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VLD [36.1K]
3 years ago
8

Triangle cde maps to triangle lmn with the transformation (x,y) —> (x+3,y-2) —> (2/3x, 2/3y)

Mathematics
1 answer:
Mamont248 [21]3 years ago
4 0
The answer is 890 best answer
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2x-y=4 and y=-2x+8.<br><br> Can anyone help me solve
Komok [63]

Answer:

The correct answer is x = 3 and y = 2.

Step-by-step explanation:

There are many ways to solve systems of equations like this, but I'm going to use substitution.  This means taking the value of y given by the second equation and plugging it into the first equation.  This is modeled below:

2x - y = 4

2x - (-2x+8) = 4

Now, we can simplify the left side of the equation.

2x + 2x - 8 = 4

4x - 8 = 4

We should add 8 to both sides as the next step.

4x = 12

Now we can divide by 4.

x = 3

To solve for y, we can substitute this value found for x back into either one of our original equations.

y = -2x + 8

y = (-2*3) + 8

y = -6 + 8

y = 2

Therefore, the correct answer is x = 3 and y = 2.

Hope this helps!

5 0
2 years ago
Read 2 more answers
If f(x) = x^3 – 2x^2, which expression is equivalent to f(i)?
arsen [322]
I=√-1 and all the steps are there

6 0
3 years ago
Read 2 more answers
Which is the solution to the inequality?
n200080 [17]

y < –12

Solution:

Step 1: Given inequality is y + 15 < 3.

To find the solution to the inequality.

Step 2: Subtract –15 on both sides to equal the expression.

⇒ y + 15 –15 < 3 –15

Step 2: Using addition identity property, any number adding with zero gives the number itself.

⇒ y + 0 < –12

⇒ y < –12

Hence the solution to the inequality is y < –12.

4 0
3 years ago
Read 2 more answers
How do i solve this?
Maksim231197 [3]
For number one is 1606, for 2 is -508, for 3 is 0, and 4 is 12
7 0
2 years ago
What is simplest form of sqrt 2 + sqrt 5 / sqrt 2 - sqrt -5? √2 + √5 / √2 - √-5
Leni [432]

Answer:

\frac{-7-2\sqrt{10}}{3}

Step-by-step explanation:

we have

\frac{\sqrt{2}+\sqrt{5}}{\sqrt{2}-\sqrt{5}}

Simplify

Multiply the expression by  \frac{\sqrt{2}+\sqrt{5}}{\sqrt{2}+\sqrt{5}}

(\frac{\sqrt{2}+\sqrt{5}}{\sqrt{2}-\sqrt{5}})(\frac{\sqrt{2}+\sqrt{5}}{\sqrt{2}+\sqrt{5}})

Apply difference of squares in the denominator

\frac{(\sqrt{2}+\sqrt{5})^2}{(\sqrt{2})^2-(\sqrt{5})^2}

\frac{2+2\sqrt{10}+5}{2-5}

\frac{7+2\sqrt{10}}{-3}

\frac{-7-2\sqrt{10}}{3}

6 0
3 years ago
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