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Ksenya-84 [330]
3 years ago
8

Write an equation that represents the following problem.

Mathematics
1 answer:
pogonyaev3 years ago
5 0
35 + 0.05m = 50 + 0.01m <== ur equation
0.05m - 0.01m = 50 - 35
0.04m = 15
m = 15/0.04
m = 375.....u need to send 375 text messages for the plans to be equal
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Jordan is stacking 2 boxes on a shelf. The bottom box measures 7 inches​ long, 6 inches​ wide, and 6 inches high. The top box is
Andre45 [30]
Answer: 377

Explanation: Length x width x height
7x6 = 42 x 6 = 252
A cube means all sides a equal. So 5x5x5 = 125
Add both together and you get 377 cubic inches.
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The spinner has eight spaces of equal size. What is the mathematical probability of the pointer's stopping in the space marked 1
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10/40

Step-by-step explanation:

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A 50 lb bag of grass seed cost $222.50 find the unit cost per lb of this grass seed
myrzilka [38]
To find the cost per one pound, we divide the total cost $222.50 by the total number of pounds (50 lb).

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3 0
3 years ago
A rectangular prism is 20 meters long, 13 meters wide, and 5 meters high. What is the surface area of the rectangular prism?
RUDIKE [14]

Answer:

850m^2

Step-by-step explanation:

Look at the formula SA=L*W*H. And plug in the numbers pertaining the question! hints on to what to add is stuff like "long" for length, "wide" for width, and "high" for height! And use a calculator, or scetch it out to find your total answer!

Hope this helps!

3 0
2 years ago
Let f(x) = 1/x^2 (a) Use the definition of the derivatve to find f'(x). (b) Find the equation of the tangent line at x=2
Verdich [7]

Answer:

(a) f'(x)=-\frac{2}{x^3}

(b) y=-0.25x+0.75

Step-by-step explanation:

The given function is

f(x)=\frac{1}{x^2}                  .... (1)

According to the first principle of the derivative,

f'(x)=lim_{h\rightarrow 0}\frac{f(x+h)-f(x)}{h}

f'(x)=lim_{h\rightarrow 0}\frac{\frac{1}{(x+h)^2}-\frac{1}{x^2}}{h}

f'(x)=lim_{h\rightarrow 0}\frac{\frac{x^2-(x+h)^2}{x^2(x+h)^2}}{h}

f'(x)=lim_{h\rightarrow 0}\frac{x^2-x^2-2xh-h^2}{hx^2(x+h)^2}

f'(x)=lim_{h\rightarrow 0}\frac{-2xh-h^2}{hx^2(x+h)^2}

f'(x)=lim_{h\rightarrow 0}\frac{-h(2x+h)}{hx^2(x+h)^2}

Cancel out common factors.

f'(x)=lim_{h\rightarrow 0}\frac{-(2x+h)}{x^2(x+h)^2}

By applying limit, we get

f'(x)=\frac{-(2x+0)}{x^2(x+0)^2}

f'(x)=\frac{-2x)}{x^4}

f'(x)=\frac{-2)}{x^3}                         .... (2)

Therefore f'(x)=-\frac{2}{x^3}.

(b)

Put x=2, to find the y-coordinate of point of tangency.

f(x)=\frac{1}{2^2}=\frac{1}{4}=0.25

The coordinates of point of tangency are (2,0.25).

The slope of tangent at x=2 is

m=(\frac{dy}{dx})_{x=2}=f'(x)_{x=2}

Substitute x=2 in equation 2.

f'(2)=\frac{-2}{(2)^3}=\frac{-2}{8}=\frac{-1}{4}=-0.25

The slope of the tangent line at x=2 is -0.25.

The slope of tangent is -0.25 and the tangent passes through the point (2,0.25).

Using point slope form the equation of tangent is

y-y_1=m(x-x_1)

y-0.25=-0.25(x-2)

y-0.25=-0.25x+0.5

y=-0.25x+0.5+0.25

y=-0.25x+0.75

Therefore the equation of the tangent line at x=2 is y=-0.25x+0.75.

5 0
3 years ago
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