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1 year ago

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Roman paid $150 to join a handball club. He pays an additional $15 every time he uses one of the club's handball courts. Write a

n equation that describes Roman's total cost for playing handball as a function of the number of times he plays. Let C = the total cost and n = the number of times he plays

2 answers:

Alekssandra [29.7K]1 year ago

4 0

Answer:

mmmmmmmmmmmmmmmmmmmm

Step-by-step explanation:

brilliants [131]1 year ago

3 0

C = 15n + 150 I think

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John Tucker is interested in purchasing stock. He wants to determine the costs.

Delvig [45]

Answer: a0 =$97.5 ; a1 =$87.96 ; a2=$282.48; a3=$1,722.5 ; a4=$1,553.96 ; a5=$4,990.48

Step-by-step explanation:

Hi, to solve this problem we have to multiply each total by the commission in decimal form (6 / 100= 0.06).

So:

$1625 x 0.06 = $97.5

$1,466 x 0.06 =$87.96

$4,708 x 0.06 =$282.48

And finally, to obtain the total costs we have to add to each total the commission of the total.

$1625 + $97.5 = $1,722.5

$1,466 +$87.96 = $1,553.96

$4,708 +$282.48 = $4,990.48

Feel free to ask for more if it´s necessary or if you did not understand something.

8 0

2 years ago

Read 2 more answers

A ball is tossed between three friends. The first toss is 8.6 feet, the second is 5.8 feet, and the third toss is 7.5 feet, whic

Rainbow [258]

angles formed by these tosses are $79.45, 59.02$ and $41.53$ degrees to the nearest hundredth.

<u>Step-by-step explanation:</u>

Here , We have a triangle with sides of length 8.6 feet, 5.8 feet and 7.5 feet.

The Law of Cosines (also called the Cosine Rule) says:

$c^2 = a^2 + b^2 - 2ab (cosx)$

Using the Cosine Rule to find the measure of the angle opposite the side of length 8.6 feet:

⇒ $c^2 = a^2 + b^2 - 2ab (cosx)$

⇒ $c^2 -a^2 - b^2 = -2ab (cosx)$

⇒ $(cosx) =\frac{ c^2 -a^2 - b^2}{ -2ab}$

⇒ $(cosx) =\frac{(8.6^2 - 5.8^2 - 7.5^2)}{ ( -2(5.8)7.5)}$

⇒ $(cosx) =0.18310$

⇒ $cos^{-1}(cosx) = cos^{-1}(0.18310)$

⇒ $x = 79.45$

The Law of Sines (or Sine Rule) is very useful for solving triangles:

$\frac{a}{sin A} = \frac{ b}{sin B} = \frac{c}{sin C}$

We can now find another angle using the sine rule:

⇒ $\frac{ 8.6 }{ sin 79.45} = \frac{7.5}{ sin Y}$

⇒ $sin Y = \frac{(7.5 (sin 79.45))}{ 8.6}$

⇒ $Y = 59.02 degrees$

So, the third angle = $180 - 79.45 - 59.02 = 41.53 degrees.$

Therefore, angles formed by these tosses are $79.45, 59.02$ and $41.53$ degrees to the nearest hundredth.

4 0

2 years ago

What’s the ratio of the distance?

Alik [6]

Answer:

B (2,8)

Step-by-step explanation:

7 0

2 years ago

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!

8 0

2 years ago

Read 2 more answers

What is the greatest common factor (GCF) of 48 and 32?

Nat2105 [25]

Answer:

The greatest common factor of 48 and 32 is 16

6 0

2 years ago

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