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lara31 [8.8K]
3 years ago
13

A= 12b1+b2h ; solve for b1

Mathematics
1 answer:
slega [8]3 years ago
8 0

Answer:

b1 = 0, b2 = 0

EXPLICACION

A = 12 b1 + b2 h

A - 12 b1 - b2 h = 0

b2!=0, h = -(12 b1)/b2

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Vlad1618 [11]

\frac{21}{20}  Should be your answer

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F(x)=2x^2+3x : evaluate the function, find f(-2)
egoroff_w [7]

Answer:

F(0) = 2

F(-1) = 7

F(2) = 4

Explanation:

Notice where the x's are in the function F(x)

For F(0) you sub in 0 wherever there is an x, ie:

F(0) = 2(0)2−3(0)+2 = 2

The same is applied for F(-1) and F(2)

HOPE THIS HELPS!! PLZ GIVE 5 STAR,  THANKS, AND BRAINLIST THXXX

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3 years ago
Please answer this question now
Lina20 [59]

Hey there!

To find the x-value of the midpoint, you add the two x-values of the endpoints and the divide the result by two.

-12+0=-12

-12/2=-6

So, the x-value of our midpoint is -6.

To find the y-value of a midpoint, you do the same thing as with the x-values but with the y-values.

2+14=16

16/2=8

Our y-value is 8.

Therefore, our midpoint is (-6,8)

I hope that this helps!

6 0
4 years ago
If f(x) = 2 - x^1/2 and g(x) = x^2 – 9, what is the domain of g(x) divided f(x)?
Inessa [10]

~Shoto Todoroki here~

Let's consider what we are asked. The domain is defined as a set of points that satisfy the equation on the x-ordinate.

So, essentially, we need to find if and what the restriction on x is.

Let's now consider just f(x) because g(x) is completely irrelevant to the question.

f(x) = 2 - x^(1/2)

Since f(x) can never be 0 for a defined function, let's consider when f(x) = 0 to find a restriction on x.

2 - x^(1/2) = 0

2 = x^(1/2)

+-4 = x

But we can only take the positive 4 because inside a square root has to always be positive (unless you're dealing with complex numbers), so the only restriction is that x cannot be equal to 4.

<h2>Therefore, our domain is:<u><em> x >= 0; x =/= 4</em></u></h2>

hope this helps :D

8 0
2 years ago
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