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4 years ago

AC = 64, AB = 7x + 8, and BC = 3x + 6, Find BC.

Mathematics

1 answer:

iVinArrow [24]4 years ago

6 0

Answer:

Step-by-step explanation:

AC = 64
AB = 7x + 8
BC = 3x + 6

Since AC = AB + BC

7x + 8 + 3x + 6 = 64
10x + 14 = 64
10x = 50
x = 50/10
x = 5

------------------

BC = 3x + 6 = 3*5 + 6 = 21

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The LCD is 10 because you can multiply 2/5 by 2 to get 4/10

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3 years ago

Greatest to least 9, 1/2, -7, 0.75, -0.75,-5, 13

Vikentia [17]

Answer:

13, 9, 0.75, 1/2, -0.75, -5, -7

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Step-by-step explanation:

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3 years ago

Read 2 more answers

Click the score on the graph where you think the balancing point would be for this data.

Ulleksa [173]

Answer:

The balancing point is 23

Step-by-step explanation:

Given

The above graph

Required

Determine the balancing point

This simply means that we calculate the mean.

From the above graph, we have:

$\begin{array}{cc}x & {f} & {21} & {3} & {22} & {5} & {28} & {1} & {29} & {1}\ \end{array}$

So, the mean is:

$\bar x = \frac{\sum fx}{\sum f}$

$\bar x = \frac{21*3+22*5+28*1+29*1}{3+5+1+1}$

$\bar x = \frac{230}{10}$

$\bar x = 23$

Hence, the balancing point is 23

3 0

3 years ago

Element X decays radioactively with a half life of 11 minutes. If there are 870 grams of Element X, how long, to the nearest ten

serg [7]

Answer:

It would take 27.5 minutes the element to decay to 154 grams.

Step-by-step explanation:

The decay equation:

$\frac {dN}{dt}\propto -N$

$\Rightarrow \R\frac {dN}{dt}=-\lambda N$

$\Rightarrow \frac {dN}N=-\lambda dt$

Integrating both sides

$\Rightarrow \int \frac {dN}N=\int-\lambda dt$

$\Rightarrow ln|N|=-\lambda t+c$

When t=0, N= $N_0$ = initial amount

$ln|N_0|=-\lambda .0+c$

$\Rightarrow c=ln|N_0|$

$ln|N|=-\lambda t+ln|N_0|$

$\Rightarrow ln|N|-ln|N_0|=-\lambda t$

$\Rightarrow ln|\frac{N}{N_0}|=-\lambda t$

Decay equation:

$ln|\frac{N}{N_0}|=-\lambda t$

Given that, the half life of of element X is 11 minutes.

For half life, $N=\frac12 N_0$ , t= 11 min.

$ln|\frac{N}{N_0}|=-\lambda t$

$\Rightarrow ln|\frac{\frac12N_0}{N_0}|=-\lambda . 11$

$\Rightarrow ln|\frac12}|=-\lambda . 11$

$\Rightarrow -\lambda . 11=ln|\frac12}|$

$\Rightarrow \lambda =\frac{ln|\frac12|}{-11}$

$\Rightarrow \lambda =\frac{ln|2|}{11}$ [ $ln|\frac12|=ln|1|-ln|2|=-ln|2|$ , since ln|1|=0]

N=154 grams, $N_0$ = 870 grams, t=?

$ln|\frac{N}{N_0}|=-\lambda t$

$\Rightarrow ln|\frac{154}{870}|=-\frac{ln|2|}{11}.t$

$\Rightarrow t= \frac{ln|\frac{154}{870}|\times 11}{-ln|2|}$

=27.5 minutes

It would take 27.5 minutes the element to decay to 154 grams.

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3 years ago

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Credit 1 and Credit 3! Just took the test :)

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4 years ago