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3 years ago

what is 2/a = a/50. These are factions if you already didn't know. I need to know what the answer is. it is a proportion.

1 answer:

3 0

2/a= a/50

Cross multiply:
a* a= 2*50
⇒ a^2= 100
⇒ a= √100 or a= -√100
⇒ a= 10 or a= -10

Final answer: a= 10 or a= -10~

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Alan has to finish an exam within 120 minutes. He finished the exam in 80 minutes. What is the percentage of the time he took fo

drek231 [11]

Answer:

66.67%

Step-by-step explanation:

80/120 = 66.67

7 0

3 years ago

The area of a triangle with base 10in and height 4 in is<br> 20in^2<br> 40in^2<br> 15in^2<br> 60in^2

natta225 [31]

1/2(BH)

=1/2(10*4)

=1/2(40)

=20

5 0

3 years ago

What is the value of x in the equation below?

belka [17]

$remember$
$ln(a^b)=bln(a)$
$ln(e)=1$
$so$
$ln(e^x)=xln(e)=x$

$1+2e^{x+1}=9$
$minus \space\ 1 \space\ both \space\ sides$
$2e^{x+1}=8$
$divide \space\ both \space\ sides \space\ by \space\ 2$
$e^{x+1}=4$
$take \space\ ln \space\ of \space\ both \space\ sides$
$ln(e^{x+1})=ln(4)$
$(x+1)ln(e)=ln(4)$
$x+1=ln(4)$
$minus \space\ 1 \space\ from \space\ both \space\ sides$
$x=ln(4)-1$
$answer \space\ is \space\ C=ln(4)-1$

5 0

3 years ago

Read 2 more answers

Through:(2,-5),slope=3/2

Rainbow [258]

Are you trying write it into slope intercept form ?

8 0

3 years ago

Solve the following equation by completing the square. 3x^2-3x-5=13

mr Goodwill [35]

we'll start off by grouping some

$\bf 3x^2-3x-5=13\implies (3x^2-3x)-5=13\implies 3(x^2-x)-5=13 \\\\\\ 3(x^2-x)=18\implies (x^2-x)=\cfrac{18}{3}\implies (x^2-x)=6\implies (x^2-x+~?^2)=6$

so we have a missing guy at the end in order to get the a perfect square trinomial from that group, hmmm, what is it anyway?

well, let's recall that a perfect square trinomial is

$\bf \qquad \textit{perfect square trinomial} \\\\ (a\pm b)^2\implies a^2\pm \stackrel{\stackrel{\text{\small 2}\cdot \sqrt{\textit{\small a}^2}\cdot \sqrt{\textit{\small b}^2}}{\downarrow }}{2ab} + b^2$

so we know that the middle term in the trinomial, is really 2 times the other two without the exponent, well, in our case, the middle term is just "x", well is really -x, but we'll add the minus later, we only use the positive coefficient and variable, so we'll use "x" to find the last term.

$\bf \stackrel{\textit{middle term}}{2(x)(?)}=\stackrel{\textit{middle term}}{x}\implies ?=\cfrac{x}{2x}\implies ?=\cfrac{1}{2}$

so, there's our fellow, however, let's recall that all we're doing is borrowing from our very good friend Mr Zero, 0, so if we add (1/2)², we also have to subtract (1/2)²

$\bf \left( x^2 -x +\left[ \cfrac{1}{2} \right]^2-\left[ \cfrac{1}{2} \right]^2 \right)=6\implies \left( x^2 -x +\left[ \cfrac{1}{2} \right]^2 \right)-\left[ \cfrac{1}{2} \right]^2=6 \\\\\\ \left(x-\cfrac{1}{2} \right)^2=6+\cfrac{1}{4}\implies \left(x-\cfrac{1}{2} \right)^2=\cfrac{25}{4}\implies x-\cfrac{1}{2}=\sqrt{\cfrac{25}{4}} \\\\\\ x-\cfrac{1}{2}=\cfrac{\sqrt{25}}{\sqrt{4}}\implies x-\cfrac{1}{2}=\cfrac{5}{2}\implies x=\cfrac{5}{2}+\cfrac{1}{2}\implies x=\cfrac{6}{2}\implies \boxed{x=3}$

6 0

3 years ago