All categories

3 years ago

If x varies inversely as y, and x = 7 when y=3, find y when x = 9.

Mathematics

2 answers:

RSB [31]3 years ago

5 0

Step-by-step explanation:

x=k/y------(1)

k=xy

k= 7×3

k=21

from equation 1

x=k/y

y=k/x

y=21/9

y = 2.33

Liono4ka [1.6K]3 years ago

5 0

Answer:

3.85714285714

Step-by-step explanation:

7 = 3

9 = y

9*3=27

27/7 = 3.85714285714

You might be interested in

C. A 40-foot ladder reaches 38 feet up the side of a house. How far from the base of the

Svet_ta [14]

Answer: 12.48

Step-by-step explanation:

38x38+bxb=40x40

1444+b=1600

1600-1444=156

$\sqrt156$ =12.48

4 0

2 years ago

D)) What is 39 x 35?<br>?<br>3° x 39 =<br>

yan [13]

Answer:

The letter "x" is often used in algebra to mean a value that is not yet known. It is called a "variable" or sometimes an "unknown". In x + 2 = 7, x is a variable, but we can work out its value if we try!

Step-by-step explanation:

8 0

2 years ago

Use the tangent identity to

Lunna [17]

Answer:

$tan(x)=\frac{44\sqrt{89}}{89}$

Step-by-step explanation:

tan is defined as: $tan(x)=\frac{opposite}{adjacent}$

sin is defined as: $sin(x)=\frac{opposite}{hypotenuse}$

cos is defined as: $cos(x)=\frac{adjacent}{hypotenuse}$

We can also define tan as: $tan(x)=\frac{sin(x)}{cos(x)}$

because plugging in the definitions of sin and cos in we get: $tan(x)=\frac{(\frac{opposite}{hypotenuse})}{(\frac{adjacent}{hypotenuse})}\\\\tan(x)=\frac{opposite}{hypotenuse}*\frac{hypotenuse}{adjacent}\\\\tan(x)=\frac{opposite}{adjacent}$

which you'll notice is the original definition of tan(x)

So using this definition of tan(x) we can use the givens sin(x) and cos(x) to find tan(x)

$sin(x)=\frac{44}{45}\\\\cos(x)=\frac{\sqrt{89}}{45}\\\\tan(x)=\frac{sin(x)}{cos(x)}$

plugging in sin(x) and cos(x) we get:

$tan(x)=\frac{\frac{44}{45}}{\frac{\sqrt{89}}{45}}\\\\tan(x)=\frac{44}{45}*\frac{45}{\sqrt{89}}\\\\tan(x)=\frac{44}{\sqrt{89}}$

We usually don't like square roots in the denominator, and from here we want to rationalize the denominator which we do by removing the square root from the denominator.

We can do this by multiplying the fraction by: $\frac{\sqrt{89}}{\sqrt{89}}$ which doesn't change the value of the fraction since it simplifies to 1, but it gets rid of the square root in the denominator