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2 years ago

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Amanda’s Training Continues…..Amanda is training to run a marathon. She must follow a strict schedule to make sure she is ready

for the race. She will start training by running two miles the first week. She wants to run one fewer mile the next week, then run three more miles the week after that. She will continue this pattern during her entire training regimen. Question: Amanda needs to run 26 miles in the final week of her training. In which week will Amanda reach her goal?

1 answer:

Paladinen [302]2 years ago

3 0

Based on her on training regiment, Amanda will be able to run 26 miles in the 25th week.

<h3 /><h3>What is the total number of miles Amanda will run every week?</h3>

Amanda will run 2 miles in the first week and then 1 less mile in the following week. Then 3 more miles in the week after.

Week 1 - 2 miles

Week 2:

= 2 - 1

= 1 mile

Week 3:

= 1 + 3

= 4 miles

Week 4:

= 4 - 1

= 3 miles

Week 5:

= 3 + 3

= 6 miles

From the given pattern, Amanda increase the number of miles ran by 2 miles every 2 weeks.

The number of miles ran every other week would therefore be:

Week 7 = 8 miles

Week 9 = 10 miles

Week 11 = 12 miles

Week 15 = 16 miles

Week 19 = 20 miles

Week 25 = 26 miles

In conclusion, she would reach her goal in 25 weeks.

Find out more on solving schedule problems at brainly.com/question/20020676.

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Solve for x in the equation 2x^2+3x-7=x^2+5x+39

Shalnov [3]

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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? A mug can hold 13.46 oz of coffee without overflowing. The radius of the mug is 3 cm. Given that 1 cm3 equals 0.034 oz, what i

Elan Coil [88]

Answer:

14 cm

Step-by-step explanation:

Given:

A mug can hold 13.46 oz of coffee.

The radius of the mug is 3 cm.

Given that 1 $cm^{3}$ equals 0.034 oz.

Question asked:

What is the height of the mug to the nearest centimeter ?

Solution:

Given that 1 $cm^{3}$ equals 0.034 oz.

By unitary method:

0.034 oz = 1 $cm^{3}$

1 oz = $\frac{1}{0.034}$

13.46 oz = $\frac{1}{0.034}\times13.46 =395.88 \ cm^{3}$

A mug can hold 13.46 oz of coffee means volume of cylinder is given which is 395.88 $cm^{3}$ . Now we can find the height of the mug by using volume of cylinder formula:

volume of cylinder = $\pi r^{2} h$

$395.88=\frac{22}{7} \times3\times3\times \ h\\395.88=\frac{198 \ h}{7}$

By cross multiplication:

$395.88\times 7 =198\ h\\2771.16= 198\ h$

By dividing both side by 198

$h = 13.99$ $cm$

Therefore, height of the mug to the nearest centimeter is 14 cm.

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creativ13 [48]

Answer:

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given

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slope is -3/4

Step-by-step explanation:

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Answer:

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