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WITCHER [35]
3 years ago
14

Evaluate the expression when, m=-2 m²-6m-5 (Thank you for your help!)

Mathematics
1 answer:
Darina [25.2K]3 years ago
7 0

Answer:

11

Step-by-step explanation:

m=-2\\m^2-6m-5\\=(-2)^2-6(-2)-5\\=4-(-12)-5\\=4+12-5\\=16-5\\=11

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Find the area of the semicircular desk. Round your answer to the nearest hundredth.
padilas [110]

Answer:

acavadods number 6.0324 10X343434

the anser is 24, 4 emc^2 æπ

its rπ^2 times avagradhdsos numer

Step-by-step explanation:

7 0
3 years ago
Read 2 more answers
How to solve -16+x=-4<br> 13
kotykmax [81]

Answer:


you want to isolate x so just add 16 to the right side (-4)

so -16 + x = -4

     +16         +16


x=12


hope this helps :)


5 0
3 years ago
Can some one help me solve for X?
Korvikt [17]

Answer:

x = 20

Step-by-step explanation:

The two angles are complementary angles, which means they add to 90 degrees

2x+7 + 43 = 90

Combine like terms

2x+50 = 90

Subtract 50 from each side

2x+50-50 = 90-50

2x = 40

Divide each side by 2

2x/2 = 40/2

x = 20

8 0
3 years ago
Read 2 more answers
Solve the equatuon<br><br> x/2-(2/(x+1))=1<br><br> x=?
Snezhnost [94]

\frac{x}{2}-\frac{2}{x+1}=1

We have 2 denominators that we need to get rid of. Whenever there are the denominators, all we have to do is multiply all whole equation with the denominators.

Our denominators are both 2 and x+1. Therefore, we multiply the whole equation by 2(x+1)

\frac{x}{2}[2(x+1)]-\frac{2}{x+1}[2(x+1)] = 1[2(x+1)]

Then shorten the fractions.

\frac{x}{2}[2(x+1)]-\frac{2}{x+1}[2(x+1)] = 1[2(x+1)]\\x(x+1)-2(2)=1(2x+2)

Distribute in all.

x^2+x-4=2x+2

We should get like this. Because the polynomial is 2-degree, I'd suggest you to move all terms to one place. Therefore, moving 2x+2 to another side and subtract.

x^2+x-4-2x-2=0\\x^2-x-6=0\\

We are almost there. All we have to do is, solving for x by factoring. (Although there are more than just factoring but factoring this polynomial is faster.)

(x-3)(x+2)=0\\x=3,-2

Thus, the answer is x = 3, -2

7 0
3 years ago
What is the equation with a slope of 0 that has a point of (6,-11)?
lianna [129]

\bf (\stackrel{x_1}{6}~,~\stackrel{y_1}{-11})~\hspace{10em} \stackrel{slope}{m}\implies 0 \\\\\\ \begin{array}{|c|ll} \cline{1-1} \textit{point-slope form}\\ \cline{1-1} \\ y-y_1=m(x-x_1) \\\\ \cline{1-1} \end{array}\implies y-\stackrel{y_1}{(-11)}=\stackrel{m}{0}(x-\stackrel{x_1}{6}) \\\\\\ y+11=0\implies y=-11

5 0
3 years ago
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