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3 years ago

Doesanyoneknowtheanswertothis???

Mathematics

1 answer:

pochemuha3 years ago

5 0

Answer:

i think it is 10.63$

Step-by-step explanation:

--lvr

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Determine the angle at the centre of a circle with radius 12.0 cm for an arc length of of 21.0 cm.

nata0808 [166]

Answer:

$\theta=\dfrac{7}{4}\ \text{radian}$

Step-by-step explanation:

We have,

Radius of the circle is 12 cm

Arc length, l = 21 cm

It is required to find the central angle of the circle. The formula for the length of an arc of a circle is :

$l=r\times \theta$

$\theta$ is central angle

$\theta=\dfrac{l}{r}\\\\\theta=\dfrac{21}{12}\\\\\theta=\dfrac{7}{4}\ \text{radian}$

So, the angle at the centre of the circle is $\dfrac{7}{4}\ \text{radian}$ .

7 0

2 years ago

3/7 g=9 how do you do it?

Lera25 [3.4K]

The answer is g = 21

just multiply the fraction.

3 × 7 is 21.

6 0

3 years ago

Write this number in expanded form 283,980

Lynna [10]

The answer is 200,000 + 80,000 + 3,000 + 900 +80.

HOPE IT HELPS ♥️

5 0

3 years ago

Read 2 more answers

a total of 440 tickets were sold for the school play. They were either adult tickets or student tickets. The number of student t

diamong [38]

Important: represent your unknowns. You might represent "number of adult tickets sold" by "a" and "number of student tickets sold" by "s."

How many tickets were sold, altogether? Express this sum algebraically, using "a" and "s."

How does the number of student tickets sold compare with the number of adult tickets sold? Express this algebraically in therms of "a" and "s."

5 0

2 years ago

Mike said last week I ran 15 miles further than Bell. Pete said last week I ran 1 mile less than three times as far as Bill. If

disa [49]

Answer:

The distance cover by Bell is 8 miles .

Step-by-step explanation:

Given as :

The distance cover by Mike = The distance cover by Pete = D miles

Let The distance cover by Bell = d miles

The distance cover by Mike = 15 miles + distance cover by Bell

i.e D = 15 + d ............1

Again

The distance cover by Pete = three time distance cover by Bell - 1 miles

i.e D = 3 × B - 1

Or, D = 3 d - 1 ............2

Now, Solving equation 1 and 2

So, 15 + d = 3 d - 1

Or, 3 d - d = 15 + 1

Or, 2 d = 16

∴ d = $\dfrac{16}{2}$

i.e d = 8 miles

So,The distance cover by Bell = d = 8 miles

Hence, The distance cover by Bell is 8 miles . Answer

6 0

2 years ago