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2 years ago

Multiply each equation by a constant that would help to eliminate the y terms.

Mathematics

2 answers:

Archy [21]2 years ago

6 0

Answer:

Step-by-step explanation:

i got it right on e2020

Zinaida [17]2 years ago

3 0

Answer:

Therefore the required resulting equation is

$6x-15y=-63\\-15x+15y=90$

6 x minus 15 y = negative 63.

Negative 15 x + 15 y = 90

Step-by-step explanation:

Given:

$2x-5y=-21$ ......................Equation ( 1 )

$3x-3y=-18$ ......................Equation ( 2 )

To Find:

Expression after multiplying to eliminate y term,

Solution:

So to eliminate 'y' term we need to multiply equation 1 by a constant 3 and equation 2 by a constant -5, such that equations becomes

$3(2x-5y)=3\times -21\\\\6x-15y=-63$ .....( 1 )

$-5(3x-3y)=-5\times -18\\\\-15x+15y=90$ .....( 2 )

so now by adding new equation one and two we can eliminate y term that means -15y and +15y will get cancel,

Therefore the required resulting equation is

$6x-15y=-63\\-15x+15y=90$

6 x minus 15 y = negative 63.

Negative 15 x + 15 y = 90

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Which of the following are identities? Check all that apply. A. cot2x + 1 = csc2x B. tan2x = 1 - sec2x C. sin2x = 1 + cos2x D. s

kirill115 [55]

I'll assume those are squares. We know D is an identity:

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Dividing through by $\sin ^2 x$

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3 years ago

If your brother has saved $60 to spend on basketball cards

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Answer:

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2 years ago

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the verticales of triangle XYZ are X(-3,-6),Y(21,-6),and Z(21,4). What is the perimeter of the triangle

andrezito [222]

Answer:

Step-by-step explanation:

The formula of a distance between two points:

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

X(-3, -6), Y(21, -6). Substitute:

$XY=\sqrt{21-(-3))^2+(-6-(-6))^2}=\sqrt{24^2+0^2}=\sqrt{24^2}=24$

X(-3, -6), Z(21, 4). Substitute:

$XZ=\sqrt{(21-(-3))^2+(4-(-6))^2}=\sqrt{24^2+10^2}=\sqrt{576+100}=\sqrt{676}=26$

Y(21, -6), Z(21, 4). Substitute:

$YZ=\sqrt{(21-21)^2+(4-(-6))^2}=\sqrt{0^2+10^2}=\sqrt{10^2}=10$

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3 years ago

Let p: A number is greater than 25.

maria [59]

We are given that:
p is greater than 25, this means that p>25
p ∈ ]25,∞[ ...........> interval I

q is less than 35, this means that q<35
q ∈ ]-∞,35[ ...........> interval II

The given condition <span>p ∧ q is true means that (p and q) is true. In other word, their intersection is true.
Therefore, the final result would be the intersection between the two intervals (interval I and interval II)

Bases on the above, the final answer would be:
</span>]-∞,35[ ∧ ]25,∞[ which is ]25,35[<span>

</span>

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2 years ago

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