Ask question

Ask question

All categories

scoundrel [369]

3 years ago

8

On a coordinate plane, triangle A B C is shifted 4 units up and 3 units to the left to form triangle A prime B prime C prime. Tr

iangle ABC is reflected over the line y = 1. What are the coordinates of B’? (–2, 3) (–2, 5) (2, –3) (4, –3)

1 answer:

VARVARA [1.3K]3 years ago

7 0

Answer:

(–2, 5)

Step-by-step explanation:

I know its late now but here is the answer.

You might be interested in

Solve for T. B=RT-4/6

lions [1.4K]

I’m not sure which of these you meant to type so I answered both ways
B=(RT-4)/6 multiply by 6
6B= RT-4. Add 4
6B+4= RT. Divide by R
(6B+4)/R = T

Or B=RT -4/6. Multiply by 6
6B=6RT -4. Add 4
6B+4= 6RT. Divide by 6R
(6B-4)/6R = T. Reduce
(3B-2)/3R = T

8 0

3 years ago

Find the volume of a cylinder with height 12 cm and diameter 7 cm.

Svet_ta [14]

<span>461.81 cm^3 .................................</span>

3 0

3 years ago

At a large nursery, a border for a rectangular garden is being built. Designers want the border’s length to be 5 feet greater th

Salsk061 [2.6K]

Let

x--------> the border’s length

y--------> the border’s width

P--------> perimeter of the border

we know that

x=5+y------> equation 1

P=2*[x+y]-----> P=2x+2y

P <=180 ft

(2x+2y) <= 180-------> equation 2

substitute the equation 1 in equation 2

2*[5+y]+2y <= 180

10+2y+2y <= 180

4y <= 180-10

4y <=170

y <=42.5 ft

so

the maximum value of the width is 42.5 ft

for y=42.5 ft

x=42.5+5------> x=47.5 ft

the answer is

the width of the border is less than or equal to 42.5 ft

6 0

3 years ago

2 tan 30°<br>II<br>1 + tan- 300

shusha [124]

Question:

$\frac{2tan30^{\circ}}{1 + tan^2(30^{\circ})}$

Answer:

$\frac{2tan30^{\circ}}{1 + tan^2(30^{\circ})}$ $= sin(60^{\circ})$

Step-by-step explanation:

Given

$\frac{2tan30^{\circ}}{1 + tan^2(30^{\circ})}$

Required

Simplify

In trigonometry:

$tan(30^{\circ}) = \frac{1}{\sqrt{3}}$

So, the expression becomes:

$\frac{2tan30^{\circ}}{1 + tan^2(30^{\circ})}$ $= \frac{2 * \frac{1}{\sqrt{3}}}{1 + (\frac{1}{\sqrt{3}})^2}$

Simplify the denominator

$\frac{2tan30^{\circ}}{1 + tan^2(30^{\circ})}$ $= \frac{2 * \frac{1}{\sqrt{3}}}{1 + \frac{1}{3}}$

$\frac{2tan30^{\circ}}{1 + tan^2(30^{\circ})}$ $= \frac{\frac{2}{\sqrt{3}}}{1 + \frac{1}{3}}$

$\frac{2tan30^{\circ}}{1 + tan^2(30^{\circ})}$ $= \frac{\frac{2}{\sqrt{3}}}{ \frac{3+1}{3}}$

$\frac{2tan30^{\circ}}{1 + tan^2(30^{\circ})}$ $= \frac{\frac{2}{\sqrt{3}}}{ \frac{4}{3}}$

Express the fraction as:

$\frac{2tan30^{\circ}}{1 + tan^2(30^{\circ})}$ $= \frac{2}{\sqrt 3} / \frac{4}{3}$

$\frac{2tan30^{\circ}}{1 + tan^2(30^{\circ})}$ $= \frac{2}{\sqrt 3} * \frac{3}{4}$

$\frac{2tan30^{\circ}}{1 + tan^2(30^{\circ})}$ $= \frac{1}{\sqrt 3} * \frac{3}{2}$

$\frac{2tan30^{\circ}}{1 + tan^2(30^{\circ})}$ $= \frac{3}{2\sqrt 3}$

Rationalize

$\frac{2tan30^{\circ}}{1 + tan^2(30^{\circ})}$ $= \frac{3}{2\sqrt 3} * \frac{\sqrt{3}}{\sqrt{3}}$

$\frac{2tan30^{\circ}}{1 + tan^2(30^{\circ})}$ $= \frac{3\sqrt{3}}{2* 3}$

$\frac{2tan30^{\circ}}{1 + tan^2(30^{\circ})}$ $= \frac{\sqrt{3}}{2}$

In trigonometry:

$sin(60^{\circ}) = \frac{\sqrt{3}}{2}$

Hence:

$\frac{2tan30^{\circ}}{1 + tan^2(30^{\circ})}$ $= sin(60^{\circ})$

3 0

3 years ago

The glass container has a length of 5 in., a width of 4.5 in., and a height of 3 in. If the container is filled to a height of 2

eduard

100(2.25/3)=75%
............................

3 0

3 years ago