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3 years ago

The graph of which function is the same as the graph of y=1/x shifted to the right 5 units and shifted up 2 units?

Mathematics

1 answer:

dybincka [34]3 years ago

5 0

Answer: I’m pretty sure it’s y=1/x+5 +2

Step-by-step explanation:

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A circle has a diameter of 14cm. What's the circumfrence?

vesna_86 [32]

I believe that when finding the radius of a circle you half the diameter, but when finding the circumference of a circle you need to multiply the diameter (14cm) by pi.

So 14xpi= 43.98229715

But it can be simplified/ rounded to 44cm.

Of course I might be incorrect so do please correct me if I am wrong.

7 0

3 years ago

Could I get some help? QUESTION: This table shows the profit for a company (in millions of dollars) in different years.

Pani-rosa [81]

Answer:

17 million

Step-by-step explanation:

-0.34( $8^{2}$ ) + 4.43(8) + 3.46

-21.76 + 35.44 + 3.46

13.68 + 3.46

17.14

Answer: C 17 million

Have an amazing day!

<u><em>PLEASE RATE!</em></u>

7 0

3 years ago

Put in order from least to greatest 65, 34, -50, 28, -64, -45

PIT_PIT [208]

-64, -50, -45, 28, 34, 65

7 0

3 years ago

Read 2 more answers

What is 178÷2???????

Nonamiya [84]

178÷2 is 89
Hope this was helpful!

8 0

4 years ago

Read 2 more answers

Find the area of the part of the plane 3x 2y z = 6 that lies in the first octant.

gavmur [86]

The area of the part of the plane 3x 2y z = 6 that lies in the first octant is mathematically given as

A=3 √(4) units ^2

<h3>What is the area of the part of the plane 3x 2y z = 6 that lies in the first octant.?</h3>

Generally, the equation for is mathematically given as

The Figure is the x-y plane triangle formed by the shading. The formula for the surface area of a z=f(x, y) surface is as follows:

$A=\iint_{R_{x y}} \sqrt{f_{x}^{2}+f_{y}^{2}+1} d x d y(1)$

The partial derivatives of a function are f x and f y.

$\begin{aligned}&Z=f(x)=6-3 x-2 y \\&=\frac{\partial f(x)}{\partial x}=-3 \\&=\frac{\partial f(y)}{\partial y}=-2\end{aligned}$

When these numbers are plugged into equation (1) and the integrals are given bounds, we get:

$&=\int_{0}^{2} \int_{0}^{3-\frac{3}{2} x} \sqrt{(-3)^{2}+(-2)^2+1dxdy} \\\\&=\int_{0}^{2} \int_{0}^{3-\frac{3}{2} x} \sqrt{14} d x d y \\\\&=\sqrt{14} \int_{0}^{2}[y]_{0}^{3-\frac{3}{2} x} d x d y \\\\&=\sqrt{14} \int_{0}^{2}\left[3-\frac{3}{2} x\right] d x \\\\$

$&=\sqrt{14}\left[3 x-\frac{3}{2} \cdot \frac{1}{2} \cdot x^{2}\right]_{0}^{2} \\\\&=\sqrt{14}\left[3-\frac{3}{2} \cdot \frac{1}{2} \cdot x^{2}\right]_{0}^{2} \\\\&=\sqrt{14}\left[3.2-\frac{3}{2} \cdot \frac{1}{2} \cdot 3^{2}\right] \\\\&=3 \sqrt{14} \text { units }{ }^{2}$

In conclusion, the area is

A=3 √4 units ^2