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3 years ago

Your Neighbor has a square-shaped pool with side lengths of 3a^5. What is the area of the pool?

Mathematics

2 answers:

lbvjy [14]3 years ago

7 0

Square Area: $A=s^2$ , with s = length of side

Plug in the side length into the square area and solve for the area as such:

$A=(3a^5)^2\\A=9a^{10}$

<u>The area of the pool is 9a^10 un^2.</u>

mrs_skeptik [129]3 years ago

4 0

if a square-shaped pool side lengths are 3a^5 the area is A= (3a^5)x(3a^5) = 9a^ 5+5= 9a^ 10. Hope this helps!

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Please help!! Quickly

irina [24]

Answer:

The answer to your question is: a = 3.5 units

Step-by-step explanation:

Right triangle

to find a we need to use the trigonometric function cosine

cosine Ф = opposite leg / hypothenemus

Clear opposite leg from the equation

opposite leg = hypothenemus x cosine Ф

a = opposite leg = 5 x cosine 45

a = 5 x 0.70

a = 3.5 units

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3 years ago

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andreev551 [17]

Answer:

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3 years ago

Read 2 more answers

What is the sign of f on the interval -2

TEA [102]

Answer:

$f$ is sometimes positive and sometimes negative.

Step-by-step explanation:

$f(x)=(x-3)(x+2)(x+4)(x+4)(x-1)(2x-9)$

Take $x=-1\in(-2,\frac{9}{2})\ as-2$

$f(-1)=(-1-3)(-1+2)(-1+4)(-1-1)(-2-9)\\\\=(-4)(1)(3)(-2)(-11)\\\\=-264\\\\f(-1)$

Take $x=2\in(-2,\frac{9}{2})\ as\ -2$

$f(2)=(2,-3)(2+2)(2+4)(2-1)(2\times2-9)\\\\=(-1)(4)(6)(1)(-5)\\\\=120\\\\f(2)>0$

Hence $f(x)$ for $x=-1$ and $f(x)>0$ for $x=2$

So $f$ is sometimes positive and sometimes negative in the interval.

3 0

3 years ago

Write an equation for each translation of Y=|x|.

Digiron [165]

$\boxed{y=\mid x \mid +14}$

<h2>Explanation:</h2>

Let's graph the equation:

$y=\mid x \mid$

Which is the red graph shown below. As you can see this is the pattern of the Absolute Value Function. In order to translate this graph 14 units up, let's add 14 to $\mid x \mid$ . In doing so we get the new function:

$y=\mid x \mid + 14$

So we get a new graph which is the blue one. In order to prove that this is the correct translation, let's take the vertex of the red graph which is the origin. To translate this graph 14 units up implies that each point is shifted 14 units up, so for the vertex of the red graph:

$Vertex \ of \ the \ red \ graph: \\ \\ V(0,0) \\ \\ \\ Vertex \ of \ the \ new \ graph: \\ \\ (0,0+14) \\ \\ (0,14)$

So the new vertex matches the vertex of the blue graph. Finally, the equation for the translation of $y=\mid x \mid$ 14 units up is $y=\mid x \mid +14$