What is the answer and how to solve?

Mathematics

1 answer:

oksian1 [2.3K]2 years ago

5 0

I’m not sure what h and t are but you would want to plug them in and solve. How I would do it is multiply h and t on the left, then multiply t squared and get an answer, then multiply that answer by -5, then multiply t by 20. Next add or subtract all of the numbers on the right and I think you would get your answer.

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Consider the following 8 numbers, where one labelled X is unknown.

Effectus [21]

Answer:

$55,-3$

Step-by-step explanation:

$We\ are\ given\ that,\\There\ are\ eight\ numbers\ in\ the\ data :\\17,18,24,x,44,48,42,4\\Where\ the\ value\ of\ x\ is\ what\ we\ need\ to\ find.\\Range\ of\ the\ data=51\\Hence,\\We\ know\ that\ the\ range\ is\ the\ difference\\ between\ the\ highest\ observation\ and\ the\ lowest\ observation\ in\ the\ data.\\We\ may\ know\ see\ two\ cases\ of\ evaluating\ x\ when:\\1. x\ is\ the\ highest\ observation\ in\ the\ data\ or\ x>48.\\Highest\ observation\ =x\\Lowest\ observation=4\\Hence,\\$

$As\ range=51,\\x-4=51\\x=51+4\ [Adding\ 4\ on\ both\ the\ sides]\\x=55 \\2. x\ is\ the\ lowest\ observation\ or\ x$

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3 years ago

Use right triangle tools to find the unknown measures. Round lengths to the nearest tenth and angle measures to the nearest degr

timama [110]

Step-by-step explanation:

OP=11.2

m<Q= 42⁰

m<R=48⁰

The tools used are....

I have used the property of Pythagoras Theorm to get the length of side QP, I have also used the property of inverse function of sin to get the measure of angle Q and the property of sum of angles of triangle to get measure of angle of R

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2 years ago

Whats the intrgral of <img src="https://tex.z-dn.net/?f=%20%5Cint%20%20%5Cfrac%7Bx%5E2%2Bx-3%7D%7B%28x%5E3%2Bx%5E2-4x-4%29%5E2%7

rosijanka [135]

$\displaystyle\int\frac{x^2+x-3}{(x^3+x^2-4x-4)^2}\,\mathrm dx$

Notice that $x^3+x^2-4x-4=x^2(x+1)-4(x+1)=(x-2)(x+2)(x+1)$ . Decompose the integrand into partial fractions:

$\dfrac{x^2+x-3}{(x-2)^2(x+2)^2(x+1)^2}$
$=\dfrac1{3(x+1)}-\dfrac{11}{32(x+2)}-\dfrac1{3(x+1)^2}-\dfrac1{16(x+2)^2}+\dfrac1{96(x-2)}+\dfrac1{48(x-2)^2}$

Integrating term-by-term, you get

$\displaystyle\int\frac{x^2+x-3}{(x^3+x^2-4x-4)^2}\,\mathrm dx$
$=-\dfrac1{48(x-2)}+\dfrac1{3(x+1)}+\dfrac1{16(x+2)}+\dfrac1{96}\ln|x-2|+\dfrac13\ln|x+1|-\dfrac{11}{32}\ln|x+2|+C$