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madam [21]
3 years ago
7

I can really use some help:)

Mathematics
2 answers:
Leona [35]3 years ago
8 0
You do -5( 3+4)

-5(7)
-35
icang [17]3 years ago
7 0
You first have to multiply the exponent into the other numbers
so -5×3=(-15)
and -5×4=(-20)
and (-20)+(-15)=(-35)
The value of the expression is A which is -35
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Step-by-step explanation:

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A rectangle box with a square base and no top needs to be made using 300ft^2 of material. Find the dimensions of the box with th
Yuri [45]

The dimensions of the box are 10 ft and 5 ft

The maximum volume is 500 ft³

Step-by-step explanation:

A rectangle box with

  • A square base and no top
  • It needs to be made using 300 ft² of material
  • It has greatest volume

Surface area of a box without top (SA) = perimeter of base × height + area of the base

Volume of a box (V) = base area × height

Assume that the length of the side of the square base is x and the height of the box is y

∵ It needs to be made using 300 ft² of material

∴ The surface area of the box is 300 ft²

∵ Its base is a square of side length x ft

∴ Perimeter of the base = 4 × x = 4 x

∴ Area of the base = x²

∵ The height of the box = y ft

∵ SA = perimeter of base × height + area of the base

∵ SA = (4x)(y) + x²

∴ SA = 4xy + x²

∵ SA of the box = 300 ft²

- Equate the two expressions of SA

∴ 4xy + x² = 300

Now let us find y in terms of x

- Subtract x² from both sides

∴ 4xy = 300 - x²

- Divide each term by 4x to find y

∴ y=\frac{75}{x}-\frac{1}{4}x

∵ V = area of the base × height

∴ V = x² × y = x²y

- Substitute y by the equation of it above

∴ V=x^{2}(\frac{75}{x}-\frac{1}{4}x)

∴ V=75x-\frac{1}{4}x^{3}

∵ The volume of the box is greatest

- That means differentiate V and equate it by 0

∵ \frac{dV}{dx}=75-\frac{3}{4}x^{2}

∵ \frac{dV}{dx}=0 ⇒ greatest volume

∴ 75-\frac{3}{4}x^{2}=0

- Subtract 75 from both sides

∴ -\frac{3}{4}x^{2}=-75

- Divide both sides by -\frac{3}{4}

∴ x² = 100

- Take √ for both sides

∴ x = 10

Substitute the value of x in the equation of y

∵ y=\frac{75}{10}-\frac{1}{4}(10)

∴ y = 5

The dimensions of the box are 10 ft and 5 ft

∵ V=75x-\frac{1}{4}x^{3}

∵ x = 10

∴ V=75(10)-\frac{1}{4}(10)^{3}

∴ V=750-\frac{1}{4}(1000)

∴ V = 750 - 250

∴ V = 500 ft³

The maximum volume is 500 ft³

Learn more:

You can learn more about the volume in brainly.com/question/6443737

#LearnwithBrainly

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3 years ago
The diagram shows the measurements of a frame into which concrete is poured to make steps. The steps are shaped like rectangular
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Find the range of the function F(x) = the integral from 0 to x of the square root of 4-t^2 dt
ioda

Note that √(4 - t²) is defined only as long as 4 - t² ≥ 0, or -2 ≤ t ≤ 2. Then the real integral exists only if -2 ≤ x ≤ 2. (Otherwise we deal with complex numbers.)

If x = 2, then the integral corresponds to the area of a quarter-circle with radius 2. This means that the integral has a maximum value of 1/4 • π • 2² = π.

On the opposite end, if x = -2, then the integral has the same value, but the integral from 0 to -2 is equal to the negative integral from -2 to 0. So the minimum value is -π.

For all x in between, we observe that the integrand is continuous over the rest of its domain, so F(x) is continuous.

Then the range of F(x) is the interval [-π, π].

8 0
2 years ago
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