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3 years ago

8

The quotient of 14 and z

1 answer:

viva [34]3 years ago

6 0

Answer:

dont know sorry

Step-by-step explanation:

sorry :(

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Vadim26 [7]

Answer:

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4 years ago

Is the product of three negative numbers positive or negative?

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Answer:

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3 years ago

Equation with a slope of 2/3 and a y intercept of -3

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Answer:

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Step-by-step explanation:

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3 years ago

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kifflom [539]

Answer:

Ummm

Step-by-step explanation:

So what do you need help with?

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3 years ago

Can someone please help with these 3 I am so confused on what to do thank you!!!!

WINSTONCH [101]

<u>Answer:</u>

26. $\frac{3}{4} \leq x$

27. $x\leq 5$

28. $0\leq b$

<u>Step-by-step explanation:</u>

26. $\sqrt{4x-3}$ :

Taking square on both the sides to get:

$(\sqrt{4x-3} )^2 < (5)^2$

$4x-3$

$4x$

$x$

$x$

For non-negative values for radical:

$4x-3\geq 0$

$x\geq \frac{3}{4}$

So solution for this: $\frac{3}{4} \leq x$

27. $2+\sqrt{4x-4}\leq 6$

Subtracting 2 from both the sides to get:

$2+\sqrt{4x-4}-2\leq 6-2$

$\sqrt{4x-4}\leq 4$

Taking square root on both sides:

$(\sqrt{4x-4})^2\leq (4)^2$

$4x-4\leq 16$

$x\leq \frac{20}{4}$

$x\leq 5$

28. $\sqrt{b+12} -\sqrt{b}>2$

Adding $\sqrt{b}$ to both the sides to get:

$\sqrt{b+12} -\sqrt{b}+\sqrt{b}>2+\sqrt{b}$

$\sqrt{b+12} >2+\sqrt{b}$

Taking square on both sides:

$(\sqrt{b+12})^2 >(2+\sqrt{b})^2$

$b+12>(2+\sqrt{b})^2$

$b+12>4+4\sqrt{b}+b$

$4+4\sqrt{b} +b$

Subtracting $b$ from both sides to get:

$4+4\sqrt{b} +b-b$

$4+4\sqrt{b}$

Subtracting 4 from both sides:

$4+4\sqrt{b}-4$

$4\sqrt{b}$

Square both sides again:

$(4\sqrt{b})^2$

$16b$

$b$

$b$

and for non-negative radical $b\geq 0$

therefore, solution is $0\leq b$ .

5 0

3 years ago