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Arturiano [62]
3 years ago
11

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.
You might be interested in
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 = -3x^{2} + 40x

Let dV / Dx = 0, we have:

-3x^{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
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