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2 years ago

7

What is the perimeter of the base, p, of this square-based pyramid? (attached image

1 answer:

Katen [24]2 years ago

8 0

Answer:

32in

Step-by-step explanation:

8in + 8in + 8in + 8in = 32in

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Please Answer ASAP!

natali 33 [55]

Population. Population is the amount of whatever is inisde

4 0

3 years ago

Read 2 more answers

One positive integer is 5 times another positive integer and their product is 320. What are the positive integers?

NeTakaya

Answer by CubeyThePenguin(3113) (Show Source): You can put this solution on YOUR website!
x = 5y
xy = 320

Substitute the first equation into the second equation
(5y)(y) = 320
5y^2 = 320
y^2 = 64
y = 8 (y must be positive)

The integers are (x, y) = (40, 8).

7 0

2 years ago

Find the magnitude of the vector <−4, −6>.<br><br> 8<br> 7.2<br> 10<br> 6.4

PolarNik [594]

The magnitude of the vectors is 7.2

The magnitude of a vector is resolved by using a formula

<h3>Magnitude of a Vector</h3>

This is the displacement between two vectors and can be calculated as

$x^2 = y^2 + z^2\\x = \sqrt{y^2 + z^2} \\$

where x is the magnitude of displacement between y and z.

In this question, the vectors are

-4
-6

Let's find the magnitude of this vector

$x = \sqrt{(-4)^2+(-6)^2}\\x = \sqrt{16+36}\\x = \sqrt{52}\\x = 7.2$

The magnitude of the vectors is 7.2

Learn more on magnitude of vectors here;

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4 0

2 years ago

PLS HELP ME solve the system of equations <br> all I need is to find x,y and z

ale4655 [162]

(x,y,z)=(-4,-2,-5)

I added some images for explanation

7 0

2 years ago

Answer the question in the picture

nekit [7.7K]

Recall the angle sum identities:

$\sin(x+y)=\sin x\cos y+\cos x\sin y$

$\cos(x+y)=\cos x\cos y-\sin x\sin y$

Now,

$\tan(x+y)=\dfrac{\sin(x+y)}{\cos(x+y)}=\dfrac{\sin x\cos y+\cos x\sin y}{\cos x\cos y-\sin x\sin y}$

Divide through numerator and denominator by $\cos x\cos y$ to get

$\tan(x+y)=\dfrac{\tan x+\tan y}{1-\tan x\tan y}$

Next, we use the fact that $x,y$ lie in the first quadrant to determine that

$\sin x=\dfrac12\implies\cos x=\sqrt{1-\sin^2x}=\dfrac{\sqrt3}2$

$\cos y=\dfrac{\sqrt2}2\implies\sin x=\sqrt{1-\cos^2x}=\dfrac1{\sqrt2}$

So we then have

$\tan x=\dfrac{\sin x}{\cos x}=\dfrac{\frac12}{\frac{\sqrt3}2}=\dfrac1{\sqrt3}$

$\tan y=\dfrac{\sin y}{\cos y}=\dfrac{\frac1{\sqrt2}}{\frac{\sqrt2}2}=1$

Finally,

$\tan(x+y)=\dfrac{\frac1{\sqrt3}+1}{1-\frac1{\sqrt3}}=\dfrac{1+\sqrt3}{\sqrt3-1}=2+\sqrt3\approx3.73$

4 0

2 years ago