Write an inequality the score in a basketball game is greater than or equal to 10 points

Mathematics

1 answer:

andreev551 [17]3 years ago

4 0

S greater than or equal to 10

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a house and lot cost $120,000. if the house cost 3 times more than the lot, how much did the house cost?

AleksandrR [38]

Answer:

$90,000

Step-by-step explanation:

We can can use substitution for this. First, we need to create two separate problems that represent this situation. The house and the lot together cost 120000, so we can make the equation h+l=120000. We also know that the house is worth 3 lots, so we can make the equation h=3l. Because h=3l, we can substitute for h, making the equation 3l+l=120000, or 4l=120000. To isolate l, we have to divide both sides by 4. If we do that, we get l=30000. Now, we have to substitute to find h. Our new equation is h+30000=120000. To isolate h, we have to subtract 30000 from each side. If we do this, we get h=90000. That's our answer.

Hope this helps!

7 0

2 years ago

Find the sum:<br> 1. 23.456 + 3.0034

Artist 52 [7]

Answer:

26.4594

Step-by-step explanation:

8 0

3 years ago

Read 2 more answers

What is the following quotient?

seropon [69]

Answer: Choice B) $\frac{5\sqrt{11}+5\sqrt{3}}{8}\\\\$

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Work Shown:

$\frac{5}{\sqrt{11}-\sqrt{3}}\\\\\\\frac{5(\sqrt{11}+\sqrt{3})}{(\sqrt{11}-\sqrt{3})(\sqrt{11}+\sqrt{3})}\\\\\\\frac{5\sqrt{11}+5\sqrt{3}}{(\sqrt{11})^2-(\sqrt{3})^2}\\\\\\\frac{5\sqrt{11}+5\sqrt{3}}{11-3}\\\\\\\frac{5\sqrt{11}+5\sqrt{3}}{8}\\\\\\$

This shows why choice B is the answer.

---------------

Explanation:

In the second step, I multiplied top and bottom by $(\sqrt{11}+\sqrt{3})$
This is so we can apply the difference of squares rule in step 3. The difference of squares rule is (x-y)(x+y) = x^2-y^2.
In step 4, the square roots cancel out with the squaring operation. The two operations are inverses of one another, which is why they cancel.

So in general, if the denominator is $\sqrt{a}-\sqrt{b}$ and you want to rationalize the denominator, then you should multiply top and bottom by $\sqrt{a}+\sqrt{b}$ . The same applies in reverse as well.