)/(2 - (-4))

=> (y - 47)/(x + 4) = -63/6

=> (y - 47)/(x + 4) = -21/2

=> 2(y - 47) = -21(x + 4)

=> 2y - 94 = -21x - 84

=> 2y + 21x = -84 + 94

=> 2y + 21x = 10

Hence, the standard form of the straight line passing through given points is 2y + 21x = 10.

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The surface area, A, is changing at a rate of 288 km/min decreasing.

Given,

Base length, l = 6 km

Height, h = 10 km

dl/dt = -9 km/min

dh/dt = 12 km/min

Surface area of a prism, A = 2 × (lw + lh + wh)

For a square prism,

Length, l = width, w = 6km

A = 2 × (l² + lh + lh)

= 2(l²) + 4lh

A = 2(l²) + 4 lh

= 2 × (l² + 2lh)

dA/dt = 2 × (2I × dI/dt + 0 × dh/dt + 2 × 2h × dl/dt + 2l × dh/dt)

dA/dt = 2 × (2l × dl/dt + 0 + 2 × 2h × dl/dt + 2l × dh/dt)

= 2 × (2(6) × (-9) + 2(10) × (-9) + 2(6) × (12))

= 2 × (- 108 + -180 + 144)

= 2 × -144

= -288 km/min

The rate of change of the surface area, A is decreasing by 288 km/min.

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Question is incomplete. Completed question is given below;

The length s(t)s(t)s, (, t, )of the side of the base of a square prism is decreasing at a rate of 9 kilometers per minute and the height h(t)h(t)h, (, t, )of the prism is increasing at a rate of 12 kilometers per minute. At a certain instant t_0t 0 ​ t, start subscript, 0, end subscript, the base's side is 6 kilometers and the height is 10 kilometers. What is the rate of change of the surface area A(t)A(t)A, (, t, )of the prism at that instant?

The way Jason's position relative to the leader changed per minute is; 6²/₅ ft/minute

We are told that Jason's position was 11 ¹/₅ feet relative to the leader after 1 ³/₄minutes.

Now, the fractions are in mixed fraction form and so let us convert them to improper fractions to get;

⁵⁶/₅ and ⁷/₄

Thus, the way Jason's position relative to the leader changed per minute is; ⁵⁶/₅ ÷ ⁷/₄

= ⁵⁶/₅ * ⁴/₇

= 32/5

= 6²/₅ ft/minute

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Answer:

NO=6

Step-by-step explanation:

If MP=16 and NP=13 that means MN=3 and MO=9 so that makes NO=6

Answer:

[tex]y=\frac{1}{2} x-2[/tex]

Answer:

The amount of goo remaining after 32 minutes is approximately 54.7 grams.

Step-by-step explanation:

The half-life of a substance is the amount of time it takes for half of the substance to decay. To find the half-life of the goo in the experiment, we need to solve the equation G ( t ) = G ( 0 ) / 2 for t , where G ( 0 ) is the initial amount of goo and G ( t ) is the amount of goo remaining after time t .In this case, G ( 0 ) = 176 grams and G ( 135 minutes ) = 11 grams, so we can write the equation as follows:11 grams = 176 grams / 2Solving for t , we get:t = 135 minutes * ( 1 / 2 ) = 67.5 minutesThis is the half-life of the goo in the experiment.To find the amount of goo remaining after time t , we can use the equation G ( t ) = G ( 0 ) * ( 1 / 2 ) ^ ( t / t1/2 ) , where G ( 0 ) is the initial amount of goo, t is the time at which we want to find the amount of goo remaining, and t1/2 is the half-life of the goo.In this case, G ( 0 ) = 176 grams, t = 32 minutes, and t1/2 = 67.5 minutes, so we can plug these values into the equation to find the amount of goo remaining after 32 minutes:G ( 32 minutes ) = 176 grams * ( 1 / 2 ) ^ ( 32 minutes / 67.5 minutes )This equation tells us that the amount of goo remaining after 32 minutes is approximately 54.7 grams.