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netineya [11]
2 years ago
14

What's equivalent to 6 to 4

Mathematics
2 answers:
xxMikexx [17]2 years ago
8 0

Answer:

Step-by-step explanation:

32 is equivalent to 64 because 3 x 4 = 2 x 6 = 12. 96 is equivalent to 64 because 9 x 4 = 6 x 6 = 36. 128 is equivalent to 64 because 12 x 4 = 8 x 6 = 48.

STatiana [176]2 years ago
6 0

Answer:

Step-by-step explanation:

6/4 =

Let's use 2 to multiply the numerator and denominator

6×2=12

4×2=8

That is 12/8

So to try it 12÷2=6

8÷2=4

Therefore, 12/8 is equivalent to 6/4

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Given a function where one x-intercept of a parabola is (-4,0), what will be the new x-intercept if h is increased by 6?
Ilia_Sergeevich [38]

the answer to your question is 2


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3 years ago
What’s a absolute deviation
Sergeeva-Olga [200]

Answer:

Absolute deviation is when you find the mean of a data set, then the difference between each number and the mean, then the average f the differences.

Step-by-step explanation:

Maddox: |10-9| = 1

Enrique: |10-13| = 3

Gloria: |10-9| = 1

McKenna: |10-10| = 0

Asher: |10-9| = 1

Hannah: |10-9| =

Danielle: |10-10| = 0

Katy:  |10-10| = 0

Timothy: |10-11| = 1

Gentry: |10-9| = 1

Now you have the numbers:  1, 3, 1, 0, 1, 1, 0, 0, 1, 1,

Now find the mean of those numbers:

\frac{1 + 3+1+0+1+1+1+0+0+1+1}{10} =\frac{10}{10} =1

The absolute deviation is 1.

6 0
3 years ago
Which equation can be used to find the volume of this solid? A rectangular prism with a length of 5 centimeters, width of 4 cent
beks73 [17]

Answer:

V= 2 times 4 times 5

4 0
3 years ago
The integral of (5x+8)/(x^2+3x+2) from 0 to 1
Gnom [1K]
Compute the definite integral:
 integral_0^1 (5 x + 8)/(x^2 + 3 x + 2) dx

Rewrite the integrand (5 x + 8)/(x^2 + 3 x + 2) as (5 (2 x + 3))/(2 (x^2 + 3 x + 2)) + 1/(2 (x^2 + 3 x + 2)):
 = integral_0^1 ((5 (2 x + 3))/(2 (x^2 + 3 x + 2)) + 1/(2 (x^2 + 3 x + 2))) dx

Integrate the sum term by term and factor out constants:
 = 5/2 integral_0^1 (2 x + 3)/(x^2 + 3 x + 2) dx + 1/2 integral_0^1 1/(x^2 + 3 x + 2) dx

For the integrand (2 x + 3)/(x^2 + 3 x + 2), substitute u = x^2 + 3 x + 2 and du = (2 x + 3) dx.
This gives a new lower bound u = 2 + 3 0 + 0^2 = 2 and upper bound u = 2 + 3 1 + 1^2 = 6: = 5/2 integral_2^6 1/u du + 1/2 integral_0^1 1/(x^2 + 3 x + 2) dx

Apply the fundamental theorem of calculus.
The antiderivative of 1/u is log(u): = (5 log(u))/2 right bracketing bar _2^6 + 1/2 integral_0^1 1/(x^2 + 3 x + 2) dx

Evaluate the antiderivative at the limits and subtract.
 (5 log(u))/2 right bracketing bar _2^6 = (5 log(6))/2 - (5 log(2))/2 = (5 log(3))/2: = (5 log(3))/2 + 1/2 integral_0^1 1/(x^2 + 3 x + 2) dx

For the integrand 1/(x^2 + 3 x + 2), complete the square:
 = (5 log(3))/2 + 1/2 integral_0^1 1/((x + 3/2)^2 - 1/4) dx

For the integrand 1/((x + 3/2)^2 - 1/4), substitute s = x + 3/2 and ds = dx.
This gives a new lower bound s = 3/2 + 0 = 3/2 and upper bound s = 3/2 + 1 = 5/2: = (5 log(3))/2 + 1/2 integral_(3/2)^(5/2) 1/(s^2 - 1/4) ds

Factor -1/4 from the denominator:
 = (5 log(3))/2 + 1/2 integral_(3/2)^(5/2) 4/(4 s^2 - 1) ds

Factor out constants:
 = (5 log(3))/2 + 2 integral_(3/2)^(5/2) 1/(4 s^2 - 1) ds

Factor -1 from the denominator:
 = (5 log(3))/2 - 2 integral_(3/2)^(5/2) 1/(1 - 4 s^2) ds

For the integrand 1/(1 - 4 s^2), substitute p = 2 s and dp = 2 ds.
This gives a new lower bound p = (2 3)/2 = 3 and upper bound p = (2 5)/2 = 5:
 = (5 log(3))/2 - integral_3^5 1/(1 - p^2) dp

Apply the fundamental theorem of calculus.
The antiderivative of 1/(1 - p^2) is tanh^(-1)(p):
 = (5 log(3))/2 + (-tanh^(-1)(p)) right bracketing bar _3^5


Evaluate the antiderivative at the limits and subtract. (-tanh^(-1)(p)) right bracketing bar _3^5 = (-tanh^(-1)(5)) - (-tanh^(-1)(3)) = tanh^(-1)(3) - tanh^(-1)(5):
 = (5 log(3))/2 + tanh^(-1)(3) - tanh^(-1)(5)

Which is equal to:

Answer:  = log(18)
5 0
3 years ago
Find the area of the triangle
makkiz [27]

Answer:

84mm

Step-by-step explanation:

Formula of a triangle:

A =bh/2

A=14(12)/2

A=168/2

A=84mm

5 0
3 years ago
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