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3 years ago

How do you eliminate the fractions on this to solve by elimination ????????

Mathematics

2 answers:

Alja [10]3 years ago

7 0

Answer:

Step-by-step explanation:

Amiraneli [1.4K]3 years ago

7 0

Good evening

Answer:

x = 6

y = 2

Step-by-step explanation:

Look at the photo below for the details.

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A distribution is symmetrical if the tails on both ends of the density curve that represents it are close to identical. Truth or

bezimeni [28]

Your answer should be true

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3 years ago

What is the quotient and remainder of (4x^3-3x^2+x+7)/(x-2)

viva [34]

Answer:

4x^2+11x+23 r53

Step-by-step explanation:

4x^3-3x^2+x+7 / x-2

2 | 4 -3 1 7

8 22 46

___________

4 11 23 53

This means 4x^2+11x+23 with a remainder of 53

7 0

3 years ago

Estimate 200 times 3

Sonja [21]

I "estimate" 600....?

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3 years ago

Read 2 more answers

Using Cramer's Rule, what are the values of x and y in the solutio

olya-2409 [2.1K]

Answer:

x = -3, y = 1

Step-by-step explanation:

To find the value of x and y, find the determinant of original matrix, which would be 21.

Then, substitute the value of x with the solutions to the equations and find the determinant of that matrix, which is -63.

Cramer's rule says that Dx ÷ D is the value of x. So, -63 ÷ 21 = -3.

So, the x-value is -3.

You can find the determinant of the y-value in the same way, and you'll find out that y = 1.

Hope this helped! :)

8 0

3 years ago

Maggie graphed the image of a 90 counterclockwise rotation about vertex A of . Coordinates B and C of are (2, 6) and (4, 3) and

Lemur [1.5K]

Answer:

A(2,2)

Step-by-step explanation:

Let the vertex A has coordinates $(x_A,y_A)$

Vectors AB and AB' are perpendicular, then

$\overrightarrow {AB}=(2-x_A,6-y_A)\\ \\\overrightarrow {AB'}=(-2-x_A,2-y_A)\\ \\\overrightarrow {AB}\perp\overrightarrow {AB'}\Rightarrow \overrightarrow {AB}\cdot \overrightarrow {AB'}=0\Rightarrow (2-x_A)(-2-x_A)+(6-y_A)(2-y_A)=0$

Vectors AC and AC' are perpendicular, then

$\overrightarrow {AC}=(4-x_A,3-y_A)\\ \\\overrightarrow {AC'}=(1-x_A,4-y_A)\\ \\\overrightarrow {AC}\perp\overrightarrow {AC'}\Rightarrow \overrightarrow {AC}\cdot \overrightarrow {AC'}=0\Rightarrow (4-x_A)(1-x_A)+(3-y_A)(4-y_A)=0$

Now, solve the system of two equations:

$\left\{\begin{array}{l}(2-x_A)(-2-x_A)+(6-y_A)(2-y_A)=0\\ \\(4-x_A)(1-x_A)+(3-y_A)(4-y_A)=0\end{array}\right.\\ \\\left\{\begin{array}{l}-4-2x_A+2x_A+x_A^2+12-6y_A-2y_A+y^2_A=0\\ \\4-4x_A-x_A+x_A^2+12-3y_A-4y_A+y_A^2=0\end{array}\right.\\ \\\left\{\begin{array}{l}x_A^2+y_A^2-8y_A+8=0\\ \\x_A^2+y_A^2-5x_A-7y_A+16=0\end{array}\right.$

Subtract these two equations:

$5x_A-y_A-8=0\Rightarrow y_A=5x_A-8$

Substitute it into the first equation:

$x_A^2+(5x_A-8)^2-8(5x_A-8)+8=0\\ \\x_A^2+25x_A^2-80x_A+64-40x_A+64+8=0\\ \\26x_A^2-120x_A+136=0\\ \\13x_A^2-60x_A+68=0\\ \\D=(-60)^2-4\cdot 13\cdot 68=3600-3536=64\\ \\x_{A_{1,2}}=\dfrac{60\pm8}{2\cdot 13}=\dfrac{34}{13},2$

Then

$y_{A_{1,2}}=5\cdot \dfrac{34}{13}-8 \text{ or } 5\cdot 2-8\\ \\=\dfrac{66}{13}\text{ or } 2$

Rotation by 90° counterclockwise about A(2,2) gives image points B' and C' (see attached diagram)

8 0

3 years ago