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QveST [7]
3 years ago
9

Couldnt find it on google so ima try here

Mathematics
1 answer:
kodGreya [7K]3 years ago
4 0

Answer:

7.0

Step-by-step explanation:

You might be interested in
14=h/5<br><br> Help me please
ValentinkaMS [17]

Answer:

14=h/5

h=14*5

h=70

<u>Check</u>:

70/5=14

Hope this helps

6 0
3 years ago
10^2+12^2+14^2+......+26^2=??
Kazeer [188]

Answer:

3156

Step-by-step explanation:

  • <em>Used formula:</em>
  • <em>(1² + 2² + 3² + ... + n²) =1/6*n(n + 1)(2n + 1)</em>

--------

  • 10²+12²+14²+......+26² =
  • (2*5)²+(2*6)² + (2*7)² + ... + (2*13)² =
  • 4*(5²+6²+7²+...+13²) =
  • 4*(1²+2²+...+13² - (1²+2²+3²+4²)) =
  • 4*(1/6*13(13+1)(2*13+1) - (1+4+9+16)) =
  • 4*(1/6*13*14*27- 30) =
  • 4*(819 - 30) =
  • 4*789 =
  • 3156
3 0
4 years ago
Michael's babysitting is 20 years old. Michael is 8 years old. How much older is his babysitter than Micheal?​
iogann1982 [59]

Answer:

12 years

Step-by-step explanation:

20-8=12

Micheal's babysitter is 12 years older than Micheal.

8 0
3 years ago
How to slove for x?​
Luba_88 [7]
Ok so you can see 2 triangles that i highlighted, flip it so it is the same direction as you can see from the drawing
using the similar triangle equation
you get
136/64 = x/136
cross multiply and get
64x=18496
x=289

hope it help

8 0
4 years ago
Max makes and sells posters. The function p(x)= -10x^2 +200x -250, graphed below, indicates how much profit he makes in a month
viktelen [127]
Here is our profit as a function of # of posters
p(x) =-10x² + 200x - 250
Here is our price per poster, as a function of the # of posters:
pr(x) = 20 - x
Since we want to find the optimum price and # of posters, let's plug our price function into our profit function, to find the optimum x, and then use that to find the optimum price:
p(x) = -10 (20-x)² + 200 (20 - x) - 250
p(x) = -10 (400 -40x + x²) + 4000 - 200x - 250
Take a look at our profit function. It is a normal trinomial square, with a negative sign on the squared term. This means the curve is a downward facing parabola, so our profit maximum will be the top of the curve.
By taking the derivative, we can find where p'(x) = 0 (where the slope of p(x) equals 0), to see where the top of profit function is.
p(x) = -4000 +400x -10x² + 4000 -200x -250
p'(x) = 400 - 20x -200
0 = 200 - 20x
20x = 200
x = 10                         
p'(x) = 0 at x=10. This is the peak of our profit function. To find the price per poster, plug x=10 into our price function:
price = 20 - x
price = 10
Now plug x=10 into our original profit function in order to find our maximum profit:
<span>p(x)= -10x^2 +200x -250
p(x) = -10 (10)</span>² +200 (10) - 250
<span>p(x) = -1000 + 2000 - 250
p(x) = 750

Correct answer is C)
</span>
7 0
3 years ago
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