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german
2 years ago
8

Find the missing side

Mathematics
1 answer:
pychu [463]2 years ago
3 0

Answer:

20

Step-by-step explanation:

Since this is a right triangle, we can use the Pythagorean theorem

a^2 + b^2 = c^2, where a and b are the sides and c is the hypotenuse

15^2 + b^2 = 25^2

225 + b^2 = 625

b^2 = 625 -225

b^2 =400

Take the square root of each side

sqrt(b^2) = sqrt(400)

b = 20

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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!
8 0
2 years ago
Read 2 more answers
because the angles in a rectangle are 90 degrees it is not a parallelogram true or false? please explain why
ra1l [238]
<span>A parallelogram is a 4-sided shape where opposites sides are parallel. A rectangle is a special case of a parallelogram. All rectangles are parallelograms.

But a rectangle is a shape where opposites sides are parallel *and* all the corners are 90 degree angles. So you can't say that all parallelograms would be rectangles. Some parallelograms would be rectangles, but not all.

Rectangles are a subset of the shapes called parallelograms. But parallelograms are *not* a subset of the shapes called rectangles.

It's similar to saying all cars are vehicles. But you can't say that all vehicles are cars. </span>
6 0
3 years ago
An amusement park charges admission plus a fee for each ride. Admission plus 2 rides costs $10. Admission plus five rides costs
LenaWriter [7]

Answer:

The charge for admission is $6 and the charge for each ride is $2

Step-by-step explanation:

Let

x ----> the charge for admission

y ----> the charge for each ride

we have that

x+2y=10 -----> equation A

x+5y=16 -----> equation B

Solve the system by elimination

Subtract equation B from equation A

x+2y=10\\-(x+5y)=-16\\---------\\2y-5y=10-16\\-3y=-6\\y=2

Find the value of x

substitute the value of y in any equation

x+2(2)=10

x+4=10

x=10-4

x=6

therefore

The charge for admission is $6 and the charge for each ride is $2

8 0
3 years ago
Nate walk 11 miles in 4 hours. At what unit rate does Nate walk?
coldgirl [10]

Answer:

2.75 or 2 3/4

Step-by-step explanation:

just divide 11/4 that's how many miles Nate walked every hour

4 0
3 years ago
Which linear inequality is graphed with y &gt;x-2 to
jek_recluse [69]

Answer:

D

Step-by-step explanation:

It's D on Edge, hope this helps

3 0
2 years ago
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