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3 years ago

Solve each equation by taking square roots. Show your work. Simplify your answers. Do not write your answers in decimal form.

Mathematics

1 answer:

natulia [17]3 years ago

6 0

Step-by-step explanation:

$\sf \hookrightarrow \: {n}^{2} = 20 \\ \sf \hookrightarrow \: \sqrt{ {n}^{2} } = \sqrt{20} \\ \sf \hookrightarrow \: n = \sqrt{20} \\ \star \bf \: n = 2 \sqrt{5} \\ \\ \sf \hookrightarrow \: {a}^{2} = 90 \\ \sf \hookrightarrow \: \sqrt{ {a}^{2} } = \sqrt{90} \\ \sf \hookrightarrow \: a = \sqrt{90} \\ \star \bf \: a = 3 \sqrt{10}$

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Consider the graph of the linear function h(x) = –x + 5. Which could you change to move the graph down 3 units?

Feliz [49]

$\bf \qquad \qquad \qquad \qquad \textit{function transformations}
\\ \quad \\\\
% left side templates
\begin{array}{llll}
f(x)=&{{ A}}({{ B}}x+{{ C}})+{{ D}}
\\ \quad \\
y=&{{ A}}({{ B}}x+{{ C}})+{{ D}}
\\ \quad \\
f(x)=&{{ A}}\sqrt{{{ B}}x+{{ C}}}+{{ D}}
\\ \quad \\
f(x)=&{{ A}}(\mathbb{R})^{{{ B}}x+{{ C}}}+{{ D}}
\\ \quad \\
f(x)=&{{ A}} sin\left({{ B }}x+{{ C}} \right)+{{ D}}
\end{array}\\\\
--------------------\\\\$

$\bf \bullet \textit{ stretches or shrinks horizontally by } {{ A}}\cdot {{ B}}\\\\
\bullet \textit{ flips it upside-down if }{{ A}}\textit{ is negative}\\
\left. \qquad \right. \textit{reflection over the x-axis}
\\\\
\bullet \textit{ flips it sideways if }{{ B}}\textit{ is negative}\\
\left. \qquad \right. \textit{reflection over the y-axis}$

$\bf \bullet \textit{ horizontal shift by }\frac{{{ C}}}{{{ B}}}\\
\left. \qquad \right. if\ \frac{{{ C}}}{{{ B}}}\textit{ is negative, to the right}\\\\
\left. \qquad \right. if\ \frac{{{ C}}}{{{ B}}}\textit{ is positive, to the left}\\\\
\bullet \textit{ vertical shift by }{{ D}}\\
\left. \qquad \right. if\ {{ D}}\textit{ is negative, downwards}\\\\
\left. \qquad \right. if\ {{ D}}\textit{ is positive, upwards}\\\\
\bullet \textit{ period of }\frac{2\pi }{{{ B}}}$

with that template in mind, let's see

$\bf h(x)=-x+5\implies h(x)=\stackrel{A}{-1}(\stackrel{B}{1}x\stackrel{C}{+0})\stackrel{D}{+5}$

a shift down by 3 units, means a vertical shift downwards, so D needs to drop by 3 units.

$\bf h(x)=\stackrel{A}{-1}(\stackrel{B}{1}x\stackrel{C}{+0})\boxed{\stackrel{D}{+5-3}}\implies h(x)=-1(1x+0)+2
\\\\\\
h(x)=-x+2$

4 0

3 years ago

n is a positive integer with exactly two different divisors greater than 1, how many positive factors does n^2 have? A. 4 B. 5 C

Evgesh-ka [11]

Answer:

C. 5

Step-by-step explanation:

Remember that for any integer $n$ , the integers $1 \text{and} n$ are both divisors (or factors) of $n$ . First, we will prove that n is a square, and then we will compute the factors of the n².

In this case, the integer n has exactly two different divisors greater than 1. It's impossible that $n=1$ , since 1 doesn't have positive factors greater than 1. Then $n>1$ , therefore $n$ itself is one of the required divisors. Denote by $a$ the other divisor greater than 1, and note that to satisfy the condition on the divisors, $a$ .

Because a divides n, there exists some integer $k$ such that $n=ak$ . We must have that $k>1$ , if not, then $k\leq 1$ , which implies that $ak=n\leq a$ , which contradicts the part above.

Now, $k>1$ and, by definition of divisibility, k divides n. Then k must be equal either to n or a, since we can't have three different divisors of this kind. If $k=n$ then $n=an$ and by cancellation, $1=a$ which is a contradiction. Therefore $k=a$ and $n=a^2$ .

We have that $n=(a^2)^2=a^4$ . We can write n as $n=a^3a=a^2a^2=a^4 \cdot 1$ . From the first equation, a divides n and a³ divides n. From the second equation, a² divides, and from the last one, 1 divides n and a⁴=n divides n. Thus n has exactly 5 positive factors.

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4 years ago

You earn $30 for washing 5 cars.how much do you want for washing 3cars

Blizzard [7]

Answer:

$18

Step-by-step explanation:

6 0

3 years ago

Read 2 more answers

With explanation pls

olga_2 [115]

Sum of all angles is 360. There are 8 angles. 360/8 is 45. Exterior and interior angles are supplementary meaning their sum is 180. 180-45 is 135.

3 0

3 years ago

A yo-yo is moving up and down a string so that its velocity at time t is given by v(t) = 3cos(t) for time t ≥ 0. The initial pos

jeka57 [31]

Part A - The average value of v(t) over the interval (0, π/2) is 6/π

Part B - The displacement of the yo-yo from time t = 0 to time t = π is 0 m

Part C - The total distance the yo-yo travels from time t = 0 to time t = π is 6 m.

<h3>Part A: Find the average value of v(t) on the interval (0, π/2)</h3>

The average value of a function f(t) over the interval (a,b) is

$f(t)_{avg} = \frac{1}{b - a} \int\limits^b_a {f(t)} \, dx$

So, since velocity at time t is given by v(t) = 3cos(t) for time t ≥ 0. Its average value over the interval (0, π/2) is given by

$v(t)_{avg} = \frac{1}{\frac{\pi }{2} - 0} \int\limits^{\frac{\pi }{2} }_0 {v(t)} \, dt$

Since v(t) = 3cost, we have

$v(t)_{avg} = \frac{1}{\frac{\pi }{2} - 0} \int\limits^{\frac{\pi }{2} }_0 {3cos(t)} \, dt\\= \frac{3}{\frac{\pi }{2}} \int\limits^{\frac{\pi }{2} }_0 {cos(t)} \, dt\\= \frac{6}{{\pi}} [{sin(t)}]^{\frac{\pi }{2} }_{0} \\= \frac{6}{{\pi}} [{sin(\frac{\pi }{2})} - sin0]\\ = \frac{6}{{\pi}} [1 - 0]\\ = \frac{6}{{\pi}} [1]\\ = \frac{6}{{\pi}}$

So, the average value of v(t) over the interval (0, π/2) is 6/π

<h3>Part B: What is the displacement of the yo-yo from time t = 0 to time t = π?</h3>

To find the displacement of the yo-yo, we need to find its position.

So, its position x = ∫v(t)dt

= ∫3cos(t)dt

= 3∫cos(t)dt

= 3sint + C

Given that at t = 0, x = 3. so

x = 3sint + C

3 = 3sin0 + C

3 = 0 + C

C = 3

So, x(t) = 3sint + 3

So, its displacement from time t = 0 to time t = π is

Δx = x(π) - x(0)

= 3sinπ + 3 - (3sin0 + 3)

= 3 × 0 + 3 - 0 - 3

= 0 + 3 - 3

= 0 + 0

= 0 m

So, the displacement of the yo-yo from time t = 0 to time t = π is 0 m

<h3>Part C: Find the total distance the yo-yo travels from time t = 0 to time t = π. (10 points)</h3>

The total distance the yo-yo travels from time t = 0 to time t = π is given by

$x(t) = \int\limits^{\pi}_0 {v(t)} \, dt\\= \int\limits^{\pi }_0 {3cos(t)} \, dt\\= 3 \int\limits^{\pi }_0 {cos(t)} \, dt\\ = 3 \int\limits^{\frac{\pi }{2} }_0 {cos(t)} \, dt + 3\int\limits^{\pi }_{\frac{\pi }{2}} {cos(t)} \, dt\\= 3 \times 2\int\limits^{\frac{\pi }{2} }_0 {cos(t)} \, dt\\= 6 [{sin(t)}]^{\frac{\pi }{2} }_{0} \\= 6[{sin\frac{\pi }{2} - sin0]\\\\= 6[1 - 0]\\= 6(1)\\= 6$

So, the total distance the yo-yo travels from time t = 0 to time t = π is 6 m.

Learn more about average value of a function here:

brainly.com/question/15870615

#SPJ1

4 0

1 year ago