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zhuklara [117]
3 years ago
6

Pls help me need help

Mathematics
1 answer:
Rashid [163]3 years ago
5 0

Answer:1,4/5

Step-by-step explanation:https://www.cymath.com/answer?q=5x%5E2%2Bx%3D4

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How would I put this into a solvable equation, m= ak/n?
12345 [234]

The given equation is

m = ak/n

We want to solve for k

The first step is to multiply both sides of the equation by n. We have

m * n = ak/n * n

mn = ak

The next step is to divide both sides of the equation by a. We have

mn/a = ak/a

mn/a = k

k = mn/a

The correct option is the second one

7 0
1 year ago
Considering the expression x/2 + y2. Which statements are true?Mark all that apply.
Svet_ta [14]

Answer:

  D.  8/2 +(-10)² = 104

Step-by-step explanation:

To find which statements are true, evaluate the expression for the given variable values. You do this by putting the numbers in place of the respective variables, then doing the arithmetic.

The respective expression values are ...

  A 83 ≠ 37 . . . . (4/2) +9² = 2 +81 = 83

  B 31 ≠ 61

  C 52 ≠ 61

  D 104 =104 . . . . true statement

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I find it less tedious to write a function into a calculator or spreadsheet and let it do the repetitive math. Examples of the calculation are shown above.

7 0
3 years ago
Please help!!<br> Find the volume of the sphere
Komok [63]

Answer:

Hi , your answer is V = 4/3 πr³.

Hope this helps

4 0
2 years ago
Which of the following is the graph of y = negative StartRoot x EndRoot + 1? On a coordinate plane, an absolute value curve curv
mestny [16]

Answer:

  • B. On a coordinate plane, an absolute value curve curves up and to the right in quadrant 4 and starts at y = 1.

Step-by-step explanation:

<u>Graph of the function:</u>

  • y = -√x + 1

The domain is x ≥ 0, the range y ≤ 1

Correct answer choice is B

  • On a coordinate plane, an absolute value curve curves up and to the right in quadrant 4 and starts at y = 1.

<em>The graph is attached</em>

7 0
3 years ago
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Divide <span>150:100 to find out what the density in g/cm3 is. So 150:100 = 1.5 g/cm3</span>
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3 years ago
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