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2 years ago

11

What kind of solution is -6(x-2)=-7(x-3)

2 answers:

sergij07 [2.7K]2 years ago

7 0

Answer:

Step-by-step explanation:

-6(x-2)=-7(x-3)

-6x + 12 = -7x + 21

-6x + 7x = 21 - 12

x = 9

Firlakuza [10]2 years ago

6 0

Answer:

-6(x-2)=-7(x-3) = False x = 9

Step-by-step explanation:

Solve for x:

-6 (x - 2) = -7 (x - 3)

Expand out terms of the left hand side:

12 - 6 x = -7 (x - 3)

Expand out terms of the right hand side:

12 - 6 x = 21 - 7 x

Add 7 x to both sides:

7 x - 6 x + 12 = (7 x - 7 x) + 21

7 x - 7 x = 0:

7 x - 6 x + 12 = 21

7 x - 6 x = x:

x + 12 = 21

Subtract 12 from both sides:

x + (12 - 12) = 21 - 12

12 - 12 = 0:

x = 21 - 12

21 - 12 = 9:

Answer: x = 9

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The least common denominator of two fractions is 20. If you subtract the two denominations, you get 1. What are the denominators

LekaFEV [45]

Steps
find the Least Common Multiple of the denominators (which is called the Least Common Denominator).

Change each fraction (using equivalent fractions) to make their denominators the same as the least common denominator.

Then add (or subtract) the fractions, as we wish!
your denominators stay the same so its 20

6 0

2 years ago

Mariulka [41]

Answer:

$A = 74.7^\circ$

$B = 42.5^\circ$

$C = 62.8^\circ$

Step-by-step explanation:

Given

$A = (-1,2) \to (x_1,y_1)$

$B = (2,8) \to (x_2,y_2)$

$C = (4,1) \to (x_3,y_3)$

Required

The measure of each angle

First, we calculate the length of the three sides of the triangle.

This is calculated using distance formula

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

For AB

$A = (-1,2) \to (x_1,y_1)$

$B = (2,8) \to (x_2,y_2)$

$d = \sqrt{(-1 - 2)^2 + (2 - 8)^2$

$d = \sqrt{(-3)^2 + (-6)^2$

$d = \sqrt{45$

So:

$AB = \sqrt{45$

For BC

$B = (2,8) \to (x_2,y_2)$

$C = (4,1) \to (x_3,y_3)$

$BC = \sqrt{(2 - 4)^2 + (8 - 1)^2$

$BC = \sqrt{(-2)^2 + (7)^2$

$BC = \sqrt{53$

For AC

$A = (-1,2) \to (x_1,y_1)$

$C = (4,1) \to (x_3,y_3)$

$AC = \sqrt{(-1 - 4)^2 + (2 - 1)^2$

$AC = \sqrt{(-5)^2 + (1)^2$

$AC = \sqrt{26$

So, we have:

$AB = \sqrt{45$

$BC = \sqrt{53$

$AC = \sqrt{26$

By representation

$AB \to c$

$BC \to a$

$AC \to b$

So, we have:

$a = \sqrt{53$

$b = \sqrt{26$

$c = \sqrt{45$

By cosine laws, the angles are calculated using:

$a^2 = b^2 + c^2 -2bc \cos A$

$b^2 = a^2 + c^2 -2ac \cos B$

$c^2 = a^2 + b^2 -2ab\ cos C$

$a^2 = b^2 + c^2 -2bc \cos A$

$(\sqrt{53})^2 = (\sqrt{26})^2 +(\sqrt{45})^2 - 2 * (\sqrt{26}) +(\sqrt{45}) * \cos A$

$53 = 26 +45 - 2 * 34.21 * \cos A$

$53 = 26 +45 - 68.42 * \cos A$

Collect like terms

$53 - 26 -45 = - 68.42 * \cos A$

$-18 = - 68.42 * \cos A$

Solve for $\cos A$

$\cos A =\frac{-18}{-68.42}$

$\cos A =0.2631$

Take arc cos of both sides

$A =\cos^{-1}(0.2631)$

$A = 74.7^\circ$

$b^2 = a^2 + c^2 -2ac \cos B$

$(\sqrt{26})^2 = (\sqrt{53})^2 +(\sqrt{45})^2 - 2 * (\sqrt{53}) +(\sqrt{45}) * \cos B$

$26 = 53 +45 -97.67 * \cos B$

Collect like terms

$26 - 53 -45= -97.67 * \cos B$

$-72= -97.67 * \cos B$

Solve for $\cos B$

$\cos B = \frac{-72}{-97.67}$

$\cos B = 0.7372$

Take arc cos of both sides

$B = \cos^{-1}(0.7372)$

$B = 42.5^\circ$

For the third angle, we use:

$A + B + C = 180$ --- angles in a triangle

Make C the subject

$C = 180 - A -B$

$C = 180 - 74.7 -42.5$

$C = 62.8^\circ$

8 0

1 year ago

5. 18 m 11 m Not drawn to scale 29 m b. 445 m 7 m d. 203 m

Luden [163]

Answer: D

Explanation: Apply the pythagorean theorem (a^2 + b^2 = c^2) to get 11^2 + b^2 = 18^2. From that you can solve for b and you get the square root of 203.

Hope this helped! :)

8 0

2 years ago

Read 2 more answers

The perimeter of a rectangular field is 74 yards. The length of the field is 27 yards. What is the width in yard of the field

KonstantinChe [14]

Answer:

width = 10 yds

Step-by-step explanation:

perimeter = 2L + 2W

74 = 2(27) + 2W

74 = 54 + 2W

subtract 54 from both sides of the equation:

2W = 20

divide both sides by 2:

W = 10

7 0

2 years ago

Help asap!!!!!!!!!!<br><br> Properties of exponents <br><br> Simplify

maksim [4K]

Answer:

C) $4a^{6}b$

Step-by-step explanation:

$\frac{(4b^{3} 3a^{4}b^{2})}{3a^{-2} b^{4}}$ Copy the original equation

$\frac{12a^{6}b^{5}}{3b^{4} }$ Combine like terms

$\frac{12a^{6}b}{3}$ Simplify the b, then divide by 3

$4a^{6}b$

If this answer is correct, please make me Brainliest!

4 0

2 years ago