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2 years ago

Here are the scores that Jenna has

Mathematics

1 answer:

amid [387]2 years ago

5 0

Answer:

the variance is 6.6

sry for not showing work

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How do u solve the system 3x+5y=8 2x-5y=-3

RSB [31]

Add both equations giving x=1
Sub back into any equation which gives you y=1

8 0

3 years ago

7 = 2g + 5<br><br> pleas help... lolz

Luden [163]

Answer:

2 times 1 = 2 and then plus 5 = 7

Hope this helped (-;

7 0

3 years ago

Solve. 1,032 divided by 24 Enter the answer in the box.<br> Response area with 1 text input box

AlexFokin [52]

Answer:

$258\quad \mathrm{Remainder}\quad \:0$

Step-by-step explanation:

$\frac{1032}{4} \\\\\begin{matrix}4\overline{|\smallspace1032}\space\space\space\space\space\space\space\space\space\space\space\space\end{matrix}\\\\\mathrm{Divide}\:10\:\mathrm{by}\:4\:\mathrm{to\:get}\:2\\\begin{matrix}\space\space\space\space\space\space\emptyspace2\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\\ 4\overline{|\smallspace1032}\space\space\space\space\space\space\space\space\space\space\space\space\end{matrix}\\\\$

$\mathrm{Multiply\:the\:quotient\:digit}\:\left(2\right)\:\mathrm{by\:the\:divisor}\:4\\\begin{matrix}\space\space\space\space\space\space\emptyspace2\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\\4\overline{|\smallspace1032}\space\space\space\space\space\space\space\space\space\space\space\space\\ \space\space\space\space\space\space\underline{\emptyspace8}\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\end$

Long story short ,

$\mathrm{The\:solution\:for\:Long\:Division\:of}\:\\\frac{1032}{4}\:\mathrm{is}\\\:258\:\mathrm{with\:remainder\:of}\:0$

6 0

3 years ago

5^(-x)+7=2x+4 This was on plato

Setler79 [48]

Answer:

Below

I hope its not too complicated

$x=\frac{\text{W}_0\left(\frac{\ln \left(5\right)}{2e^{\frac{3\ln \left(5\right)}{2}}}\right)}{\ln \left(5\right)}+\frac{3}{2}$

Step-by-step explanation:

$5^{\left(-x\right)}+7=2x+4\\\\\mathrm{Prepare}\:5^{\left(-x\right)}+7=2x+4\:\mathrm{for\:Lambert\:form}:\quad 1=\left(2x-3\right)e^{\ln \left(5\right)x}\\\\\mathrm{Rewrite\:the\:equation\:with\:}\\\left(x-\frac{3}{2}\right)\ln \left(5\right)=u\mathrm{\:and\:}x=\frac{u}{\ln \left(5\right)}+\frac{3}{2}\\\\1=\left(2\left(\frac{u}{\ln \left(5\right)}+\frac{3}{2}\right)-3\right)e^{\ln \left(5\right)\left(\frac{u}{\ln \left(5\right)}+\frac{3}{2}\right)}$

$Simplify\\\\\mathrm{Rewrite}\:1=\frac{2e^{u+\frac{3}{2}\ln \left(5\right)}u}{\ln \left(5\right)}\:\\\\\mathrm{in\:Lambert\:form}:\quad \frac{e^{\frac{2u+3\ln \left(5\right)}{2}}u}{e^{\frac{3\ln \left(5\right)}{2}}}=\frac{\ln \left(5\right)}{2e^{\frac{3\ln \left(5\right)}{2}}}$

$\mathrm{Solve\:}\:\frac{e^{\frac{2u+3\ln \left(5\right)}{2}}u}{e^{\frac{3\ln \left(5\right)}{2}}}=\frac{\ln \left(5\right)}{2e^{\frac{3\ln \left(5\right)}{2}}}:\quad u=\text{W}_0\left(\frac{\ln \left(5\right)}{2e^{\frac{3\ln \left(5\right)}{2}}}\right)\\\\\mathrm{Substitute\:back}\:u=\left(x-\frac{3}{2}\right)\ln \left(5\right),\:\mathrm{solve\:for}\:x$

$\mathrm{Solve\:}\:\left(x-\frac{3}{2}\right)\ln \left(5\right)=\text{W}_0\left(\frac{\ln \left(5\right)}{2e^{\frac{3\ln \left(5\right)}{2}}}\right):\\\quad x=\frac{\text{W}_0\left(\frac{\ln \left(5\right)}{2e^{\frac{3\ln \left(5\right)}{2}}}\right)}{\ln \left(5\right)}+\frac{3}{2}$

3 0

3 years ago

Select "Growth" or "Decay" to classify each function

natulia [17]

Answer:

decay

growth

decay

i'm sure im in k12 for 9 th grade please tell me if i'm right

Step-by-step explanation:

4 0

3 years ago