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3 years ago

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7/8 of a number is 63

1 answer:

ololo11 [35]3 years ago

6 0

$\frac{7}{8} \times x = 63 \: so \\ x = 63 \times \frac{8}{7} = 9 \times 8 = 72$

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Prove that: (b²-c²/a)CosA+(c²-a²/b)CosB+(a²-b²/c)CosC = 0

IRISSAK [1]

<u>Prove that:</u>

$\:\:\sf\:\:\left(\dfrac{b^2-c^2}{a}\right)\cos A+\left(\dfrac{c^2-a^2}{b}\right)\cos B +\left(\dfrac{a^2-b^2}{c}\right)\cos C=0$

<u>Proof: </u>

We know that, by Law of Cosines,

$\sf \cos A=\dfrac{b^2+c^2-a^2}{2bc}$
$\sf \cos B=\dfrac{c^2+a^2-b^2}{2ca}$
$\sf \cos C=\dfrac{a^2+b^2-c^2}{2ab}$

<u>Taking</u><u> </u><u>LHS</u>

$\left(\dfrac{b^2-c^2}{a}\right)\cos A+\left(\dfrac{c^2-a^2}{b}\right)\cos B +\left(\dfrac{a^2-b^2}{c}\right)\cos C$

<em>Substituting</em> the value of <em>cos A, cos B and cos C,</em>

$\longmapsto\left(\dfrac{b^2-c^2}{a}\right)\left(\dfrac{b^2+c^2-a^2}{2bc}\right)+\left(\dfrac{c^2-a^2}{b}\right)\left(\dfrac{c^2+a^2-b^2}{2ca}\right)+\left(\dfrac{a^2-b^2}{c}\right)\left(\dfrac{a^2+b^2-c^2}{2ab}\right)$

$\longmapsto\left(\dfrac{(b^2-c^2)(b^2+c^2-a^2)}{2abc}\right)+\left(\dfrac{(c^2-a^2)(c^2+a^2-b^2)}{2abc}\right)+\left(\dfrac{(a^2-b^2)(a^2+b^2-c^2)}{2abc}\right)$

$\longmapsto\left(\dfrac{(b^2-c^2)(b^2+c^2)-(b^2-c^2)(a^2)}{2abc}\right)+\left(\dfrac{(c^2-a^2)(c^2+a^2)-(c^2-a^2)(b^2)}{2abc}\right)+\left(\dfrac{(a^2-b^2)(a^2+b^2)-(a^2-b^2)(c^2)}{2abc}\right)$

$\longmapsto\left(\dfrac{(b^4-c^4)-(a^2b^2-a^2c^2)}{2abc}\right)+\left(\dfrac{(c^4-a^4)-(b^2c^2-a^2b^2)}{2abc}\right)+\left(\dfrac{(a^4-b^4)-(a^2c^2-b^2c^2)}{2abc}\right)$

$\longmapsto\dfrac{b^4-c^4-a^2b^2+a^2c^2}{2abc}+\dfrac{c^4-a^4-b^2c^2+a^2b^2}{2abc}+\dfrac{a^4-b^4-a^2c^2+b^2c^2}{2abc}$

<em>On combining the fractions,</em>

$\longmapsto\dfrac{(b^4-c^4-a^2b^2+a^2c^2)+(c^4-a^4-b^2c^2+a^2b^2)+(a^4-b^4-a^2c^2+b^2c^2)}{2abc}$

$\longmapsto\dfrac{b^4-c^4-a^2b^2+a^2c^2+c^4-a^4-b^2c^2+a^2b^2+a^4-b^4-a^2c^2+b^2c^2}{2abc}$

<em>Regrouping the terms,</em>

$\longmapsto\dfrac{(a^4-a^4)+(b^4-b^4)+(c^4-c^4)+(a^2b^2-a^2b^2)+(b^2c^2-b^2c^2)+(a^2c^2-a^2c^2)}{2abc}$

$\longmapsto\dfrac{(0)+(0)+(0)+(0)+(0)+(0)}{2abc}$

$\longmapsto\dfrac{0}{2abc}$

$\longmapsto\bf 0=RHS$

LHS = RHS proved.

7 0

3 years ago

Determine the value of f(2) that will lead to an average rate of change of 15 over the interval [2,6]. (HELP PLEASE)

Fofino [41]

Average Rate of Change/Slope Formula: $\frac{y_2-y_1}{x_2-x_1}$

So using points (6,71) and (2,y), and the slope of 15, plug them into the slope formula as such:

$\frac{71-y}{6-2}=15$

From here we can solve for y. Firstly, solve the subtraction:

$\frac{71-y}{4}=15$

Next, multiply both sides by 4:

$71-y=60$

Next, subtact both sides by 71:

$-y=-11$

Lastly, multiply both sides by -1, and <u>your final answer will be $y=11$ , or A.</u>

6 0

4 years ago

Find the distance between the points given. (-3, -6) and (3, 2) 4 10

DanielleElmas [232]

Answer:

10

Step-by-step explanation:

6 0

4 years ago

Read 2 more answers

According to USA Today, customers are not settling for automobiles straight off the production lines. As an example, those who

MrMuchimi

Answer: $(24980.25,\ 25419.75)$

Step-by-step explanation:

The confidence interval for population mean is given by :-

$\overline{x}\pm z_{\alpha/2}\dfrac{\sigma}{\sqrt{n}}$

Given : Sample size : $n= 179$ , which is a large sample , so we apply z-test .

Sample mean : $\overline{x}=20200+5000=25200$

Standard deviation : $\sigma= 1500$

Significance level : $\alpha=1-0.95=0.05$

Critical value : $z_{\alpha/2}=1.96$

Now, a confidence interval at the 95% level of confidence will be :-

$25200\pm(1.96)\dfrac{1500}{\sqrt{179}}\\\\\approx25200\pm219.75\\\\=(25200-219.75,\ 25200+219.75)\\\\=(24980.25,\ 25419.75)$

8 0

3 years ago

Plz solve the problem.

adell [148]

Answer:
If this is what your looking for, the answer may be:
x=7
Hope this helps!

3 0

3 years ago

Read 2 more answers