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3 years ago

13

Which transformation represents a reflection over the y=x line?

1 answer:

levacccp [35]3 years ago

6 0

Answer:

can you give me a picture of the graph?

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If the measure of F is 46 degrees. What does angle C measure?<br><br> A<br> E<br> B<br> D<br> C

melomori [17]

Answer:

46 degrees, they are equal. Hope that helped! (would appreciate being brainliest)

Step-by-step explanation:

3 0

3 years ago

A book is 6 inches wide and 9 inches tall.

Makovka662 [10]

Answer:

54

Step-by-step explanation:

Leave the 2.5 out

Multiply 9 × 6

So.....9 × 6 = 54

So the area will be 54 of the front of the book after the book is enlarged.

Hope this helps!

4 0

3 years ago

It takes Natalie five hours to tar a roof. Imani can tar the same roof in ten hours. How long would it take them if they worked

tresset_1 [31]

Answer:

C. 3.33 hours

Step-by-step explanation:

Use the algebra work equation:

$\frac{1}{tb}$ = $\frac{1}{t1}$ + $\frac{1}{t2}$ , where tb is the time to work together, t1 is the time it takes one person, and t2 is the time it takes the other person

Plug in the values we know:

$\frac{1}{tb}$ = $\frac{1}{5}$ + $\frac{1}{10}$

$\frac{1}{tb}$ = $\frac{4}{20}$ + $\frac{2}{20}$

$\frac{1}{tb}$ = $\frac{6}{20}$

20 = 6tb

3.33 = tb

So, it would take them 3.33 hours when working together.

5 0

3 years ago

The height of a stuntperson jumping off a building that is 20 m high is modeled by the equation h = 20 -57, where t is the time

cupoosta [38]

A stuntman jumping off a 20-m-high building is modeled by the equation h = 20 – 5t2, where t is the time in seconds. A high-speed camera is ready to film him between 15 m and 10 m above the ground. For which interval of time should the camera film him?

Answer:

$1\leq t\geq \sqrt{2}$

Step-by-step explanation:

Given:

A stuntman jumping off a 20-m-high building is modeled by the equation

$h =20-5t^{2}$ -----------(1)

A high-speed camera is ready to making film between 15 m and 10 m above the ground

when the stuntman is 15m above the ground.

height $h = 15m$

Put height value in equation 1

$15 =20-5t^{2}$

$5t^{2} =20-15$

$5t^{2} =5$

$t^{2} =1$

$t =\pm1$

We know that the time is always positive, therefore $t=1$

when the stuntman is 10m above the ground.

height $h = 10m$

Put height value in equation 1

$10 =20-5t^{2}$

$5t^{2} =20-10$

$5t^{2} =10$

$t^{2} =\frac{10}{5}$

$t^{2} =2$

$t=\pm\sqrt{2}$

$t=\sqrt{2}$

Therefore ,time interval of camera film him is $1\leq t\geq \sqrt{2}$

7 0

3 years ago

If triangle RST is within Quadrant 2 and csc T = 13/12, what is the value of cotT?

zhuklara [117]

Answer:

$\boxed {cot T = -\frac{5}{12}}$

Step-by-step explanation:

<u>Finding the missing side</u> :

x² = 13² - 12²
x² = 169 - 144
x = √25
x = 5

<u>Taking the cot value</u> :

cot T = adjacent / opposite
cot T = -5/12 (As cot is negative in Quadrant 2)

5 0

2 years ago

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