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fgiga [73]
3 years ago
10

the circumference, or perimeter, of a circle can be found by multiplying its diameter by 3.14. what is the circumference of a ci

rcle that has a diameter of 1/2 inch? Express your answer as a decimal
Mathematics
1 answer:
Dahasolnce [82]3 years ago
8 0

Answer:

4.71238898

Step-by-step explanation:

The formula used to calculate circle circumference is:

C = π · ø

Enter the diameter of a circle. The diameter of a circle is the length of a straight line drawn between two points on a circle where the line also passes through the centre of a circle, or any two points on the circle as long as they are exactly 180 degrees apart

This is the total length of the edge around the circle with the specified diameter, if it was straightened out.

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maria walks 22 feet, then goes another 40 feet on a path that makes a 70 degree angle with the path she was on. how far is she,
hammer [34]

Answer:

132 feet

Step-by-step explanation:

22+60+70

7 0
3 years ago
If the sum of a number and seven is doubled, the result is one less than the number. find the number
lawyer [7]
(n + 7)2 = n - 1
2n +14 = n - 1
2n - n +14 = n - n - 1
n + 14 -14 = -1 - 14
n = -15
(-15 + 7)2 = -15 - 1
(-8)2 = -16
-16 = -16
7 0
3 years ago
Need help please don't know this
vesna_86 [32]
Perimeter = 2w + 2l
2(8x+2) + 2(6x + 6)
Use distributive property
16x + 4 + 12x + 12
Combine like terms
28x + 16

Solution: 28x + 16
3 0
3 years ago
Read 2 more answers
Given that 1 x2 dx 0 = 1 3 , use this fact and the properties of integrals to evaluate 1 (4 − 6x2) dx. 0
Debora [2.8K]

So, the definite integral  \int\limits^1_0 {(4 - 6x^{2} )} \, dx= - 74

Given that

\int\limits^1_0 {x^{2} } \, dx = 13

We find

\int\limits^1_0 {(4 - 6x^{2} )} \, dx

<h3>Definite integrals </h3>

Definite integrals are integral values that are obtained by integrating a function between two values.

So, Integral \int\limits^1_0 {(4 - 6x^{2} )} \, dx

So, \int\limits^1_0 {(4 - 6x^{2} )} \, dx = \int\limits^1_0 {4} \, dx - \int\limits^1_0 {6x^{2} } \, dx \\=  4[x]^{1}_{0}    - \int\limits^1_0 {6x^{2} } \, dx \\=  4[x]^{1}_{0}    - 6\int\limits^1_0 {x^{2} } \, dx \\= 4[1 - 0]    - 6\int\limits^1_0 {x^{2} } \, dx\\= 4[1]    - 6\int\limits^1_0 {x^{2} } \, dx\\= 4    - 6\int\limits^1_0 {x^{2} } \, dx

Since

\int\limits^1_0 {x^{2} } \, dx = 13,

Substituting this into the equation the equation, we have

\int\limits^1_0 {(4 - 6x^{2} )} \, dx = 4 - 6\int\limits^1_0 {x^{2} } \, dx\\= 4 - 6 X 13 \\= 4 - 78\\= -74

So, \int\limits^1_0 {(4 - 6x^{2} )} \, dx= - 74

Learn more about definite integrals here:

brainly.com/question/17074932

4 0
2 years ago
4 Solve the system to find the y-value of the
ser-zykov [4K]

Answer: B

Step-by-step explanation:

7(-2) - 4(1) = -18

-14 - 4 = -18

-18 = -18

TRUE

(-2) + 12(1) = 10

-2 + 12 = 10

10 = 10

TRUE

3 0
3 years ago
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