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3 years ago

Find the product. (y^3)^2 y^7

Mathematics

2 answers:

JulsSmile [24]3 years ago

8 0

Power rule

y^3(2)
y^6y^7

product rule, add them

y^13

Sholpan [36]3 years ago

4 0

$Use\ (a^n)^m=a^{n\cdot m}\ and\ a^n\cdot a^m=a^{n+m}\\\\(y^3)^2y^7=y^{3\cdot2}y^7=y^6y^7=y^{6+7}=y^{13}\\\\Answer:\boxed{y^{13}}$

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Find the surface area of the tool box. Round your answer to the nearest tenth and explain your answer. Pls I NEED THE ANSWER NO

Andrej [43]

Answer:

807.8 in^2

Step-by-step explanation:

The total area of the box is the sum of the areas of all faces of the box. The top, bottom, front, and back faces are rectangles 18 in long. The end faces each consist of a rectangle and a triangle. We can compute the sum of these like this:

The areas of top, bottom, front, and back add up to be 18 inches wide by the length that is the perimeter of the end: 2·5in +2·8 in + 9.6 in = 35.8 in. That lateral area is ...

(18 in)(35.6 in) = 640.8 in^2

The area of the triangle on each end is equivalent to the area of a rectangle half as high, so we can compute the area of each end as ...

(9.6 in)(8.7 in) = 83.52 in^2

Then the total area is the lateral area plus the area of the two ends:

640.8 in^2 + 2·83.52 in^2 = 807.84 in^2 ≈ 807.8 in^2

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3 years ago

Which of the following equations below is a linear inequality?

Ket [755]

Answer:

Step-by-step explanation:

A <u>linear inequality</u> is an inequality which involves a linear function. A linear inequality contains one of the symbols of inequality: <, >, ≤, ≥.

Consider all options:

A. The inequality $\dfrac{y}{x}>3\cdot x+2$ is not linear linear inequality, because x is in the left part denominator and in the right part numerator.

B. This option shows linear equation, not inequality.

C. This option shows quadratic equation, not linear inequality.

D. This option shows linear inequality $y\le 5x-5$

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3 years ago

9. Which number is divisible by 2,3, and 6?

ANEK [815]

12, 18 , 24, 30, 36, 42….

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3 years ago

Read 2 more answers

Water flows from a faucet at a constant rate. Assume that 6 gallons of water are already in a tub by the time we notice the fauc

Thepotemich [5.8K]

The equation would be y=3.2x+6

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3 years ago

1. Let f(x, y) be a differentiable function in the variables x and y. Let r and θ the polar coordinates,and set g(r, θ) = f(r co

Olenka [21]

Answer:

$g_{r}(\sqrt{2},\frac{\pi}{4})=\frac{\sqrt{2}}{2}\\$

Step-by-step explanation:

First, notice that:

$g(\sqrt{2},\frac{\pi}{4})=f(\sqrt{2}cos(\frac{\pi}{4}),\sqrt{2}sin(\frac{\pi}{4}))\\$

$g(\sqrt{2},\frac{\pi}{4})=f(\sqrt{2}(\frac{1}{\sqrt{2}}),\sqrt{2}(\frac{1}{\sqrt{2}}))\\$

$g(\sqrt{2},\frac{\pi}{4})=f(1,1)\\$

We proceed to use the chain rule to find $g_{r}(\sqrt{2},\frac{\pi}{4})$ using the fact that $X(r,\theta)=rcos(\theta)\ and\ Y(r,\theta)=rsin(\theta)$ to find their derivatives:

$g_{r}(r,\theta)=f_{r}(rcos(\theta),rsin(\theta))=f_{x}( rcos(\theta),rsin(\theta))\frac{\delta x}{\delta r}(r,\theta)+f_{y}(rcos(\theta),rsin(\theta))\frac{\delta y}{\delta r}(r,\theta)\\$

Because we know $X(r,\theta)=rcos(\theta)\ and\ Y(r,\theta)=rsin(\theta)$ then:

$\frac{\delta x}{\delta r}=cos(\theta)\ and\ \frac{\delta y}{\delta r}=sin(\theta)$

We substitute in what we had:

$g_{r}(r,\theta)=f_{x}( rcos(\theta),rsin(\theta))cos(\theta)+f_{y}(rcos(\theta),rsin(\theta))sin(\theta)$

Now we put in the values $r=\sqrt{2}\ and\ \theta=\frac{\pi}{4}$ in the formula:

$g_{r}(\sqrt{2},\frac{\pi}{4})=f_{r}(1,1)=f_{x}(1,1)cos(\frac{\pi}{4})+f_{y}(1,1)sin(\frac{\pi}{4})$

Because of what we supposed:

$g_{r}(\sqrt{2},\frac{\pi}{4})=f_{r}(1,1)=-2cos(\frac{\pi}{4})+3sin(\frac{\pi}{4})$

And we operate to discover that:

$g_{r}(\sqrt{2},\frac{\pi}{4})=-2\frac{\sqrt{2}}{2}+3\frac{\sqrt{2}}{2}$

$g_{r}(\sqrt{2},\frac{\pi}{4})=\frac{\sqrt{2}}{2}$

and this will be our answer

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3 years ago