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ryzh [129]
3 years ago
13

What is 190×76=? Please help!

Mathematics
2 answers:
mart [117]3 years ago
6 0
190*76 is equal to 14440.
vazorg [7]3 years ago
6 0
The answer to this problem is 14440.


Hope this helped you :D
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Write the equations included in the same set of related facts as 6x8=48
Rus_ich [418]
8x6=48, 1x48, 2x24=48, 3x16=48, 4x12, and 6x8
I hope this helps!!!
6 0
3 years ago
50 points
ycow [4]

Answer:

4 out of 10 or 4/10 is my answer

5 0
2 years ago
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Which is greater 5/6 7/12 7/10
Makovka662 [10]

5/6=0.8

7/12=0.5

7/10=0.7


so ... 0.8 is greater than 0.7 and 7.0 also greater than 0.5

0.8>0.7>0.5

I hope that's help !

8 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
How to find the hypotenuse of a triangle?
ExtremeBDS [4]
To find the hypotenuse of the triangle
you use the length of the legs
Plug it into the Pythagorean Theorem

so a^{2} +  b^{2} =  c^{2}

a^{2}  is one of the legs

and 

b^{2}  is the other leg

then 

c^{2}  is the hypothenuse
5 0
3 years ago
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