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2 years ago

What is polynomial in factored form for p(x) = (x+5) (x

Mathematics

1 answer:

aev [14]2 years ago

7 0

Answer:

p(x) = x^2 + 5x

Step-by-step explanation:

p(x) = (x+5) (x <====== something may be missing here?

p(x) = (x+5) (x Use distributive property to expand R side

p(x) = x^2 + 5x

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After spending 5/9 of his money, David has<br>$36 left. How much money did he have at first?

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3 years ago

A car dealership sells suv's and passenger cars. For a recent year, 90 more suv's were sold than passenger cars. If a total of 5

Lunna [17]

Answer:

300 SUVs were sold.

210 Passenger cars were sold.

Step-by-step explanation:

Let S denote the amount of SUVs sold and let P denote the amount of passenger cars sold.

The total amount of vehicles sold was 510, so:

$S+P=510$

90 more SUVs were sold than passenger cars. In other words:

$S=90+P$

We know have a system of equations. We can solve it buy substituting the second equation into the first. Thus:

$90+P+P=510$

Combine like terms:

$2P+90=510$

Subtract 90 from both sides:

$2P=420$

Divide both sides by 2:

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So, the 210 passenger cars were sold.

This means that 210+90=300 SUVs were sold.

And we're done!

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3 years ago

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malfutka [58]

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3 years ago

<img src="https://tex.z-dn.net/?f=%5Csqrt%5B4%5D%7B5x%2F8y%7D" id="TexFormula1" title="\sqrt[4]{5x/8y}" alt="\sqrt[4]{5x/8y}" al

Furkat [3]

Answer: $\frac{\sqrt[4]{10xy^3}}{2y}$

where y is positive.

The 2y in the denominator is not inside the fourth root

==================================================

Work Shown:

$\sqrt[4]{\frac{5x}{8y}}\\\\\\\sqrt[4]{\frac{5x*2y^3}{8y*2y^3}}\ \ \text{.... multiply top and bottom by } 2y^3\\\\\\\sqrt[4]{\frac{10xy^3}{16y^4}}\\\\\\\frac{\sqrt[4]{10xy^3}}{\sqrt[4]{16y^4}} \ \ \text{ ... break up the fourth root}\\\\\\\frac{\sqrt[4]{10xy^3}}{\sqrt[4]{(2y)^4}} \ \ \text{ ... rewrite } 16y^4 \text{ as } (2y)^4\\\\\\\frac{\sqrt[4]{10xy^3}}{2y} \ \ \text{... where y is positive}\\\\\\$

The idea is to get something of the form $a^4$ in the denominator. In this case, $a = 2y$

To be able to reach the $16y^4$ , your teacher gave the hint to multiply top and bottom by $2y^3$

For more examples, search out "rationalizing the denominator".

Keep in mind that $\sqrt[4]{(2y)^4} = 2y$ only works if y isn't negative.

If y could be negative, then we'd have to say $\sqrt[4]{(2y)^4} = |2y|$ . The absolute value bars ensure the result is never negative.

Furthermore, to avoid dividing by zero, we can't have y = 0. So all of this works as long as y > 0.

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2 years ago