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3 years ago

a line passes through the point -2 and five and has a slope of 2/3 points A(x,3) and B(-2,y)lie on the line

1 answer:

7 0

A and B lie on the line, yes, but what specifically are you supposed to do? Looks like your problem statement was cut off before you'd finished typing it in.

You say your line passes thru (-2,5) and has a slope of 2/3? Then, using the point-slope formula,

y-5 = (2/3)(x+2) This is the general equation for your line.

Now let's play around with B(-2,y). Suppose we subst. the x-coordinate of B, which is -2, into the equation y-5 = (2/3)(x+2); we get y-5 = (2/3)(-2+2) = 0. This tells us that y must be 5. But we already knew that!!

So, please review the original problems with its instructions and this discussion and tell me what you need to know from this point on.

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Dmitry_Shevchenko [17]

Answer:

It will be 2.83 inches long on the drawing

Step-by-step explanation:

The scale we have is;

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This;

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3 years ago

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Step-by-step explanation:

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2 years ago

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lakkis [162]

Answer:

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3 years ago

Does the function satisfy the hypotheses of the Mean Value Theorem on the given interval? f(x) = 3x2 − 2x + 1, [0, 2] Yes, it do

expeople1 [14]

Answer:

Yes , function is continuous in [0,2] and is differentiable (0,2) since polynomial function are continuous and differentiable

Step-by-step explanation:

We are given the Function

f(x) = $3x^2 -2x +1$

The two basic hypothesis of the mean valued theorem are

The function should be continuous in [0,2]

The function should be differentiable in (1,2)

upon checking the condition stated above on the given function

f(x) is continuous in the interval [0,2] as the functions is quadratic and we can conclude that from its graph

also the f(x) is differentiable in (0,2)

f'(x) = 6x - 2

Now the function satisfies both the conditions

so applying MVT

6x-2 = f(2) - f(0) / 2-0

6x-2 = 9 - 1 /2

6x-2 = 4

6x=6

x=1

so this is the tangent line for this given function.

4 0

3 years ago