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mote1985 [20]
4 years ago
7

A baseball must weigh between 5 ounces and 5.25 ounces. Ten baseball would wight between ________ounces and ________ ounces.

Mathematics
1 answer:
sergey [27]4 years ago
5 0

Ten baseballs would weigh between 50 ounces and 52.5 ounces.

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If tan(theta) is 5/12 and theta is an acute angle, find sin(theta)
navik [9.2K]

Comment

The tangent represents the sides making up the right angle in a right angle triangle. To find out the answer to what you have asked, the simplest way is to assume that you are talking about a right angle triangle.

Step One

Find the hypotenuse. Use a variation of the Pythagorean Theorem.

opposite^2 + adjacent^2 = hypotenuse^2

5^2 + 12^2 = x^2

25 + 144 = x^2

x^2 = 169

sqrt(x^2) = sqrt(169)

x = 13

Step Two

Find the sine.

When the tangent is given, the opposite in this case is 5.

That's because the opposite is always the numerator of a tangent.

Sin(theta) = 5/13 Answer

7 0
4 years ago
Denzel is dividing 23 by 11. If he continues the process, what will keep repeating in the quotient?
VLD [36.1K]

Answer:

1 will be the repeating in the quotient if he continues the process

Step-by-step explanation:

divide 23 by 11 and see the remainder

5 0
3 years ago
Complete the pattern
romanna [79]
1.) 376.2
2.) 37620
3.)376200. 
3.) 3762000

hope that helped
7 0
4 years ago
a can do.a piece of work in 6 days and B in 8 days each working 7hours a day In what time will they finish it together, working
Irina-Kira [14]
<h2>A can complete the work in 7*6 hrs = 42 hrs. </h2><h2> </h2><h2>A can complete in 1 hr = 1/42 part. </h2><h2> </h2><h2>B can complete the work in 7*8 hr = 56 hrs. </h2><h2> </h2><h2>A&B together do in 1 hr = 1/42 + 1/56 = 7/168 = 1/24 part. So they will do in 8 hrs = 8*1/24 = 1/3 part. </h2><h2> </h2><h2>Hence they need 1÷1/3 = 3 days to complete the work.</h2>

6 0
3 years ago
What are the solutions of x^2 = -7x-8
Vilka [71]

Solution, solve\:for\:x,\:x^2=-7x-8\quad :\quad x=\frac{-7+\sqrt{17}}{2},\:x=\frac{-7-\sqrt{17}}{2}

Steps:

x^2=-7x-8

\mathrm{Add\:}8\mathrm{\:to\:both\:sides},\\x^2+8=-7x-8+8

\mathrm{Simplify},\\x^2+8=-7x

\mathrm{Add\:}7x\mathrm{\:to\:both\:sides},\\x^2+8+7x=-7x+7x

\mathrm{Simplify},\\x^2+7x+8=0

Solve\:with\:the\:quadratic\:formula,\\\mathrm{For\:}\quad a=1,\:b=7,\:c=8:\quad x_{1,\:2}=\frac{-7\pm \sqrt{7^2-4\cdot \:1\cdot \:8}}{2\cdot \:1},\\x=\frac{-7+\sqrt{7^2-4\cdot \:1\cdot \:8}}{2\cdot \:1}:\quad \frac{-7+\sqrt{17}}{2},\\x=\frac{-7-\sqrt{7^2-4\cdot \:1\cdot \:8}}{2\cdot \:1}:\quad \frac{-7-\sqrt{17}}{2}

\mathrm{The\:final\:solutions\:to\:the\:quadratic\:equation\:are:},\\x=\frac{-7+\sqrt{17}}{2},\:x=\frac{-7-\sqrt{17}}{2}

\mathrm{Hope\:This\:Helps!!!}

\mathrm{-Austint1414}

7 0
4 years ago
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