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Naily [24]
3 years ago
8

Solve for angle Z PLEASE HELP

Mathematics
1 answer:
Katena32 [7]3 years ago
4 0

Answer:

60°

Step-by-step explanation:

I'm 100% sure good luck!

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How many students were asked? How do you know?<br><br><br> Find the median for these data.
kkurt [141]

Answer:

a=10 students because there are 10 answers.

b=5

Step-by-step explanation:

a=10 students because there are 10 answers.

b, you need to list the numbers all out and rearrange them from smallest to largest.

0, 0, 1, 2, 3, 7, 9, 10, 13, 15

The median means the middle which is 3 and 7

Because we have two numbers, we need to find the middle number of 3 and 7 which is <u><em>5</em></u>

4 0
3 years ago
Which of the following numbers is a COUNTEREXAMPLE to the statement "All odd numbers greater than 2 are prime numbers" ? 2 3 5 7
GuDViN [60]
9 would be ur answer....because 9 is an odd number greater then 2, but it is not a prime number
5 0
4 years ago
Convert 6 2/3 cups to pints. Express your answer in simplest form.
Dmitry_Shevchenko [17]
First off, let's convert the mixed fraction to "improper", keeping in mind that, there are 2 cups in 1 pint.

\bf \stackrel{mixed}{6\frac{2}{3}}\implies \cfrac{6\cdot 3+2}{3}\implies \stackrel{improper}{\cfrac{20}{3}}&#10;\\\\\\&#10;\begin{array}{ccll}&#10;cups&pints\\&#10;\text{\textemdash\textemdash\textemdash}&\text{\textemdash\textemdash\textemdash}\\&#10;2&1\\\\&#10;\frac{20}{3}&p&#10;\end{array}\implies \cfrac{\quad 2\quad }{\frac{20}{3}}=\cfrac{1}{p}\implies \cfrac{\quad \frac{2}{1}\quad }{\frac{20}{3}}=\cfrac{1}{p}

\bf \cfrac{2}{1}\cdot \cfrac{3}{20}=\cfrac{1}{p}\implies \cfrac{3}{10}=\cfrac{1}{p}\implies p=\cfrac{10\cdot 1}{3}\implies p=\cfrac{10}{3}\implies p=3\frac{1}{3}
5 0
3 years ago
Find the missing side of the right triangle
o-na [289]

Answer:

<h2>1</h2>

Step-by-step explanation:

Use the Pythagorean theorem:

leg^2+leg^2=hypotenuse^2

We have

leg=3,\ leg=x,\ hypotenuse=\sqrt{10}

Substitute:

3^2+x^2=(\sqrt{10})^2

9+x^2=10             <em>subtract 9 from both sides</em>

9-9+x^2=10-9

x^2=1\to x=\sqrt1\\\\x=1

7 0
4 years ago
Find f. f ″(x) = x^−2, x &gt; 0, f(1) = 0, f(6) = 0
marin [14]

If you do in fact mean f(1)=f(6)=0 (as opposed to one of these being the derivative of f at some point), then integrating twice gives

f''(x) = -\dfrac1{x^2}

f'(x) = \displaystyle -\int \frac{dx}{x^2} = \frac1x + C_1

f(x) = \displaystyle \int \left(\frac1x + C_1\right) \, dx = \ln|x| + C_1x + C_2

From the initial conditions, we find

f(1) = \ln|1| + C_1 + C_2 = 0 \implies C_1 + C_2 = 0

f(6) = \ln|6| + 6C_1 + C_2 = 0 \implies 6C_1 + C_2 = -\ln(6)

Eliminating C_2, we get

(C_1 + C_2) - (6C_1 + C_2) = 0 - (-\ln(6))

-5C_1 = \ln(6)

C_1 = -\dfrac{\ln(6)}5 = -\ln\left(\sqrt[5]{6}\right) \implies C_2 = \ln\left(\sqrt[5]{6}\right)

Then

\boxed{f(x) = \ln|x| - \ln\left(\sqrt[5]{6}\right)\,x + \ln\left(\sqrt[5]{6}\right)}

3 0
2 years ago
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