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3 years ago

Substitute for y=4x-1;y=2x-5

2 answers:

6 0

<span>y=4x−1;y=2x−5
Step: Solve y=4x−1 for y:
Step: Substitute 4x−1 for y in y=2x−5:
y=2x−5
4x−1=2x−5
4x−1+−2x=2x−5+−2x(Add -2x to both sides)
2x−1=−5
2x−1+1=−5+1(Add 1 to both sides)
2x=−4
2x/2=−4/2(Divide both sides by 2)
x=−2
Step: Substitute−2 for x in y=4x−1:
y=4x−1
y=(4)(−2)−1
y=−9(Simplify both sides of the equation)
<u>Answer:</u>y=−9 and x=−2</span>

Darya [45]3 years ago

3 0

You can say 2x - 5 = 4x - 1 ⇒ 2x - 4x = 5 - 1 ⇒ -2 x = 4 ⇒ x= -2 and y gonna be
y = -4 -5 = -9 :D

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The volume of the cone is 12936 cubic mm.

Step-by-step explanation:

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4 years ago

Find the area of each parallelogram. What is the relationship between the areas?

castortr0y [4]

Answer:

Area of parallelogram is given by:

$A = bh$

where b is the base and h is the height of parallelogram.

In parallelogram TQRS.

Coordinate of TQRS are;

T(8, 16), Q(4, 4), R(16, 4) and S(20, 16)

Coordinate of T'Q'R'S' are;

T'(2, 4), Q'(1, 1), R'(4, 1) and S'(5, 4)

Find the length of QR and PT:

Using distance(D) formula:

$D = \sqrt{(x_1-x_2)^2+(y_1-y_2)^2}$

$QR = \sqrt{(4-16)^2+(4-4)^2} =\sqrt{(-12)^2+0}= \sqrt{144}= 12$ units

Similarly;

For PT:

From the graph:

P(8, 4) and T(8, 16), then

$PT = \sqrt{(8-8)^2+(4-16)^2} =\sqrt{(-0)^2+(-12)^2}= \sqrt{144}= 12$ units

In parallelogram TQRS

PT represents the height and QR represents the base of the parallelogram respectively.

then;

Area of parallelogram TQRS = $QR \cdot PT$

⇒Area of parallelogram TQRS = $12 \cdot 12 = 144$ unit square.

Now, in parallelogram T'Q'R'S'

Q'R' represents the base and P'T' represents the height of the parallelogram respectively.

here, P'(2, 1)

Find the length of Q'R' and P'T':

$Q'R' = \sqrt{(1-4)^2+(1-1)^2} =\sqrt{(-3)^2+0}= \sqrt{9}= 3$ units

$P'T' = \sqrt{(2-2)^2+(1-4)^2} =\sqrt{(-0)^2+(-3)^2}= \sqrt{9}= 3$ units

Then;

Area of parallelogram T'Q'R'S' = $Q'R' \cdot P'T'$

Area of parallelogram T'Q'R'S' = $9 \cdot 9= 81$ unit square.

Now, we have to find the relationship between the areas.

$\frac{\text{Area of parallelogram TQRS}}{\text{Area of parallelogram T'Q'R'S'}} = \frac{144}{81}$

then;

the relationship between the areas of TQRS and T'Q'R'S' is:

$\text{Area of parallelogram TQRS} = \frac{144}{81} \cdot {\text{Area of parallelogram T'Q'R'S'}$

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Answer:

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Step-by-step explanation:

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3 years ago

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