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1 year ago

Point A bisects LS BC. If LS BA=6x-3 and LS AC= 3x+7. Solve for X. PLS ANSWER QUICKLY!!!

Mathematics

1 answer:

denis23 [38]1 year ago

8 0

Based on the bisection of line BC by point A, the value of x is 10/3

<h3>How to determine the value of x?</h3>

The given parameters are:

Point A bisects BC
BA = 6x - 3
AC= 3x + 7

Because the point A bisects BC, then

BA = AC

Substitute the known values in the above equation

So, we have

6x - 3 = 3x + 7

Collect the like terms

6x - 3x = 3 + 7

Evaluate the like terms

3x = 10

Divide both sides of the equation by 3

x = 10/3

Hence, based on the bisection of line BC by point A, the value of x is 10/3

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Answer:

$\boxed {m = -\frac{5}{7}}$

Step-by-step explanation:

Use the <u>Slope Formula</u> to determine the slope of two given points:

$m = \frac{y_{2} - y_{1}}{x_{2} - x_{1}}$

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Second Point: $(x_{2}, y_{2})$

-Substitute both points:

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2 years ago

Simplify: 18q2−45q+25/9q2−25.

Svet_ta [14]

Answer:

$\frac{(6q-5)}{(3q+5)}$

Step-by-step explanation:

Given: $\frac{18q^{2}-45q+25 }{9q^{2}-25 }$

Factorizing the numerator and the denominator, we have:

18 $q^{2}$ - 45q + 25 = 18 $q^{2}$ - 30q - 15q + 25

= (3q - 5)(6q - 5)

and

9 $q^{2}$ - 25 = (3q - 5)(3q + 5)

So that,

$\frac{18q^{2}-45q+25 }{9q^{2}-25 }$ = $\frac{(3q-5)(6q-5)}{(3q-5)(3q+5)}$

= $\frac{(6q-5)}{(3q+5)}$

Therefore,

$\frac{18q^{2}-45q+25 }{9q^{2}-25 }$ = $\frac{(6q-5)}{(3q+5)}$

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