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3 years ago

How do you solve the system of equations 4 x − 2 y = 1 and -4 + 4 y = -2

Mathematics

1 answer:

Dmitrij [34]3 years ago

4 0

Answer: y= -2 - 1/2

Step-by-step explanation:

-4x -2y = 1

+4x +4x

-2y = 4x + 1

/-2 /-2 /-2

y= -2 - 1/2

I think, i may be incorrect but the verbal explanation is; -4x is needed to be moved so you add 4x to both sides and 4x cancels out on the left which leaves -2y on the left (when it was moved "x" is always first). then divide all numbers by -2 to get y by itself. this gives you y= -2 - 1/2 if my calculations are correct.

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Need help on this question

Sedaia [141]

Answer:

Step-by-step explanation:

Hey There!

So these triangles are congruent so segment CD is equal to segment DB

so to find x we use the equation

7x+10=10x+7

now we solve for x

step 1 subtract each side by 7

10-7=3

7-7 cancels out

now we have 7x+3=10x

step 2 subtract each side by 7 x

10x-7x=3x

7x-7x= cancels out

now we have

3x=3

step 3 divide each side by 3

3/3=1

3/3 cancels out

now we have

x=1

3 0

2 years ago

Read 2 more answers

Hiw ti factorise x*-16

ra1l [238]

Im assuming the problem is x^2-16, which would be factored to (x-4)(x+4)

7 0

2 years ago

Evaluate the following double integral: xy dA D where the region D is the triangular region whose vertices are (0, 0), (0, 3), (

natulia [17]

Answer:

I= 84

Step-by-step explanation:

for

$I=\int\limits^{}_{} \int\limits^{}_D {x*y} \, dA = \int\limits^{}_{} \int\limits^{}_D {x*y} \, dx*dy$

since D is the rectangle such that 0<x<3 , 0<y<3

$I=\int\limits^{}_{} \int\limits^{}_D {x*y} \, dA = \int\limits^{3}_{0} \int\limits^{3}_{0} {x*y} \, dx*dy = \int\limits^{3}_{0} {x} \, dx\int\limits^{3}_{0} {y} \, dy = x^{2} /2*y^{2} /2 = (3^{2} /2 - 0^{2} /2)* (3^{2} /2 - 0^{2} /2) = 3^{4} /4 = 81/4$