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Lelu [443]
3 years ago
8

Find the circumference of the circle. Round your answer to two decimal places, if necessary.

Mathematics
2 answers:
babymother [125]3 years ago
4 0

Answer:

Trust me! I got this on a test!

C≈12.57in

SSSSS [86.1K]3 years ago
3 0
I don’t fell like answering this
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Find the value of x. Round to the<br> nearest tenth.<br> 35<br> 12<br> х<br> X<br> ?
Anuta_ua [19.1K]

9514 1404 393

Answer:

  6.9

Step-by-step explanation:

The relevant relation is ...

  Sin = Opposite/Hypotenuse

Substituting given values, we have ...

  sin(35°) = x/12

  x = 12·sin(35°) ≈ 6.883

The value of x is about 6.9.

6 0
3 years ago
What is the measure of ∠XBC?
Angelina_Jolie [31]

Answer:

m∠XBC = 135

Step-by-step explanation:

What is the measure of ∠XBC?

m∠XBC = m∠BAC + m∠BCA

3p – 6 = p + 4 + 84

3p – 6 = p + 88

2p – 6 = 88

2p = 94

From the simplified expression, we need to find the value of p first

From the last line we can see that;

2p =94

Divide both sides by 2:

2p/2 =94/2

p = 47

From the expression, m∠XBC =3p-6

m∠XBC = 3(47)-6

m∠XBC = 141 -6

m∠XBC = 135

Hence the measure of m∠XBC is 135

8 0
3 years ago
When Marcus weighed the watermelon in his garden a week ago, it weighed 2.8 pounds. Yesterday, the watermelon weighed 4.67 pound
CaHeK987 [17]
The answer is 1.87 pounds heavier. This is found out by subtracting 2.8 from 4.67.
Hope this helps!
5 0
3 years ago
Read 2 more answers
PLEASE HELP ME I WILL GIVE YOU EXTRA POINTS.
katrin [286]

Answer:irdk im aorry i wish i could help because im in the same boat.

Step-by-step explanation:

3 0
3 years ago
Read 2 more answers
Jorge is asked to build a box in the shape of a rectangular prism. The maximum girth of the box is 20 cm. What is the width of t
MariettaO [177]

Answer:

The width of the box is 6.7 cm

The maximum volume is 148.1 cm³

Step-by-step explanation:

The given parameters of the box Jorge is asked to build are;

The maximum girth of the box = 20 cm

The nature of the sides of the box = 2 square sides and 4 rectangular sides

The side length of square side of the box = w

The length of the rectangular side of the box = l

Therefore, we have;

The girth = 2·w + 2·l = 20 cm

∴ w + l = 20/2 = 10

w + l = 10

l = 10 - w

The volume of the box, V = Area of square side × Length of rectangular side

∴ V = w × w × l = w × w × (10 - w)

V = 10·w² - w³

At the maximum volume, we have;

dV/dw = d(10·w² - w³)/dw = 0

∴ d(10·w² - w³)/dw = 2×10·w - 3·w² = 0

2×10·w - 3·w² = 20·w - 3·w² = 0

20·w - 3·w² = 0 at the maximum volume

w·(20 - 3·w) = 0

∴ w = 0 or w = 20/3 = 6.\overline 6

Given that 6.\overline 6 > 0, we have;

At the maximum volume, the width of the block, w = 6.\overline 6 cm ≈ 6.7 cm

The maximum volume, V_{max}, is therefore given when w = 6.\overline 6 cm = 20/3 cm  as follows;

V = 10·w² - w³

V_{max} = 10·(20/3)² - (20/3)³ = 4000/27 = 148.\overline {148}

The maximum volume, V_{max} = 148.\overline {148} cm³ ≈ 148.1 cm³

Using a graphing calculator, also, we have by finding the extremum of the function V = 10·w² - w³, the coordinate of the maximum point is (20/3, 4000/27)

The width of the box is;

6.7 cm

The maximum volume is;

148.1 cm³

5 0
3 years ago
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