Ask question

Ask question

All categories

3 years ago

7

A county has 90 schools. The county received 992 new computers. Are there enough computers so that each school can get 11 comput

ers? Explain

1 answer:

Ray Of Light [21]3 years ago

4 0

Answer:

there will be enough for each school to get 11 computers

Step-by-step explanation:

use a calculator and put in 992 divided by 11. your answer should appear as 90.1818182

You might be interested in

Easy Multiple Choice Questions about expressions and linear equations / ASAP / Will mark brainliest / 25 Points

nata0808 [166]

Answer:

1.C

2.A

3.D

4.C

5.B

6.A

7.B

8.C

9.A

10. D

Step-by-step explanation:

For Problem 1, the work is as follows:

$2(3x-1)+x\\Distribute\\6x-2+x\\7x-2$

For Problem 2, the work is as follows:

$(3x^{2} +6x+4) - (-x^{2} +2x-1)\\Distribute -\\(3x^{2} +6x+4) + (x^{2} -2x+1)\\Combine\\4x^{2} +4x+5$

For Problem 3, the work is as follows:

$5b-(a+c)^{2} \\5(3) - (-1+-4)^{2} \\15 - (-5)^{2} \\15-25\\-10$

For Problem 4, the work is as follows:

$4x-8 = -12\\+8\\4x = -4\\-4/4 = -1\\x = -1$

For Problem 5, the work is as follows:

$\frac{1}{2} (12x-6) = 2x-15\\Distribute\\\\6x - 3 = 2x -15\\+3\\6x = 2x -12\\-2x\\4x = -12\\-12/4\\x = -3$

For Problem 6, the work is as follows:

$17 *0.06 = 1.02\\17+1.02 = 18.02$

For Problem 7, the work is as follows:

$x = (2,0)\\y = (0,-1)\\\frac{rise}{run} \\y = \frac{1}{2} x-1$

For Problem 8, the work is as follows:

$y_{2} -y_{1} \\x_{2} -x_{1} \\-1-1 = -2\\0-(-2) = 2\\Slope = -1$

For Problem 9, the work is as follows:

$3y - 2x +6 = 0\\-3y\\(-3y - -2x+6)/-3\\y = \frac{2}{3} x -2$

For Problem 10, the work is as follows:

$(y-3) = -2(x-1)\\Distribute\\(y-3) = -2x+2\\+3\\y = -2x +5$

Hope this Helps!

4 0

3 years ago

10. A In the adjoining Figure, ab is perpendicular to bc and ed is to cd while bc=cd . prove i) triangle abc is corresponding to

xxTIMURxx [149]

<u>Part</u><u> </u><u>(</u><u>i</u><u>)</u>

1) AB is perpendicular to BC, ED is perpendicular to CD, BC = CD (given)

2) Angles ABC and CDE are right angles (perpendicular lines form right angles)

3) Angles ABC and CDE are equal (all right angles are equal)

4) Angles ACB and DCE are equal (vertical angles are equal)

5) Triangles ABC and EDC are congruent (ASA)

<u>Part</u><u> </u><u>(</u><u>ii</u><u>)</u>

6) AB = DE (corresponding parts of congruent triangles are equal)

4 0

2 years ago

In 1980, the median age of the U.S. population was 30.0; in 2000, the median age was 35.3.

miss Akunina [59]

Answer:

y = 0.265x - 494.7

Step-by-step explanation:

Let median age be represent by 'a' and time be represent by 't'

In 1980, median age is given 30

which means that

a₁ = 30

t₁ = 1980

In 2000, the median age is given 35.3

which means that.

a₂ = 35.3

t₂ = 2000

The slope 'm' of the linear equation can be found by:

m = (a₂ - a₁) /(t₂ - t₁)

m = (35.3 - 30)/(2000-1980)

m = 0.265

General form of linear equation is given by:

y = mx + c

y = 0.265x +c

Substitute point (1980,30) in the equation.

30 = 0.265(1980) + c

c = -494.7

Hence the the linear equation can be written as:

y = mx + c

y = 0.265x - 494.7

4 0

3 years ago

Can someone help me solve this and put this on a graph. Just one problem may help.

Sauron [17]

I thought of it as answering a regular equation and imagine it's equal signs but when your done solving you with have like ex x<5 and where ever the butt of the arrow is pointing is where the lines goes if there's ano equal sign below it then it is includive

3 0

4 years ago

Whats the intrgral of <img src="https://tex.z-dn.net/?f=%20%5Cint%20%20%5Cfrac%7Bx%5E2%2Bx-3%7D%7B%28x%5E3%2Bx%5E2-4x-4%29%5E2%7

rosijanka [135]

$\displaystyle\int\frac{x^2+x-3}{(x^3+x^2-4x-4)^2}\,\mathrm dx$

Notice that $x^3+x^2-4x-4=x^2(x+1)-4(x+1)=(x-2)(x+2)(x+1)$ . Decompose the integrand into partial fractions:

$\dfrac{x^2+x-3}{(x-2)^2(x+2)^2(x+1)^2}$
$=\dfrac1{3(x+1)}-\dfrac{11}{32(x+2)}-\dfrac1{3(x+1)^2}-\dfrac1{16(x+2)^2}+\dfrac1{96(x-2)}+\dfrac1{48(x-2)^2}$

Integrating term-by-term, you get

$\displaystyle\int\frac{x^2+x-3}{(x^3+x^2-4x-4)^2}\,\mathrm dx$
$=-\dfrac1{48(x-2)}+\dfrac1{3(x+1)}+\dfrac1{16(x+2)}+\dfrac1{96}\ln|x-2|+\dfrac13\ln|x+1|-\dfrac{11}{32}\ln|x+2|+C$

5 0

4 years ago