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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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2 years ago

Solve 2x + 4 > 16<br><br> A. x > 10<br> B. x < 6<br> C. x < 10<br> D. x > 6

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Hey there! :)

2x + 4 > 16

We're simply just trying to get x onto it's own side, so let's get rid of any numbers on the same side of it.

We can start out by subtracting both sides by 4.

2x + 4 - 4 > 16 - 4

Simplify.

2x > 12

Then, divide both sides by 2.

2x ÷ 2 > 12 ÷ 2

Simplify.

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Therefore, our answer is D. x > 6

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3 years ago

The sum of 4 consecutive integers is 354.<br> What is the second number in this sequence?

Ulleksa [173]

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so2 numbers below that and 2 numbers above that:

87 + 88 + 89 +90 =

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3 years ago

PLEASE HELP

dsp73

Answer:

All events are independent

Step-by-step explanation:

You are given the table

$\begin{array}{cccc}&\text{Chocolate}&\text{Vanilla}&\text{Total}\\\text{Adults}&0.21&0.39&0.60\\\text{Children}&0.14&0.26&0.40\\\text{Total}&0.35&0.65&1.00\end{array}$

Two events A and B are independent when

$Pr(A\cap B)=Pr(A)\cdot Pr(B)$

a) A="Chocolate"

B="Adults"

A and B="Chocolate and Adults"

$Pr(A)=0.35\\ \\Pr(B)=0.60\\ \\Pr(A\cap B)=0.21$

Since $0.35\cdot 0.60=0.21$ events are independent

b) A="Children"

B="Chocolate"

A and B="Children and Chocolate"

$Pr(A)=0.40\\ \\Pr(B)=0.35\\ \\Pr(A\cap B)=0.14$

Since $0.40\cdot 0.35=0.14$ events are independent

c) A="Vanilla"

B="Children"

A and B="Vanilla and Children"

$Pr(A)=0.65\\ \\Pr(B)=0.40\\ \\Pr(A\cap B)=0.26$

Since $0.65\cdot 0.40=0.26$ events are independent

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