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2 years ago

What is the value of the expression a2 + b3 when a= 1/3 and b= 1/2 ?

Mathematics

1 answer:

Aleksandr-060686 [28]2 years ago

6 0

Answer:

B. $\frac{17}{72}$

Step-by-step explanation:

${a}^{2} + {b}^{3} \\ \\ = \bigg( \frac{1}{3} \bigg)^{2} + \bigg( \frac{1}{2} \bigg)^{3} \\ \\ = \frac{1}{9} + \frac{1}{8} \\ \\ = \frac{8 + 9}{9 \times 8} \\ \\ = \frac{17}{72} \\$

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Solve for m please. No explanation needed thank you.

liq [111]

Answer is below.

(Note to asker: Lack of explanations can get my answer deleted)

Step 1

We'll to isolate the variable, but to do that, get rid of the coefficient that multiplies it.

To do that, divide both sides by 22.

$\frac{22m}{22}=\frac{22}{10}$

Step 2.

Now we have divided both sides by 22, but we'll need to reduce the fraction $\frac{22m}{22}$ .

We can reduce the fraction by dividing the factors that are in the numerator and denominator.

That being said, 22 appears both in the numerator and denominator.

$\frac{22m}{22}=m=\frac{2\times5}{2\times11}$

Step 3

Now, for $\frac{2\times5}{2\times11}$ , we'll need to reduce it to the lowest terms. Like in the 2nd step, we'll divide factors that are in the numerator and denominator.

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3 years ago

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Answer:

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Step-by-step explanation:

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2 years ago

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3 years ago

What is the slope of the line passing through the points (2, -3) and (4, 3)? (Make sure to reduce your answer.)

Gnoma [55]

Answer:

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4 0

3 years ago

Read 2 more answers

Give a third degree polynomial that has zeros of 13, 5i, and −5i, and has a value of −680 when x=3. Write the polynomial in stan

Vikki [24]

Answer:

$\dfrac{17x^3-221x^2+425x-5525}{9}\\$

Step-by-step explanation:

hello

we need to find something like, a being real

$a(x-13)(x-5i)(x+5i)$

with a so that

$a(3-13)(3-5i)(3+5i)=-680\\\\a(-10)(3^2-(5i)^2)= -10(9+25)a=-360a=-680\\ a = \dfrac{680}{360}=\dfrac{17*40}{9*40}=\dfrac{17}{9}$

so the third degree polynomial that we are looking for is

$\dfrac{17}{9}(x-13)(x-5i)(x+5i)\\=\dfrac{17}{9}(x-13)(x^2+25)\\\\=\dfrac{17}{9}(x^3+25x-13x^2-13*25)\\\\\\=\dfrac{17}{9}(x^3-13x^2+25x-325)\\\\\\\\=\dfrac{17}{9}x^3-\dfrac{17*13}{9}x^2+\dfrac{17*25}{9}x-\dfrac{17*325}{9}\\=\dfrac{17}{9}x^3-\dfrac{221}{9}x^2+\dfrac{425}{9}x-\dfrac{5525}{9}\\$

hope this helps

3 0

3 years ago