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IRINA_888 [86]
3 years ago
12

HELP WITH MATH PLEASE! ASAP!!! Annika borrowed a book from the library and forgot to return it on time. The library charges $0.2

5 per day for the first four days and $0.30 per day for all of the remaining days of Annika had to pay the library $4.00. How many days late was her book? THANKS
Mathematics
2 answers:
ser-zykov [4K]3 years ago
6 0
14 days
.25 x 4days = $1.00
0.30x 10 days = $3.00

$3 + $1 = $4
mario62 [17]3 years ago
5 0
If she had to pay $4.00, then we know she was late 4 days for sure. 4*0.25 = 1.

Then we know she was an extra 10 days late because 10*0.30=3.

3+1= $4.00

She was late 14 days altogether.
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Hey there, hope I can help!

\mathrm{Subtract\:}x^2+5x+39\mathrm{\:from\:both\:sides}
2x^2+3x-7-\left(x^2+5x+39\right)=x^2+5x+39-\left(x^2+5x+39\right)

Assuming you know how to simplify this, I will not show the steps but can add them later on upon request
x^2-2x-46=0

Lets use the quadratic formula now
\mathrm{For\:a\:quadratic\:equation\:of\:the\:form\:}ax^2+bx+c=0\mathrm{\:the\:solutions\:are\:}
x_{1,\:2}=\frac{-b\pm \sqrt{b^2-4ac}}{2a}

\mathrm{For\:} a=1,\:b=-2,\:c=-46: x_{1,\:2}=\frac{-\left(-2\right)\pm \sqrt{\left(-2\right)^2-4\cdot \:1\left(-46\right)}}{2\cdot \:1}

\frac{-\left(-2\right)+\sqrt{\left(-2\right)^2-4\cdot \:1\cdot \left(-46\right)}}{2\cdot \:1} \ \textgreater \  \mathrm{Apply\:rule}\:-\left(-a\right)=a \ \textgreater \  \frac{2+\sqrt{\left(-2\right)^2-4\cdot \:1\cdot \left(-46\right)}}{2\cdot \:1}

Multiply the numbers 2 * 1 = 2
\frac{2+\sqrt{\left(-2\right)^2-\left(-46\right)\cdot \:1\cdot \:4}}{2}

2+\sqrt{\left(-2\right)^2-4\cdot \:1\cdot \left(-46\right)} \ \textgreater \  \sqrt{\left(-2\right)^2-4\cdot \:1\cdot \left(-46\right)}

\mathrm{Apply\:rule}\:-\left(-a\right)=a \ \textgreater \  \sqrt{\left(-2\right)^2+1\cdot \:4\cdot \:46} \ \textgreater \  \left(-2\right)^2=2^2, 2^2 = 4

\mathrm{Multiply\:the\:numbers:}\:4\cdot \:1\cdot \:46=184 \ \textgreater \  \sqrt{4+184} \ \textgreater \  \sqrt{188} \ \textgreater \  2 + \sqrt{188}
\frac{2+\sqrt{188}}{2} \ \textgreater \  Prime\;factorize\;188 \ \textgreater \  2^2\cdot \:47 \ \textgreater \  \sqrt{2^2\cdot \:47}

\mathrm{Apply\:radical\:rule}: \sqrt[n]{ab}=\sqrt[n]{a}\sqrt[n]{b} \ \textgreater \  \sqrt{47}\sqrt{2^2}

\mathrm{Apply\:radical\:rule}: \sqrt[n]{a^n}=a \ \textgreater \  \sqrt{2^2}=2 \ \textgreater \  2\sqrt{47} \ \textgreater \  \frac{2+2\sqrt{47}}{2}

Factor\;2+2\sqrt{47} \ \textgreater \  Rewrite\;as\;1\cdot \:2+2\sqrt{47}
\mathrm{Factor\:out\:common\:term\:}2 \ \textgreater \  2\left(1+\sqrt{47}\right) \ \textgreater \  \frac{2\left(1+\sqrt{47}\right)}{2}

\mathrm{Divide\:the\:numbers:}\:\frac{2}{2}=1 \ \textgreater \  1+\sqrt{47}

Moving on, I will do the second part excluding the extra details that I had shown previously as from the first portion of the quadratic you can easily see what to do for the second part.

\frac{-\left(-2\right)-\sqrt{\left(-2\right)^2-4\cdot \:1\cdot \left(-46\right)}}{2\cdot \:1} \ \textgreater \  \mathrm{Apply\:rule}\:-\left(-a\right)=a \ \textgreater \  \frac{2-\sqrt{\left(-2\right)^2-4\cdot \:1\cdot \left(-46\right)}}{2\cdot \:1}

\frac{2-\sqrt{\left(-2\right)^2-\left(-46\right)\cdot \:1\cdot \:4}}{2}

2-\sqrt{\left(-2\right)^2-4\cdot \:1\cdot \left(-46\right)} \ \textgreater \  2-\sqrt{188} \ \textgreater \  \frac{2-\sqrt{188}}{2}

\sqrt{188} = 2\sqrt{47} \ \textgreater \  \frac{2-2\sqrt{47}}{2}

2-2\sqrt{47} \ \textgreater \  2\left(1-\sqrt{47}\right) \ \textgreater \  \frac{2\left(1-\sqrt{47}\right)}{2} \ \textgreater \  1-\sqrt{47}

Therefore our final solutions are
x=1+\sqrt{47},\:x=1-\sqrt{47}

Hope this helps!
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