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Rus_ich [418]
3 years ago
11

X squared +10x-24

x {?}^{2} + 10x - 24" alt="x {?}^{2} + 10x - 24" align="absmiddle" class="latex-formula">
Mathematics
1 answer:
insens350 [35]3 years ago
3 0
Solve for x using factorizing:
x^2 + 10 x - 24 = 0

The left hand side factors into a product with two terms:
(x - 2) (x + 12) = 0

Split into two equations:
x - 2 = 0 or x + 12 = 0

Add 2 to both sides:
x = 2 or x + 12 = 0
Subtract 12 from both sides:
Answer:  x = 2 or x = -12
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elena-14-01-66 [18.8K]

First derisive the equation for C

\\ \sf\longmapsto F=\dfrac{9}{5}C+32

\\ \sf\longmapsto C+32=\dfrac{5}{9}F

\\ \sf\longmapsto C=\dfrac{5}{9}F-32

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\\ \sf\longmapsto C=\dfrac{5}{9}101-32

\\ \sf\longmapsto C=\dfrac{505}{9}-32

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Given: N is the midpoint of LQ,
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Answer:

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Step-by-step explanation:

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3 years ago
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damaskus [11]
The answer is 2.667.
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3 years ago
Please help it's for a test!
Tanya [424]

Answer:

0.2322   or    23.22 %

Step-by-step explanation:

We have to solve and find the area out of these limits

μ + 0,3 =  210 + 0,3   ⇒    210,3  and

μ - 0,3  =  210 - 0,3   ⇒    209.7

z(l)   = ( x - 210 ) / (2.8/√84)   ⇒  z(l) =  - (0.3 * 9,17)/ 2.8

z (l)  = - 1.195  

We need to interpole from z table

               1.19         ⇒     0.1170

              1.20       ⇒      0.1151

   Δ  ⇒     0.01     ⇒      0.0019

And between our point   1,195 and 1,19   the difference is  0.005

then        0.01          ⇒   0.0019

              0.005      ⇒      ?? (x)

we find x =  0.00095

to get the area for poin  z (l)  - 1.195 up to final left tail is from  z table

0,1170 - 0.00095 =  0.1161

And by symmetry  to the right is the same

So   0.1161 * 2   =  0.2322

We find the area out of the above indicated limits  the area we were looking for. This is the probability of finding shafts over and below the population mean and 0.3 inches

Step-by-step explanation:

5 0
2 years ago
Hello help me with this question thanks in advance​
AfilCa [17]
<h2>Question #22 Answer</h2>

B. 2 in.

<h3>Explanation:</h3>

\frac{3\times4}{6}

Cross out the common factor

\frac{4}{2}\\ 2

________________________________________________________

<h2>Question #23 Answer (Picture attached)</h2>

D. proportional, equal

<h3>Explanation:</h3>

Two\ figure\ are\ sand\ to\ \\ be\ similar\ if\ the\\ Conesesponding\ sides\ we\\ proportional\ and\ the\\ Corresponding\ ongles\ are\\ equal.

If\ triangle\ ΔABC\ is\ similar\ to
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\frac{AB}{xy}=\frac{AC}{xz}=\frac{BC}{yz}

and

< A= < X,\ < B= < y\ and\ < C= < z

5 0
2 years ago
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