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denpristay [2]
3 years ago
8

Some one please help me!

Mathematics
1 answer:
pashok25 [27]3 years ago
7 0
Really ok wait hold up
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Collected 20 cans this week, 25 cans next week what is the percent of increase
BartSMP [9]
Hello!!

25 - 20 = 5 

5/20 = 0.25

0.25 * 100 = 25%

Good luck :)
5 0
2 years ago
I need help can someone help me
Marta_Voda [28]

Answer:

the answer is definitely d

8 0
3 years ago
In the diagram below, triangle ABC is drawn with altitude BD. If AD = 30 and AC = 36, determine the length of AB rounded to the
denis-greek [22]

9514 1404 393

Answer:

  32.9

Step-by-step explanation:

The ratio of long side to hypotenuse is the same for all of the triangles in the figure.

  AD/AB = AB/AC

  AB^2 = AD·AC = 30·36

  AB = 6√30 ≈ 32.863

To the nearest tenth, AB ≈ 32.9.

5 0
3 years ago
HELP!!!
marusya05 [52]
The equation of an ellipse is x²/a²  + y²/b² =1

a being the distance between the center & x intercept =7
b being the distance between the center & y intercept =5

Hence the equation of this ellipse is:

x²/49 + y²/25 =1
3 0
3 years ago
Find all values of c in the open interval (a, b) such that f'(c)=(f(b)-f(a))/(b-a)
timama [110]
<h3>Answer:   c = 7/4</h3>

================================================

Work Shown:

Compute the function value at the endpoints

f(x) = \sqrt{4-x}\\\\f(-5) = \sqrt{4-(-5)} = 3\\\\f(4) = \sqrt{4-4} = 0\\\\

With a = -5 and b = 4, we have

f'(c) = \frac{f(b)-f(a)}{b-a}\\\\f'(c) = \frac{f(4)-f(-5)}{4-(-5)}\\\\f'(c) = \frac{0-3}{9}\\\\f'(c) = -\frac{1}{3}\\\\

So,

f(x) = \sqrt{4-x}\\\\f'(x) = -\frac{1}{2\sqrt{4-x}}\\\\f'(c) = -\frac{1}{3}\\\\-\frac{1}{2\sqrt{4-c}} = -\frac{1}{3}\\\\

Use algebra to solve for c

-\frac{1}{2\sqrt{4-c}} = -\frac{1}{3}\\\\\frac{1}{2\sqrt{4-c}} = \frac{1}{3}\\\\3 = 2\sqrt{4-c}\\\\2\sqrt{4-c} = 3\\\\\sqrt{4-c} = \frac{3}{2}\\\\4-c = \frac{9}{4}\\\\c = 4-\frac{9}{4}\\\\c = \frac{16-9}{4}\\\\c = \frac{7}{4}\\\\

6 0
2 years ago
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