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scoundrel [369]
3 years ago
8

11 ft = ao in. Please help I am in neeed!

Mathematics
1 answer:
Vedmedyk [2.9K]3 years ago
3 0

Answer:

132 inches

Step-by-step explanation:

multiply 11(ft) by 12(inches)

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If m∠1 =63° , find m∠2 and m∠3 by calculation
Neko [114]
Since ∠1 is 63 than ∠2 is 63 too because they are alternate interior. To find ∠3 you subtract 180 by the measure of ∠2 because both of these angles create a straight line.

180 - 63 = ∠3
117 = ∠3

Hope this helps :)
6 0
2 years ago
Read number 5 and help me out plz add an explanation
maks197457 [2]

Answer:

the new values could be 3 and one half and 5 and one half.

Step-by-step explanation:

This is correct because they are asking for a new one. Also, one that has 3 as a difference in between.

PLEASE LOOK AT ATTACTHMENT!!!

Download pdf
<span class="sg-text sg-text--link sg-text--bold sg-text--link-disabled sg-text--blue-dark"> pdf </span>
1377a08b6d21b745b22e15b9654dc988.png
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2 years ago
Solve irrational equation pls
rusak2 [61]
\hbox{Domain:}\\&#10;x^2+x-2\geq0 \wedge x^2-4x+3\geq0 \wedge x^2-1\geq0\\&#10;x^2-x+2x-2\geq0 \wedge x^2-x-3x+3\geq0 \wedge x^2\geq1\\&#10;x(x-1)+2(x-1)\geq 0 \wedge x(x-1)-3(x-1)\geq0 \wedge (x\geq 1 \vee x\leq-1)\\&#10;(x+2)(x-1)\geq0 \wedge (x-3)(x-1)\geq0\wedge x\in(-\infty,-1\rangle\cup\langle1,\infty)\\&#10;x\in(-\infty,-2\rangle\cup\langle1,\infty) \wedge x\in(-\infty,1\rangle \cup\langle3,\infty) \wedge x\in(-\infty,-1\rangle\cup\langle1,\infty)\\&#10;x\in(-\infty,-2\rangle\cup\langle3,\infty)


&#10;\sqrt{x^2+x-2}+\sqrt{x^2-4x+3}=\sqrt{x^2-1}\\&#10;x^2-1=x^2+x-2+2\sqrt{(x^2+x-2)(x^2-4x+3)}+x^2-4x+3\\&#10;2\sqrt{(x^2+x-2)(x^2-4x+3)}=-x^2+3x-2\\&#10;\sqrt{(x^2+x-2)(x^2-4x+3)}=\dfrac{-x^2+3x-2}{2}\\&#10;(x^2+x-2)(x^2-4x+3)=\left(\dfrac{-x^2+3x-2}{2}\right)^2\\&#10;(x+2)(x-1)(x-3)(x-1)=\left(\dfrac{-x^2+x+2x-2}{2}\right)^2\\&#10;(x+2)(x-3)(x-1)^2=\left(\dfrac{-x(x-1)+2(x-1)}{2}\right)^2\\&#10;(x+2)(x-3)(x-1)^2=\left(\dfrac{-(x-2)(x-1)}{2}\right)^2\\&#10;(x+2)(x-3)(x-1)^2=\dfrac{(x-2)^2(x-1)^2}{4}\\&#10;4(x+2)(x-3)(x-1)^2=(x-2)^2(x-1)^2\\
&#10;4(x+2)(x-3)(x-1)^2-(x-2)^2(x-1)^2=0\\&#10;(x-1)^2(4(x+2)(x-3)-(x-2)^2)=0\\&#10;(x-1)^2(4(x^2-3x+2x-6)-(x^2-4x+4))=0\\&#10;(x-1)^2(4x^2-4x-24-x^2+4x-4)=0\\&#10;(x-1)^2(3x^2-28)=0\\&#10;x-1=0 \vee 3x^2-28=0\\&#10;x=1 \vee 3x^2=28\\&#10;x=1 \vee x^2=\dfrac{28}{3}\\&#10;x=1 \vee x=\sqrt{\dfrac{28}{3}} \vee x=-\sqrt{\dfrac{28}{3}}\\

There's one more condition I forgot about
-(x-2)(x-1)\geq0\\&#10;x\in\langle1,2\rangle\\

Finally
x\in(-\infty,-2\rangle\cup\langle3,\infty) \wedge x\in\langle1,2\rangle \wedge x=\{1,\sqrt{\dfrac{28}{3}}, -\sqrt{\dfrac{28}{3}}\}\\&#10;\boxed{\boxed{x=1}}
3 0
2 years ago
Help me please and the last option is 9cm on the left and 12 on bottom
Rasek [7]
The last one is the answer
8 0
3 years ago
Plz help! Ive been stuck on this for hours!!!!
Marysya12 [62]

Answer:

h\geq8

Step-by-step explanation:

h-(-2)\geq10=h+2\geq10

subtract 2 on both sides to get h\geq8

3 0
3 years ago
Read 2 more answers
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