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3 years ago

Can someone help me QUICK lol sorry!! B is the midpoint of AC. If AB=8x+2, and BC = 4x+18. Find AC

Mathematics

2 answers:

dalvyx [7]3 years ago

4 0

Answer:

AC = 12x + 20

Step-by-step explanation:

Combine the terms:

AB + BC = AC

AB = 8x + 2

BC = 4x + 18

Plug in the corresponding terms to the corresponding variables:

(8x + 2) + (4x + 18) = AC

AC = 8x + 4x + 18 + 2

AC = 12x + 20

12x + 20 is your answer for AC.

masha68 [24]3 years ago

4 0

Answer:

Step-by-step explanation:

8x+2=4x+18

8x-4x=18-2

4x=16

x=4

Now we substitute it back in:

8(4)+2+4(4)+18-

32+2+16+18=

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Answer:

The piecewise functions accurately represents charges based on Ben's cell phone plan is :

$f(x)= 39, x \leq 200\\f(x)=39 + 0.35(x - 200), x > 200}$

Step-by-step explanation:

Let x be the minutes over cell phone

Let f(x) represents the piecewise function that represents the charges based on Ben's cell phone plan.

We are given that Ben has a cell phone plan that provides 200 free minutes each month for a flat rate of $39.

So, $f(x)= 39, x \leq 200$

We are also given that For any minutes over 200, Ben is charged $0.35 per minute.

So, Minutes over 200 = x-200

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3 years ago

The average score of all golfers for a particular course has a mean of 61 and a standard deviation of 3.5 . Suppose 49 golfers p

Nadya [2.5K]

Answer:

The probability that the average score of the 49 golfers exceeded 62 is 0.3897

Step-by-step explanation:

The average score of all golfers for a particular course has a mean of 61 and a standard deviation of 3.5

$\mu = 61$

$\sigma = 3.5$

We are supposed to find he probability that the average score of the 49 golfers exceeded 62.

Formula : $Z=\frac{x-\mu}{\sigma}$

$Z=\frac{62-61}{3.5}$

$Z=0.285$

Refer the z table for p value

p value = 0.6103

P(x>62)=1-P(x<62)=1-0.6103=0.3897

Hence the probability that the average score of the 49 golfers exceeded 62 is 0.3897

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3 years ago

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3 years ago

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3 years ago

For what value of k is there one solution to the given quadratic equation (k+1)x²+4kx+2=0.

andriy [413]

Answer:

If $k = 1$ or $k = -\frac{1}{2}$ , there is only one solution to the given quadratic equation.

Step-by-step explanation:

Given a second order polynomial expressed by the following equation:

$ax^{2} + bx + c, a\neq0$

This polynomial has roots $x_{1}, x_{2}$ such that $ax^{2} + bx + c = (x - x_{1})*(x - x_{2})$ , given by the following formulas:

$x_{1} = \frac{-b + \sqrt{\bigtriangleup}}{2*a}$

$x_{2} = \frac{-b - \sqrt{\bigtriangleup}}{2*a}$

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The signal of $\bigtriangleup$ determines how many real roots an equation has:

$\bigtriangleup > 0$ : Two real and different solutions

$\bigtriangleup = 0$ : One real solution

$\bigtriangleup < 0$ : No real solutions

In this problem, we have the following second order polynomial:

$(k+1)x^{2} + 4kx + 2 = 0$ .

This means that $a = k+1; b = 4k; c = 2$

It has one solution if

$\bigtriangleup = 0$

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$16k^{2} -8(k+1) = 0$

$16k^{2} - 8k - 8 = 0$

We can simplify by 8

$2k^{2} - k - 1 = 0$

The solution is:

$k = 1$ or $k = -\frac{1}{2}$

So, if $k = 1$ or $k = -\frac{1}{2}$ , there is only one solution to the given quadratic equation.

4 0

3 years ago