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2 years ago

Double the difference of 25 and 10 is

Mathematics

2 answers:

Gennadij [26K]2 years ago

7 0

Answer:

Step-by-step explanation:

25-10 = 15

15 x 2 = 30

Oduvanchick [21]2 years ago

6 0

<h2>Double the difference of 25 and 10 is:</h2>

------------------

First we must get the difference between 25 and 10.

$25 - 10 = \green{15}$

Next, we must double the answer.

$15 \times 2 = \bold \blue{30} \\ \tt or \\ {15}^{2} = \purple {\sf{30}}$

Therefore, the answer is 30.

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3. A jump rope is 9 feet long. How long is the jump 4 rope in yards?

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17. A bag contains 2x yellow counters, 4x + 6 red counters and 6x - 10 blue counters. a. Write an expression, in terms of x, for

Natalija [7]

Answer: 12x-4, 14 counters, 8 more counters

Step-by-step explanation: There are 2x yellow, 4x+6 red, and 6x-10 blue counters. We need to add all of this up to get the total number of counters. We first add the x's : 2x+4x+6x = 12x. Then we add the numbers. 6+ (-10) is -4. So, our expression is 12x-4. Next, we need to find out how many blue counters are in the bag. We know that there are 44 total counters and we need to find x because all the counters have x in them. 12x-4 = 44. We first add 4 to both sides to get 12x= 48 and x = 4. blue has 6x-10 counters so, blue has 24-10 = 14 counters. Red has 16+6 = 22 counters. 22-14 = 8 more counters

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1 year ago

A rectangular solid has width w, a length of 7 more than the width, and a height that is equivalent to 15 decreased by 3 times t

dexar [7]

Answer:

For a rectangular solid with:

width = w

length = l

height = h

The volume is equal to:

V = w*l*h

in this case we know that:

width = w.

"length of 7 more than the width"

l = w + 7.

"a height that is equivalent to 15 decreased by 3 times the width."

h = 15 - 3*w

Then the volume will be:

V = w*l*h = w*(w + 7)*(15 - 3w) = (w^2 + 7*w)*(15 - 3*w)

V = ( -3*w^3 + 15*w^2 + 105*w - 21*w^2)

V = (-3*w^3 - 6*w^2 + 105*w)

Now, the maximum volume will be for the value of w such that:

V'(w) = 0.

and:

V''(W) < 0

Where:

dV/dw = V'(w).

dV'/dw = V''(w)

Then first we need to differentiate the equation for the volume.

V'(w) = dV/dw = ( 3*(-3*w^2) + 2*(-6*w) + 105)

V'(w) = -9*w^2 - 12*w + 105.

Then we need to find the solution for:

-9*w^2 - 12*w + 105 = 0.

We can use the Bhaskara formula, and we will get:

$w = \frac{+12 +-\sqrt{(-12)^2 - 4*(-9)*105} }{2*-9} = \frac{+12 +- 62.6}{-18}$

Then the two solutions are:

w₁ = (+12 - 62.6)/(-18) = 2.81

w₂ = (+12 + 62.6)/(-18) = -15.5

But we can not have a negative width, so we can just discard the second solution.

Now let's check the second condition for the maximum, we must have:

V''(2.81) < 0.

V'' = dV'/dw = 2*(-9*w) - 12 = -18*w - 12

V''(2.81) = -18*2.81 - 12 = -62.58 < 0 .

Then the volume is maximized when w = 2.81, and the maximum volume will be:

V(2.81) = (-3*(2.81)^3 - 6*(2.81)^2 + 105*2.81) = 180.1

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