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juin [17]
3 years ago
6

PLEASE HELP I ONT GET THIS QUESTION

Mathematics
1 answer:
tatuchka [14]3 years ago
5 0

Here's a trick.

Don't solve all the equations, you're given the answer to x so just substitute.

5/2*8+7/2=3/4*8+14

23.5=20 INCORRECT

5/4*8-9=3/2*8-21

1=-9 INCORRECT

5/4*8-2=3/2*8-4 <- YOUR ANSWER

8=8 CORRECT


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Solve for p: -9 + p &lt; 15 <br> A. p &gt; 24 B. -p &gt; 6 C. p &lt; 24 D. -p &lt; -6
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Answer:

C.  In the given expression -9 + p < 15 , the value of p <  24.

Step-by-step explanation:

Here, the given expression is :

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If equals are added to both sides of inequality, the inequality remains unchanged.

Now, -9 + p < 15

⇒  -9 + p  + 9 < 15  + 9                                  (adding +9 on both sides)

or,  p <  24

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\bf ~~~~~~~~~~~~\textit{middle point of 2 points }&#10;\\\\&#10;\begin{array}{ccccccccc}&#10;&&x_1&&y_1&&x_2&&y_2\\&#10;%  (a,b)&#10;&&(~ -2 &,& -4~) &#10;%  (c,d)&#10;&&(~ 3 &,& 8~)&#10;\end{array}\qquad&#10;%   coordinates of midpoint &#10;\left(\cfrac{ x_2 +  x_1}{2}\quad ,\quad \cfrac{ y_2 +  y_1}{2} \right)&#10;\\\\\\&#10;\left( \cfrac{3-2}{2}~~,~~\cfrac{8-4}{2} \right)\implies \left( \frac{1}{2}~,~2 \right)\impliedby center

\bf ~~~~~~~~~~~~\textit{distance between 2 points}&#10;\\\\&#10;\begin{array}{ccccccccc}&#10;&&x_1&&y_1&&x_2&&y_2\\&#10;%  (a,b)&#10;&&(~ -2 &,& -4~) &#10;%  (c,d)&#10;&&(~ 3 &,& 8~)&#10;\end{array}~~~ &#10;%  distance value&#10;d = \sqrt{( x_2- x_1)^2 + ( y_2- y_1)^2}&#10;\\\\\\&#10;d=\sqrt{[3-(-2)]^2+[8-(-4)]^2}\implies d=\sqrt{(3+2)^2+(8+4)^2}&#10;\\\\\\&#10;d=\sqrt{25+144}\implies d=\sqrt{169}\implies d=13\qquad\qquad \qquad  \stackrel{radius}{\frac{13}{2}}

\bf \textit{equation of a circle}\\\\ &#10;(x- h)^2+(y- k)^2= r^2&#10;\qquad &#10;center~~(\stackrel{\frac{1}{2}}{ h},\stackrel{2}{ k})\qquad \qquad &#10;radius=\stackrel{\frac{13}{2}}{ r}&#10;\\\\\\&#10;\left( x-\frac{1}{2} \right)^2+(y-2)^2=\left( \frac{13}{2} \right)^2\implies \left( x-\frac{1}{2} \right)^2+(y-2)^2=\frac{169}{4}

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