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2 years ago

If (3a + 3b + 1) (3a + 3b - 1) = 80, then a + b =

Mathematics

1 answer:

seropon [69]2 years ago

8 0

Answer:

a+b = 4 $\sqrt{3}$

Step-by-step explanation:

3(a + b)^2 = 81

(a + b)^2 = 27

a + b = sq rt 27

a + b = $\sqrt{9}$ · $\sqrt{3}$

a + b = 3 $\sqrt{3}$

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Bas_tet [7]

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Step-by-step explanation:

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3 years ago

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2 years ago

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spayn [35]

Answer:

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1500 : 1060

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2 years ago

Can someone please help me solve these two problems and explain them to me? About the one with proportions though... I've always

NeTakaya

Answers:

A) $-\frac{11}{4}$

C) $m_{1}=\frac{Fr^{2}}{G m_{2}}$

Step-by-step explanation:

We have the followig equation:

$\frac{x-1}{3}-\frac{x+4}{5}=\frac{4x-1}{8}$

Calculating the least common multiple (l.c.m) in the denominator in the left side of the equation, being l.c.m=15:

$\frac{5(x-1)-3(x+4)}{15}=\frac{4x-1}{8}$

Solving for the left part of the equation:

$\frac{2x-17}{15}=\frac{4x-1}{8}$

Operating with cross product:

$8(2x-17)=15(4x-1)$

Applying the distributive property:

$16x-136=60x-15$

Isolating $x$ :

$x=-\frac{121}{44}$

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$x=-\frac{11}{4}$ Hence, the correct option is A

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$F=\frac{G m_{1} m_{2}}{r^{2}}$

Operating with cross product:

$Fr^{2}=G m_{1} m_{2}$

Isolating $m_{1}$ :

$m_{1}=\frac{Fr^{2}}{G m_{2}}$ Hence, the correct option is C

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3 years ago