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Vlada [557]
3 years ago
6

Need help asp please I ready

Mathematics
2 answers:
Otrada [13]3 years ago
8 0

Answer:

-4m - 3 , B is your answer

Step-by-step explanation:

-0.25 x 16m = -4m

0.25 x 12 = -3

Whenever you multiply a negative by a positive number you will get a negative number using the negative sign from -3 it would be -4m - 3 or you could also write with its number -4m -3.

Natalka [10]3 years ago
3 0

Answer:

-4m-3

Step-by-step explanation:

-0.25 times 16 is the same thing as dividing 16 by -4. 16m divided by -4 is -4m

-0.25 times 12 is the same thing as dividing 12 by -4. 12 divided by -4 is -3

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Find the smallest number by which 248 should be subtracted to make them perfect square​
dimaraw [331]

Answer:

21.

Step-by-step explanation:

The next perfect square below 248 i s 225 ( = 15^2).

So the answer is 248 - 225 = 21.

4 0
3 years ago
What’s the answer for a, and b, and how would you solve it?
xxTIMURxx [149]

Answer:

a) negative

b) -7/6

Step-by-step explanation:

The x value increased, but the y value decreased so the line goes down and has a negative slope

(y₁-y₂)/(x₁-x₂) is the slope where (x₁,y₁) and (x₂,y₂) are points on the line

(4-(-3))/(-1-5)=7/-6

6 0
2 years ago
Please help 2х -3y =18
siniylev [52]

Answer:

x = 3(6 + y)/2

Step-by-step explanation:

Solving for x

Add 3y to both sides.

2x = 18 + 3y

Divide both sides by 2.

x = 18 + 3y/2

Factor out the common term 3.

x = 3(6 + y)/2

3 0
3 years ago
Which of the following represents 4x^5/6 in radical form
mash [69]
Remember, order of operations
exponent before multiply

so 4x^5/6=4 times x^5/6
simplify the x^5/6 first

remember
x^\frac{m}{n}=\sqrt[n]{s^m}

so
x^\frac{5}{6}=\sqrt[6]{x^5}

so
4x^\frac{5}{6}=4\sqrt[6]{x^5}
<span />
6 0
4 years ago
Prove that <img src="https://tex.z-dn.net/?f=cos%209x-cos%205x%2Fsin17x-sin3x%3D-sin2x%2Fcos10x" id="TexFormula1" title="cos 9x-
maria [59]
Cos9x - cos5x = -2sin2x.sin7x
sin17x - sin3x = 2sin7x.cos10x 
so f(x) = \frac{-2sin2x.sin7x}{2sin7x.cos10x} =\frac{-sin2x}{cos10x}

8 0
3 years ago
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