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3 years ago

Which of the following expressions could be used to represent "the difference between -17 and -8"?

Mathematics

1 answer:

musickatia [10]3 years ago

7 0

The last bc the + and - are different that means it will be subtracted.

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The class treasury contained 30$ in nickels and dimes. If there were 500 coins than how many coins of each type are there?

Alenkasestr [34]

Equation 1. N+D=500

equation 2 5N+10D=3000
This is for equation 1
N=500-d
Substitute into equations 2 and solve for D
5N+10D=3000
5(500-D)+10D=3000
2500-5D+10D=3000
5D=500
D=100
Lets go back the equation 1
N=500-D
N=500-100
N=400
there are 400 Nickles and 100 Dimes

7 0

2 years ago

A rectangular box has a square base. The combined length of a side of the square base, and the height is 20 in. Let x be the len

aniked [119]

Answer:

a. V = (20-x) $x^{2}$ $in^{3}$

b . 1185.185 $in^{3}$

Step-by-step explanation:

Given that:

The height: 20 - x (in )
Let x be the length of a side of the base of the box (x>0)

a. Write a polynomial function in factored form modeling the volume V of the box.

As we know that, this is a rectangular box has a square base so the Volume of it is:

V = h * $x^{2}$ $in^{3}$

<=> V = (20-x) $x^{2}$ $in^{3}$

b. What is the maximum possible volume of the box?

To maximum the volume of it, we need to use first derivative of the volume.

<=> dV / Dx = -3 $x^{2}$ + 40x

Let dV / Dx = 0, we have:

-3 $x^{2}$ + 40x = 0

<=> x = 40/3

=>the height h = 20/3

So the maximum possible volume of the box is:

V = 20/3 * 40/3 *40/3

= 1185.185 $in^{3}$

7 0

2 years ago

Determine the mass of a solid steel ball that has a diameter of 10 cm. The density of pure steel is 8.09 g/cm3

Basile [38]

Answer:

<em>The mass of the steel ball is 4,235.9 gr</em>

Step-by-step explanation:

<u>Density</u>

The density ρ of a substance is a measure of its mass per unit volume:

$\displaystyle \rho=\frac{m}{V}$

If the density and the volume are given, the mass can be calculated by solving the above formula for m:

$m=\rho.V$

We know the density of pure steel ρ=8.09 gr/cm3 and the diameter of a solid steel ball d=10 cm.

We need to calculate the volume of the sphere:

The volume of a sphere of radius r is given by:

$\displaystyle V=\frac{4}{3}\cdot \pi\cdot r^3$

The radius is half the diameter: r= 10/2 = 5 cm. Thus:

$\displaystyle V=\frac{4}{3}\cdot \pi\cdot 5^3$

Calculating:

$V=523.6\ cm^3$

The mass is:

$m=8.09 gr/cm^3 \cdot 523.6\ cm^3$

m=4,235.9 gr

The mass of the steel ball is 4,235.9 gr

4 0

2 years ago

An architect is designing square windows with an area of(x2 + 16x +64) ft?. The dimensions of the

Darina [25.2K]

A) The dimensions are (x+10) by (x+10).

B) The perimeter is given by 4x+40.

C) The perimeter when x is 4 is 56.

The quadratic can be factored by finding factors of c, the constant, that sum to b, the coefficient of x. Our c is 100 and our b is 20; we want factors of 100 that sum to 20. 10*10=100 and 10+10=20, so those are what we need. This gives us (x+10)(x+10 for the factored form.

Since the dimensions are all (x+10), and there are 4 sides, the perimeter is given by 4(x+10). Using the distributive property we have 4*x+4*10=4x+40.

To find the perimeter when x=4, substitute 4 into our perimeter expression:

4*4+40=16+40=56.

6 0

2 years ago

Read 2 more answers

The base of a triangle is 10 cm

bonufazy [111]

I’m not sure about this question

4 0

2 years ago