All categories

3 years ago

What is the translation for B’ (6,-5) B(-5,-2)

Mathematics

1 answer:

Elena L [17]3 years ago

5 0

Answer:

-7

Step-by-step explanation:

You might be interested in

Suppose △PKN≅△BGH△PKN≅△BGH. Which other congruency statements are correct? Select each correct answer. △NKP≅△HBG△NKP≅△HBG △KPN≅△

NeTakaya

Answer:

△NPK≅△HBG; and △KNP≅△GHB

Explanation:

Using the given congruence statement, we can determine corresponding pieces of the triangles:

P corresponds to B; K corresponds to G; and N corresponds to H.

This means that △NKP≅△HGB, not HBG.

△KPN≅△GBH, not BHG.

△NPK≅△HBG; this is correct.

△KNP≅△GHB; this is correct.

8 0

3 years ago

In a circle of radius 10 cm, a sector has an area of 40 (pi) sq. cm. What is the degree measure of the arc of the sector?

Rashid [163]

Answer:

144°

Step-by-step explanation:

First, find the area of the circle, with the formula A = $\pi$ r²

Plug in 10 as the radius, and solve

A = $\pi$ r²

A = $\pi$ (10²)

A = 100 $\pi$

Using this, create a proportion that relates the area of the sector to the degree measure of the arc.

Let x represent the degree measure of the arc of the sector:

$\frac{40\pi }{100\pi }$ = $\frac{x}{360}$

Cross multiply and solve for x:

100 $\pi$ x = 14400 $\pi$

x = 144

So, the degree measure of the sector arc is 144°

6 0

3 years ago

Read 2 more answers

What is the area of a sector that has a radius of 25 mm and has an angle measure of 50°?

Nimfa-mama [501]

The area of the sector is 86.81π mm²

<u>Explanation:</u>

Given:

Radius of the sector, 25mm

Angle, α = 50°

Area of the sector, A = ?

We know:

Area of the sector = $\frac{\alpha }{360} X \pi r^2$

On substituting the value, we get:

$A = \frac{50}{360} X \pi X (25)^2\\\\A = 86.81 \pi mm^2$

Therefore, the area of the sector is 86.81π mm²

3 0

3 years ago

Use distributive property to rewrite 3×546

sergij07 [2.7K]

Answer:

1638

Step-by-step explanation:

3(500)+3(40)+3(6)=1500+120+18=1638

4 0

3 years ago

Suppose that the functions r and s are defined for all real numbers x as follows.

mixas84 [53]

ANSWER

Given

$r(x) = x - 1$

and

$s(x) = 3 {x}^{2}$
ANSWER TO QUESTION 1

$(r + s)(x) = r(x) + s(x)$

$(r + s)(x) = x - 1 + 3 {x}^{2}$

$(r + s)(x) = 3 {x}^{2} + x - 1$

ANSWER TO QUESTION 2

$(r \times s)(x) = r(x) \times s(x)$

$(r \times s)(x) = (x - 1) \times 3 {x}^{2}$

$(r \times s)(x) = 3 {x}^{3} - 3 {x}^{2}$

ANSWER TO QUESTION 3

$( r - s)(x) = r(x) - s(x)$

$( r - s)(x) = x - 1 - 3 {x}^{2}$

$( r - s)( - 3) = (- 3) - 1 - 3 ({ - 3}^{2} )$

$( r - s)( - 3) = - 4 - 3 \times 9$

$( r - s)( - 3) = - 4 - 27 = - 31$

4 0

3 years ago