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Pepsi [2]
3 years ago
12

What is 24% as a fraction?

Mathematics
2 answers:
nata0808 [166]3 years ago
5 0

Answer: 6/25 or 0.24! :)

IgorLugansk [536]3 years ago
4 0

24

___

100

yesssa above is right

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Question 3 (2 points)
Sveta_85 [38]
I believe the answer is a
5 0
2 years ago
-64+ x = -90.3<br><br><br> PLEASE HELP ASAP!!:))
ludmilkaskok [199]

Answer:

add 64 to both sides

Step-by-step explanation:

x = -26.3

7 0
2 years ago
Write an equation in slope-intercept form for the line with slope<br> -1/4<br> and y-intercept 4.
Setler [38]

Answer:

4

Step-by-step explanation:

The equation of the line is written in the slope-intercept form, which is: y = mx + b, where m represents the slope and b represents the y-intercept. In our equation, y = − 7 x + 4 , we see that the y-intercept of the line is 4.

Add me brainiest

8 0
2 years ago
Read 2 more answers
Problem page a delivery truck is transporting boxes of two sizes: large and small. the large boxes weigh 45 pounds each, and the
marta [7]
S= number of small boxes
l= number of large boxes

equation 1:  s+l=120
equation 2:  15s+45l=3300

solve by elimination, multiply equation 1 by -15.

-15(s+l=120) =  -15s-15l=-1800  add to equation 2.

-15s+15s-15l+45l=-1800+3300  = 30l=1500

30l=1500 , l=50

s+l=120, s+50=120 --> s=70

7 0
3 years ago
HELP ASAP!!! 40 POINTS
ICE Princess25 [194]

Using derivatives, it is found that the x-values in which the slope belong to the interval (-1,1) are in the following interval:

(-15,-10).

<h3>What is the slope of the tangent line to a function f(x) at point x = x0?</h3>

It is given by the derivative at x = x0, that is:

m = f^{\prime}(x_0).

In this problem, the function is:

f(x) = 0.2x^2 + 5x - 12

Hence the derivative is:

f^{\prime}(x) = 0.4x + 5

For a slope of -1, we have that:

0.4x + 5 = -1

0.4x = -6

x = -15.

For a slope of 1, we have that:

0.4x + 5 = 1.

0.4x = -4

x = -10

Hence the interval is:

(-15,-10).

More can be learned about derivatives and tangent lines at brainly.com/question/8174665

#SPJ1

8 0
2 years ago
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