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mariarad [96]
3 years ago
12

I don't understand this question can you help me please

Mathematics
2 answers:
larisa86 [58]3 years ago
8 0
Errrrmmmm I hope you figure it out soon :-)
Maslowich3 years ago
5 0
I believe the answer is A because the formula for area is length times width. 
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Solve: q + 24 = 24 A. q = 8 B. q = 2 C. q = 3 D. q = 0
vekshin1

Answer:

Hi! The answer to your question is D. q = 0

Step-by-step explanation:

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Hope this helps!!

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3 0
3 years ago
A football stadium sells regular and box seating. There are twelve times as many
Vera_Pavlovna [14]

Answer:

There are 801 box seats and 9612 regular seats.

explanation:

Let the number of box seats be x,

Then the regular seats is 12x

The sum of seats equals to 10,413

<u>Solve:</u>

12x + x ➙ 10,413

13x ➙ 10413

x ➙ 801

There are 801 box seats.

<u>Find for regular seats:</u>

➙ 12(x)

➙ 12(801)

➙ 9612

There are 9612 regular seats.

6 0
2 years ago
Read 2 more answers
Order from least to greatest 52,701, 54,025, 5, 206
andre [41]

First: 5.

Second: 206.

Third: 52,701.

Fourth: 54,025.

5 0
3 years ago
Two vectors A⃗ and B⃗ are at right angles to each other. The magnitude of A⃗ is 5.00. What should be the length of B⃗ so that th
creativ13 [48]

Answer:

Length of B is 7.4833

Step-by-step explanation:

The vector sum of A and B vectors in 2D is

C=A+B=(a_1+b_1,a_2+b_2)

And its magnitude is:

C=\sqrt{(a_1+b_1)^2+(a_2+b_2)^2} =9

Where

a_1=Asinx

a_2=Acosx

b_1=Bsin(x+90)

b_2=Bcos(x+90)

Using the properties of the sum of two angles in the sin and cosine:

b_1=Bsin(x+90)=B(sinx*cos90+sin90*cosx)=Bcosx

b_2=Bcos(x+90)=B(cosx*cos90-sinx*sin90)=-Bsinx

Sustituying in the magnitud of the sum

C=\sqrt{(Asinx+Bcosx)^2+(Acosx-Bsinx)^2} =9

C=\sqrt{A^2sin^2x+2ABsinxcosx+B^2cos^2x+A^2cos^2x-2ABsinxcosx+B^2sin^2x} =9

C=\sqrt{A^2(sin^2x+cos^2x)+B^2(cos^2x+sin^2x)}

C=\sqrt{A^2+B^2} =9

Solving for B

A^2+B^2 =9^2

B^2 =9^2-A^2

Sustituying the value of the magnitud of A

B^2=81-5^2=81-25=56

B= 7. 4833

8 0
3 years ago
f the centripetal and thus frictional force between the tires and the roadbed of a car moving in a circular path were reduced, w
LUCKY_DIMON [66]
The frictional force between the tires and the road prevent the car from skidding off the road due to centripetal force.

If the frictional force is less than the centripetal force, the car will skid when it navigates a circular path.

The diagram below shows that when the car travels at tangential velocity, v, on a circular path with radius, r, the centripetal acceleration of v²/ r acts toward the center of the circle.

The resultant centripetal force is (mv²)/r, which should be balanced by the frictional force of μmg, where μ =  coefficient of kinetic friction., and mg is the normal reaction on a car with mass, m.

This principle is applied on racing tracks, where the road is inclined away from the circle to give the car an extra restoring force  to overcome the centripetal force.
8 0
2 years ago
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