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2 years ago

Help picture below please

Mathematics

1 answer:

s2008m [1.1K]2 years ago

3 0

Answer:

B. NRS

Step-by-step explanation:

You have to put the letters left to right.

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A^2−8ab+16b^2=81, find a–4b

neonofarm [45]

Answer:

-9, 9

Step-by-step explanation:

a² − 8ab + 16b² = 81

(a - 4b)² = 81 Wrote the left-hand side as a square

a - 4b = ±9 Took the square root of each side

All values of a - 4b are -9 and +9.

4 0

2 years ago

There is less gravity on the planet Trang then on earth. In fact, you could jump about 2 2/3 times as high on Trang as planet ea

nasty-shy [4]

1/4 times 2 2/3 = 2/3

Change from a mixed number to improper fraction: 1/4 times 8/3

Multiply across so 1x8/4x3 to get 8/12

Simplify to 2/3

8 0

2 years ago

What is the value of x? Leave your answer in simplest radical form.

BigorU [14]

the correct answer should be 2radical 13

8 0

2 years ago

Read 2 more answers

Find the are of the SHADED region.

Mazyrski [523]

First find the area of the parallelogram

Area of parallelogram = b x h
= 12 x 7
= 84

Then find area of kite

Area of kite = (xy)/2
= (7 x 12)/2
= 84/2
= 42

Now we are gonna subtract the area of the kite from the area of the parallelogram.

84 - 42 = 42

Therefore the area of the shaded region is 42.

7 0

2 years ago

A plane flying horizontally at an altitude of 3 miles and a speed of 500 mi/h passes directly over a radar station. Find the rat

konstantin123 [22]

Answer:

The rate at which the distance from the plane to the station is increasing is 331 miles per hour.

Step-by-step explanation:

We can find the rate at which the distance from the plane to the station is increasing by imaging the formation of a right triangle with the following dimensions:

a: is one side of the triangle = altitude of the plane = 3 miles

b: is the other side of the triangle = the distance traveled by the plane when it is 4 miles away from the station and an altitude of 3 miles

h: is the hypotenuse of the triangle = distance between the plane and the station = 4 miles

First, we need to find b:

$a^{2} + b^{2} = h^{2}$ (1)

$b = \sqrt{h^{2} - a^{2}} = \sqrt{(4 mi)^{2} - (3 mi)^{2}} = \sqrt{7} miles$

Now, to find the rate we need to find the derivative of equation (1) with respect to time:

$\frac{d}{dt}(a^{2}) + \frac{d}{dt}(b^{2}) = \frac{d}{dt}(h^{2})$

$2a\frac{da}{dt} + 2b\frac{db}{dt} = 2h\frac{dh}{dt}$

Since "da/dt" is constant (the altitude of the plane does not change with time), we have:

$0 + 2b\frac{db}{dt} = 2h\frac{dh}{dt}$

And knowing that the plane is moving at a speed of 500 mi/h (db/dt):

$\sqrt{7} mi*500 mi/h = 4 mi*\frac{dh}{dt}$

$\frac{dh}{dt} = \frac{\sqrt{7} mi*500 mi/h}{4 mi} = 331 mi/h$

Therefore, the rate at which the distance from the plane to the station is increasing is 331 miles per hour.

I hope it helps you!

4 0

2 years ago