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Digiron [165]
3 years ago
9

How much greater is the perimeter of a square with an area of 144 square meters than the perimeter of a square with a side lengt

h of 9 meters?
Mathematics
1 answer:
musickatia [10]3 years ago
6 0

Answer:

12 meters

Step-by-step explanation:

Area of square = 144 Sq m

side \: of \: square =  \sqrt{144}  \\side \: of \: square    = 12 \: m \\  \\ perimeter \: of \: square \:   \\ = 4 \times 12  \\ = 48 \: m \\  \\ perimeter \: of \: square \: with \:  \\ side \: 9 \: m \\  = 4 \times 9 \\  = 36 \: m \\  \\ diference \: in \: perimeters \\  = 48 - 36 = 12 \: m

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Kareem cannot decide which of two washing machines to buy. The selling price of each is ​$650. The first is marked down by ​40%.
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Each washing machine is $390 and the first washing machine is better because its marked down all the way in stead of a discount, meaning that the tax will be lower (I think)
4 0
3 years ago
A number k is greater than 3. Which of the following is true about 승 ? 1 It is between 3 3 It is equal to 1 It is greater than 1
Alekssandra [29.7K]

If the number k is greater than 3, we have the following inequality:

k>3

In order to evaluate k/3, let's divide the inequality by 3:

\begin{gathered} k>3 \\ \frac{k}{3}>\frac{3}{3} \\ \frac{k}{3}>1 \end{gathered}

So the expression k/3 is greater than 1, therefore the correct option is D.

3 0
10 months ago
Find the measures of the angles of the triangle whose vertices are A = (-3,0) , B = (1,3) , and C = (1,-3).A.) The measure of ∠A
alekssr [168]

Answer:

\theta_{CAB}=128.316

\theta_{ABC}=25.842

\theta_{BCA}=25.842

Step-by-step explanation:

A = (-3,0) , B = (1,3) , and C = (1,-3)

We're going to use the distance formula to find the length of the sides:

r= \sqrt{(x_1-x_2)^2+(y_1-y_2)^2+(z_1-z_2)^2}

AB= \sqrt{(-3-1)^2+(0-3)^2}=5

BC= \sqrt{(1-1)^2+(3-(-3))^2}=9

CA= \sqrt{(1-(-3))^2+(-3-0)^2}=5

we can use the cosine law to find the angle:

it is to be noted that:

the angle CAB is opposite to the BC.

the angle ABC is opposite to the AC.

the angle BCA is opposite to the AB.

to find the CAB, we'll use:

BC^2 = AB^2+CA^2-(AB)(CA)\cos{\theta_{CAB}}

\dfrac{BC^2-(AB^2+CA^2)}{-2(AB)(CA)} =\cos{\theta_{CAB}}

\cos{\theta_{CAB}}=\dfrac{9^2-(5^2+5^2)}{-2(5)(5)}

\theta_{CAB}=\arccos{-\dfrac{0.62}}

\theta_{CAB}=128.316

Although we can use the same cosine law to find the other angles. but we can use sine law now too since we have one angle!

To find the angle ABC

\dfrac{\sin{\theta_{ABC}}}{AC}=\dfrac{\sin{CAB}}{BC}

\sin{\theta_{ABC}}=AC\left(\dfrac{\sin{CAB}}{BC}\right)

\sin{\theta_{ABC}}=5\left(\dfrac{\sin{128.316}}{9}\right)

\theta_{ABC}=\arcsin{0.4359}\right)

\theta_{ABC}=25.842

finally, we've seen that the triangle has two equal sides, AB = CA, this is an isosceles triangle. hence the angles ABC and BCA would also be the same.

\theta_{BCA}=25.842

this can also be checked using the fact the sum of all angles inside a triangle is 180

\theta_{ABC}+\theta_{BCA}+\theta_{CAB}=180

25.842+128.316+25.842

180

6 0
3 years ago
Read 2 more answers
Am I right or if not what is it. :)
dangina [55]

Answer:

Step-by-step explanation:

That is correct.

P=4s=4(12)=48in

A=s^2=12^2=144in^2

8 0
2 years ago
Read 2 more answers
I need help ASAP! I am offering brainiest to whoever answers the question first!!!
lara [203]
-3:1 :) have a nice day!
8 0
3 years ago
Read 2 more answers
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