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Nutka1998 [239]
2 years ago
6

What is the solution of StartRoot negative 4 x EndRoot = 100?

Mathematics
2 answers:
Brilliant_brown [7]2 years ago
6 0

Answer:

A

Step-by-step explanation:

slava [35]2 years ago
3 0

Answer:

x=-2500

Step-by-step explanation:

\sqrt{-4x} =100\\\\(\sqrt{-4x})^2=100^2\\\\-4x=10,000\\\\x=-\frac{10000}{4}\\\\x=-2500

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Find the sum of (x+5), (-4x-2), and (2x-1)<br><br> Show your work please
Papessa [141]
Sum means add
(x+5) + (-4x-2) + (2x-1)
=-3x+3+(2x-1)
=-x+2
7 0
3 years ago
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What is the volume of a rectangular prism with the base area of 15m2 and height of 5cm?
koban [17]

Answer:

The correct answer is option C. 75 m³

Step-by-step explanation:

<u>Points to remember</u>

Volume of rectangular prism = Base area * Height

<u>To find the volume of given prism</u>

Here Base area = 15 m² and

Height = 5 m

Volume = base area * height

 = 15 * 5

 = 75 m³

Volume of prism = 75 m³

Therefore the correct answer is option C. 75 m³

3 0
3 years ago
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1/2 + 56.5 = ? what is the answer reward 10 pts for it.
Nat2105 [25]
I'm assuming it's 57. Because 1/2 can equal .5 and 56.5+.5=57
7 0
3 years ago
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student randomly receive 1 of 4 versions(A, B, C, D) of a math test. What is the probability that at least 3 of the 5 student te
alexdok [17]

Answer:

1.2%

Step-by-step explanation:

We are given that the students receive different versions of the math namely A, B, C and D.

So, the probability that a student receives version A = \frac{1}{4}.

Thus, the probability that the student does not receive version A = 1-\frac{1}{4} = \frac{3}{4}.

So, the possibilities that at-least 3 out of 5 students receive version A are,

1) 3 receives version A and 2 does not receive version A

2) 4 receives version A and 1 does not receive version A

3) All 5 students receive version A

Then the probability that at-least 3 out of 5 students receive version A is given by,

\frac{1}{4}\times \frac{1}{4}\times \frac{1}{4}\times \frac{3}{4}\times \frac{3}{4}+\frac{1}{4}\times \frac{1}{4}\times \frac{1}{4}\times \frac{1}{4}\times \frac{3}{4}+\frac{1}{4}\times \frac{1}{4}\times \frac{1}{4}\times \frac{1}{4}\times \frac{1}{4}

= (\frac{1}{4})^3\times (\frac{3}{4})^2+(\frac{1}{4})^4\times (\frac{3}{4})+(\frac{1}{4})^5

= (\frac{1}{4})^3\times (\frac{3}{4})[\frac{3}{4}+\frac{1}{4}+(\frac{1}{4})^2]

= (\frac{3}{4^4})[1+\frac{1}{16}]

= (\frac{3}{256})[\frac{17}{16}]

= 0.01171875 × 1.0625

= 0.01245

Thus, the probability that at least 3 out of 5 students receive version A is 0.0124

So, in percent the probability is 0.0124 × 100 = 1.24%

To the nearest tenth, the required probability is 1.2%.

4 0
3 years ago
PLS HELP WITH 3 THROUGH 8!!!!! Plzzz helpppppp I'll give you brainliestttttttttttttt
Masteriza [31]

Answer:

3=%20

4=%14.17

5=%48.8

6=%165

7=%75

8=%15

3 0
3 years ago
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