Ask question

Ask question

All categories

2 years ago

7

In a recent survey ,5 out of every 8 middle school students use a computer to complete school projects .franklin middle school h

as 520 students .based on the results of the survey ,how many students at franklin middle school are expected to use a computer school projects

1 answer:

NeX [460]2 years ago

3 0

The answer i think its 104

You might be interested in

A fence is guarding off a vegetable garden in the form of a rectangle. It has one side that is 10m greater than the other side.

DanielleElmas [232]

Answer:

The length of the fence needed to surround this garden is 188 meters.

Step-by-step explanation:

Given : A fence is guarding off a vegetable garden in the form of a rectangle. It has one side that is 10 m greater than the other side.

To find : The length of the fence needed to surround this garden if the area of the vegetable garden is 2184 m² ?

Solution :

Let the one side of rectangle be 'x'.

Then the other side is 'x+10'.

The area of the rectangle is 2184 m²,

i.e. $x(x+10)=2184$

$x^2+10x-2184=0$

Solve by middle term split,

$x^2+52x-42x-2184=0$

$x(x+52)-42(x+52)=0$

$(x+52)(x-42)=0$

$x=-52,42$

Reject negative value,

The side of the rectangle is 42 m.

The other side is 42+10=52 m

The perimeter of the rectangle is $P=2(l+b)$

$P=2(42+52)$

$P=2(94)$

$P=188$

Therefore, the length of the fence needed to surround this garden is 188 meter.

5 0

1 year ago

Alexeev081 [22]

<em>so</em><em> </em><em>the</em><em> </em><em>answer</em><em> </em><em>is</em><em> </em><em>3</em><em>6</em><em>.</em>

<em>Hope</em><em> </em><em>this</em><em> </em><em>will</em><em> </em><em>help</em><em> </em><em>u</em><em>.</em><em>.</em><em>.</em>

7 0

2 years ago

Read 2 more answers

What's the area of the above figure which has sides of length 6 and 8

san4es73 [151]

Area means times so wouldn’t it just be 6 times 8 which is 48?!

3 0

1 year ago

Read 2 more answers

Sharon and Jacob started at the same place. Jacob walked 3 m north and then 4 m west. Sharon walked 5 m south and 12 m east. How

Ilya [14]

Consider the coordinate plane:

1. The origin is the point where Sharon and Jacob started - (0,0).

2. North - positive y-direction, south - negetive y-direction.

3. East - positive x-direction, west - negative x-direction.

Then,

if Jacob walked 3 m north and then 4 m west, the point where he is now has coordinates (-4,3);
if Sharon walked 5 m south and 12 m east, the point where she is now has coordinates (12,-5).

The distance between two points with coordinates $(x_1,y_1)$ and $(x_2,y_2)$ can be calculated using formula

$D=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}.$

Therefore, the distance between Jacob and Sharon is

$D=\sqrt{(12-(-4))^2+(-5-3)^2}=\sqrt{16^2+8^2}=\sqrt{256+64}=\sqrt{320}=8\sqrt{5}\approx 11.18\ m.$

7 0

2 years ago

In △ABC, m∠A=15 °, a=10 , and b=11 . Find c to the nearest tenth.

pshichka [43]

Answer:

The answer is:

$\bold{c\approx 20.2\ units}$

Step-by-step explanation:

Given:

In △ABC:

m∠A=15°

a=10 and

b=11

To find:

c = ?

Solution:

We can use cosine rule here to find the value of third side c.

Formula for cosine rule:

$cos A = \dfrac{b^{2}+c^{2}-a^{2}}{2bc}$

Where

a is the side opposite to $\angle A$

b is the side opposite to $\angle B$

c is the side opposite to $\angle C$

Putting all the values.

$cos 15^\circ = \dfrac{11^{2}+c^{2}-10^{2}}{2\times 11 \times c}\\\Rightarrow 0.96 = \dfrac{121+c^{2}-100}{22c}\\\Rightarrow 0.96 \times 22c= 121+c^{2}-100\\\Rightarrow 21.25 c= 21+c^{2}\\\Rightarrow c^{2}-21.25c+21=0\\\\\text{solving the quadratic equation:}\\\\c = \dfrac{21.25+\sqrt{21.25^2-4 \times 1 \times 21}}{2}\\c = \dfrac{21.25+\sqrt{367.56}}{2}\\c = \dfrac{21.25+19.17}{2}\\c \approx 20.2\ units$

The answer is:

$\bold{c\approx 20.2\ units}$

4 0

2 years ago