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Triss [41]
3 years ago
9

Can someone help me with factoring by grouping please show step by step

Mathematics
1 answer:
ValentinkaMS [17]3 years ago
4 0
Answer is 3-110x

3 times 1 squared - 11 times 10
1 raised to any power equals 1

calculate the product
3 times 1 -110x

any expression multiplied by one stays the same
3-110x

i hope this helped! thanks!
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Solve the equation 6t-1/6=9
gavmur [86]
6t=9 + 1/6
6t=9.17
t=1.53
7 0
3 years ago
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Pls help ASAP first correct answer gets brainleist ​
AnnZ [28]

Answer:

4.58257569 meters (Could probably get away with rounding it to 4.6 meters)

Step-by-step explanation:

Pythag theorem

a^2 +b^2 =c^2

Insert your numbers

10^2+b^2=11^2

Simplify squares

100+b^2=121

subtract 100 from both sides

b^2=21

sqrt both sides

sqrt(b^2)=sqrt(21)

simplify sqrts to get b

b= 4.58257569

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2 years ago
Michael can bike 2 miles in 12 minutes how long will it take him to bike 7 miles
stich3 [128]

42 minutes for 7 miles

6 0
2 years ago
Rationalize the numerator or denominator, and simplify:
allochka39001 [22]
\dfrac{\sqrt[3]{x+h}-\sqrt[3]x}h\times\dfrac{\sqrt[3]{(x+h)^2}+\sqrt[3]{x(x+h)}+\sqrt[3]{x^2}}{\sqrt[3]{(x+h)^2}+\sqrt[3]{x(x+h)}+\sqrt[3]{x^2}}=\dfrac{(\sqrt[3]{x+h})^3-(\sqrt[3]x)^3}\cdots
=\dfrac{x+h-x}\cdots=\dfrac h\cdots

The hs then cancel, leaving you with the \cdots=\sqrt[3]{(x+h)^2}+\sqrt[3]{x(x+h)}+\sqrt[3]{x^2} term.

If it's not clear what I did above, consider the substitution a=\sqrt[3]{x+h} and b=\sqrt[3]x. Then

a^3-b^3=(a-b)(a^2+ab+b^2)
\implies\dfrac{a-b}h=\dfrac{a-b}h\times\dfrac{a^2+ab+b^2}{a^2+ab+b^2}=\dfrac{a^3-b^3}{h(a^2+ab+b^2)}
4 0
3 years ago
CAN SOMEONE PLEASE SOLVE THIS????????
WARRIOR [948]

Answer:

27; C; B; D

Step-by-step explanation:

1. B = 120 degrees, and C = 60 degrees. Q = 90 degrees, because B + C = 180 degrees, and the triangle BQC has angle B and angle C bisected. If B = 120 degrees, then the height is equal to 9 sqrt 3, which makes BC equal to 27.

2.C

3. B (You don't actually know if any of the sides are parallel.)

4. D

4 0
3 years ago
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