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3 years ago

PLEASE HELP!!

Mathematics

1 answer:

goldenfox [79]3 years ago

8 0

<h3>Answers:</h3>

Problem 5) x = 4
Problem 6) y = 8
Problem 7) z = 4

=======================================================

Explanations:

Problem 5)

Assuming HEFG is a parallelogram, this means the opposite sides are the same length.

EH = GF

x-3 = 4x-15

x-4x = -15+3

-3x = -12

x = -12/(-3)

x = 4

--------------------------

Problem 6)

We'll use the same idea from problem 5. The opposite sides EF and HG are congruent

EF = HG

3y = 3x+12

3y = 3*4+12

3y = 24

y = 24/3

y = 8

--------------------------

Problem 7)

For any parallelogram, the diagonals always cut each other in half.

Therefore HK = 2z+3 is exactly half that of segment HF = 22

Put another way, HF is twice as long as HK

2*(HK) = HF

2*(2z+3) = 22

4z+6 = 22

4z = 22-6

4z = 16

z = 16/4

z = 4

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3 years ago

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2 years ago

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Given:

x and y are both differentiable functions of t.

$4x^2+3y^3=28$

$x=-1\text{ and }\dfrac{dy}{dt}=8$

To find:

The value of $\dfrac{dx}{dt}$ .

Solution:

We have,

$4x^2+3y^3=28$ ...(i)

At x=-1,

$4(-1)^2+3y^3=28$

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Divide both sides by 3.

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$4(2x\dfrac{dx}{dt})+3(3y^2\dfrac{dy}{dt})=0$

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Divide both sides by -8.

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Therefore, the value of $4x^2+3y^3=28$ is 36.

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3 years ago