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kap26 [50]
3 years ago
14

If f(x) = 6x – 4, what is f(x) when x = 8?

Mathematics
1 answer:
RSB [31]3 years ago
3 0
The answer should be 6(8)-4
F(x) is 44
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PLEASE HELP ME!!!!!!!!
hjlf
The Answer Is 60
2.5 x 8 The Number Of Sides On That Shape Was 8 
2.5 x 8 = 20
20 times the radius 3
20 x 3 = 60
60cm^2

~Hope This Helps :)
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Also Rate, Give Thanks, And Mark As Brainiliest. 
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4 0
3 years ago
Read 2 more answers
How many groups of 6 are in 95?
NikAS [45]

Answer:

15

Step-by-step explanation:

It does not divide evenly, but there are 15 groups of 6 in 95.

8 0
3 years ago
Which of the following groups of measurements will form a triangle?<br>Help please.​
erastovalidia [21]

Answer:

It's C

Step-by-step explanation:

A triangles mesurement has to add up to 180. C is the only one that does.

3 0
2 years ago
Read 2 more answers
Find two unit vectors in 2-space that make an angle of 45◦ with 4i+3j.
inn [45]
First, lets transform the given vector into an unit vector (dividing by its module)


UnitVec = 4/5 i + 3/5 j

Then lets change this vector into a polar form

UnitVec = 1. with angle of 36.869 degrees taking as a reference the i vector

Then, the probem tells us that the vectors u and v make an angle of 45 degrees with UnitVec, so lets add+-45 to the vector in polar form

U = 1*[cos(
36.869 +45)i + sin(36.869 +45)j] = 0.1414 i + 0.9899 j


V = 1*[cos(
36.869 -45)i + sin(36.869 -45)j] = 0.9899 i - 0.1414 j




4 0
3 years ago
Sketch the region enclosed by y=3x and y=x^2 . Then find the area of the region.
kolezko [41]

Answer:

A=4.5u^2

Step-by-step explanation:

The integral of a function gives you the area under the curve, the subtraction of one of the areas from the other will give you the area in between.

The limits of integration are the points where the curves intersect each other(take the curves has a system of equation and solve for x and y):

y=3x, y=x^2\\3x=x^2\\x^2-3x=0\\x(x-3)=0\\x_1=0\\x-3=0\\x_2=3

y_1=3x_1=3(0)=0\\y_2=3x_2=3(3)=9

The integral will be the subtraction of the curve y=3x and y=x^2  (In the graph you can see y=3x is the upper curve):

\int\limits^3_0 {3x-x^2} \, dx =\left\frac{3x^2}{2}-\frac{x^3}{3}\right|^3_0=\frac{3(3)^2}{2}-\frac{(3)^3}{3}-\left(\frac{3(0)^2}{2}-\frac{(0)^3}{3}\right)\\\int\limits^3_0 {3x-x^2} \, dx =\frac{27}{2}-\frac{27}{3} =13.5-9=4.5u^2

6 0
2 years ago
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