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docker41 [41]
3 years ago
12

Problem #2: Ethan bought plates and cups for his upcoming birthday party. He

Mathematics
1 answer:
podryga [215]3 years ago
3 0
Create an equation.

2c + c = 102

3c = 102

c = 34

Ethan buys 34 cups and 68 plates.
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What is the area of r=5
Vlad1618 [11]

Answer:

if you need the circumferance of a circle with a radious of 5, then it should be 31.42.

Step-by-step explanation:

8 0
3 years ago
Read 2 more answers
What is x. Please show *all* the steps.
Zolol [24]

The equation 5/2 - x + x - 5/x + 2 + 3x + 8/x^2 - 4 = 0 is a quadratic equation

The value of x is 8 or 1

<h3>How to determine the value of x?</h3>

The equation is given as:

5/2 - x + x - 5/x + 2 + 3x + 8/x^2 - 4 = 0

Rewrite as:

-5/x - 2 + x - 5/x + 2 + 3x + 8/x^2 - 4 = 0

Take the  LCM

[-5(x + 2) + (x -5)(x- 2)]\[x^2 - 4 + [3x + 8]/[x^2 - 4] = 0

Expand

[-5x - 10 + x^2 - 7x + 10]/[x^2 - 4] + [3x + 8]/[x^2 - 4] = 0

Evaluate the like terms

[x^2 - 12x]/[x^2 - 4] + [3x + 8]/[x^2 - 4 = 0

Multiply through by x^2 - 4

x^2 - 12x+ 3x + 8 = 0

Evaluate the like terms

x^2 -9x + 8 = 0

Expand

x^2 -x - 8x + 8 = 0

Factorize

x(x -1) - 8(x - 1) = 0

Factor out x - 1

(x -8)(x - 1) = 0

Solve for x

x = 8 or x = 1

Hence, the value of x is 8 or 1

Read more about equations at:

brainly.com/question/2972832

8 0
2 years ago
Look at the picture below?
FinnZ [79.3K]

Answer:

well, okay. its basically what you wrote. assuming there was missing information.

let's dive right into it.

we'll assume that the problem asks for integers, whole numbers.

if x would be 0,

we would get 9 numbers out, 0, -1, -2, -3, -4, -5, -6, -7 and -8

if x would one, the range would be from 7 to -8, giving us 16 numbers

if x would be 2, the range would be from 14 to -8, giving us 23 numbers (7 more each time we increasex by one)

so the answer isn't a fixed value, but a function.

7x+9

the plus nine is true when x=0 and is still relevant for every other scenario

6 0
3 years ago
Kylo begins to fill up a bathtub at a steady rate. When the bathtub reaches 50 gallons at 15 minutes, Kylo
attashe74 [19]

Answer:

Please find a visual representation of the scenario generated with MS Excel

Step-by-step explanation:

The given data on how Kylo fills up the bathtub with water are;

The time it takes to fill the bathtub with 50 gallons of water = 15 minutes

The time it takes to drain 30 gallons from the bathtub = 5 minutes

The time it takes to add 5 gallons of hot water = 2 minutes

The time it takes for all the water to drain out = 4 minutes

The given scenario can be represented visually on the coordinate plane, as follows;

The y-axis represents the amount of water in the bathtub in gallons

The x-axis represents the time from the when Kylo begins to fill the bathtub

The coordinates of points in the visual representation of the given scenario are therefore;

At the start, the bathtub is empty;

y = 0, x = 0

Then we have;

\begin{array}{ccc}The \ Number \  of  \ Gallons, \,  y &Time &Cumulative  \  Time, \, x\\0&0&0\\50&15&15\\20&5&20\\25&2&22\\0&4&26\end{array}

Plotting the number of gallons and the cumulative time values on MS Excel gives a visual representation of the scenario

5 0
3 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
3 years ago
Read 2 more answers
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