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3 years ago

BC = 4x + 2 AB = 8x + 8 and AC = 22 Find BC

Mathematics

1 answer:

Scilla [17]3 years ago

8 0

Answer:

BC = 6.8

Step-by-step explanation:

The distance from A to B, then from B to C, is the same distance as from A to C.

Thus,

AB + BC = AC

AB = 8x + 8

BC = 4x + 2

AC = 22

AB + BC = AC

(8x + 8) + (4x + 2) = 22

Expanding the parenthesis:

8x + 8 + 4x + 2 = 22

8x + 4x + 8 + 2 = 22

12x + 12 = 22

Subtracting 12 from both sides:

12x = 10

Dividing both sides by 12:

x = 10/12 = 1.2

x = 1.2

We're looking for BC. As we know,

BC = 4x + 2

Let's now put in x = 1.2 into this equation:

BC = 4x + 2

BC = 4 * 1.2 + 2

Since 4 * 1.2 is 4.8:

BC = 4.8 + 2

BC = 6.8

Answer: BC = 6.8

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‼️10 points‼️

Tamiku [17]

Answer:

<h3>The answer is option D.</h3>

Step-by-step explanation:

Equation of a line is y = mx + c

where

m is the slope

c is the y intercept

Slope of the line using points ( 5 , - 1) and

( - 5 , 2) is

m = 2+1/ -5-5 = - 3 / 10

Equation of the line using point ( 5 , - 1) is

y + 1 = -3/10(x - 5)

y = -3/10x + 3/2 - 1

The final answer is

Hope this helps you.

4 0

3 years ago

5x+7y=31 , 3x-5y=21??

nikklg [1K]

Answer:

yes

Step-by-step explanation:

5 0

3 years ago

Which of the following constants can be added to x2 + x to form a perfect square trinomial?

solmaris [256]

The constant should be added to form a perfect square trinomial will be 1/4. Then the correct option is D.

<h3>What is a quadratic equation?</h3>

It's a polynomial with a value of zero. There exist polynomials of variable power 2, 1, and 0 terms. A quadratic equation is an equation with one statement in which the degree of the parameter is a maximum of 2.

The expression is x² + x.

Then the constant should be added to form a perfect square trinomial.

Then the constant will be

The square of the half of the coefficient of the variable x is to be added to make a perfect square.

Then the constant will be 1/4.

Then the perfect square will be

$\rm \rightarrow x^2 + x + \dfrac{1}{4}\\\\\\\rightarrow x^2 + 2 \times \dfrac{1}{2} x + \left (\dfrac{1}{2} \right )^2\\\\\\\rightarrow \left (x + \dfrac{1}{2} \right)^2$