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Anton [14]
3 years ago
12

What is 4% of 52???? WILL MARK BRAINLLEST

Mathematics
2 answers:
Nimfa-mama [501]3 years ago
6 0

Answer:

2.08

Step-by-step explanation:

4%=.04

52*.04= 2.08

Hope this helps .-.

KATRIN_1 [288]3 years ago
3 0

The answer to 4% of 52 is 2.08.

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Slove the following:​
Aleksandr [31]

Answer:

let required no be x

\frac{1}{7}  =  \frac{x}{42}

by doing crisscrossed multiplication

42 \times 1 = 7 \times x

42=7x

dividing both side by 7

\frac{42}{7}  = 7x \div 7

6=x

therefore x=6

therefore value of remaining no.is 6

7 0
3 years ago
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Where does the graph of y = 4x + 24 intersect the x-axis?
ehidna [41]

Answer:

(-6, 0)

Step-by-step explanation:

y = 4x + 24

Plug y = 0

0 = 4x + 24

4x = - 24

x = - 24/4

x = - 6

So, the graph of y = 4x + 24 intersect the x-axis at (-6, 0)

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3 years ago
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8x + y = 20<br> x - 2y = -6
OleMash [197]

Answer: the answer is

Step-by-step explanation:

6 0
2 years ago
Solve for w.<br> 4 = 0.25(w-4.3)
Snezhnost [94]

W=20.3

solve what's in the parenthesis, move the terms, calculate, then divide both sides.

8 0
4 years ago
Sara is working on a Geometry problem in her Algebra class. The problem requires Sara to use the two quadrilaterals below to ans
zloy xaker [14]
Part A:

Given a square with sides 6 and x + 4. Also, given a rectangle with sides 2 and 3x + 4

The perimeter of the square is given by 4(x + 4) = 4x + 16

The area of the rectangle is given by 2(2) + 2(3x + 4) = 4 + 6x + 8 = 6x + 12

For the perimeters to be the same

4x + 16 = 6x + 12
4x - 6x = 12 - 16
-2x = -4
x = -4 / -2 = 2

The value of x that makes the <span>perimeters of the quadrilaterals the same is 2.



Part B:

The area of the square is given by

Area=(x+4)^2=x^2+8x+16

The area of the rectangle is given by 2(3x + 4) = 6x + 8

For the areas to be the same

x^2+8x+16=6x+8 \\  \\ \Rightarrow x^2+8x-6x+16-8=0 \\  \\ \Rightarrow x^2+2x+8=0 \\  \\ \Rightarrow x= \frac{-2\pm\sqrt{2^2-4(8)}}{2}  \\  \\ = \frac{-2\pm\sqrt{4-32}}{2} = \frac{-2\pm\sqrt{-28}}{2}  \\  \\ = \frac{-2\pm2i\sqrt{7}}{2} =-1\pm i\sqrt{7}

Thus, there is no real value of x for which the area of the quadrilaterals will be the same.
</span>
7 0
4 years ago
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