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2 years ago

Find all cube roots of -1000i

Mathematics

1 answer:

NemiM [27]2 years ago

3 0

Answer:

10i

Step-by-step explanation:

Since i^2 = -1, and 1000 = 10^3, we can instinctively guess that the answer is 10i. We can validate this by cubing it, yielding 1000i^3, or -1000i.

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What is 7√11√ in simplest radical form? Enter your answer in the box.

yaroslaw [1]

23.21 is the answer i believe

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3 years ago

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An expression that is equivalent to (-8a-1)-(-7a+5)

horrorfan [7]

Answer:

Step-by-step explanation:

(-8a - 1)- (-7a + 5) = -8a - 1 - 7a*(-1) + 5*(-1) { (-1) is disturbed to all the terms in (-7a +5)}

= -8a - 1 + 7a - 5

= -8a + 7a -1 - 5 {Combine like terms}

= -a - 6

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3 years ago

The swim team has 41 people on their team they need new swim gear the swim caps cost $5 and the swim suit cost $ 16 dollars how

IRISSAK [1]

Answer:

The money they will have to pay is $861.

Step-by-step explanation:

Given:

The swim team has 41 people on their team.

They need new swim gear the swim caps cost $5 and the swim suit cost $16.

Now, to find the money they will have to pay.

Number of people on the swim team = 41.

Cost of swim caps = $5.

Cost of swim suit = $16.

Now, to get the total money they will have to pay:

Number of people on the swim team× (cost of swim caps + cost of swim suit)

$=41\times (5 +16)$

$=41\times 21$

$=\$861.$

Therefore, the money they will have to pay is $861.

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3 years ago

For one month Siera calculated her home town’s average high temperature in degrees Fahrenheit. She wants to convert that tempera

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C(F) is the degrees Celsius given the degrees Fahrenheit.

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3 years ago

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Solve the following equation by completing the square. 3x^2-3x-5=13

mr Goodwill [35]

we'll start off by grouping some

$\bf 3x^2-3x-5=13\implies (3x^2-3x)-5=13\implies 3(x^2-x)-5=13 \\\\\\ 3(x^2-x)=18\implies (x^2-x)=\cfrac{18}{3}\implies (x^2-x)=6\implies (x^2-x+~?^2)=6$

so we have a missing guy at the end in order to get the a perfect square trinomial from that group, hmmm, what is it anyway?

well, let's recall that a perfect square trinomial is

$\bf \qquad \textit{perfect square trinomial} \\\\ (a\pm b)^2\implies a^2\pm \stackrel{\stackrel{\text{\small 2}\cdot \sqrt{\textit{\small a}^2}\cdot \sqrt{\textit{\small b}^2}}{\downarrow }}{2ab} + b^2$

so we know that the middle term in the trinomial, is really 2 times the other two without the exponent, well, in our case, the middle term is just "x", well is really -x, but we'll add the minus later, we only use the positive coefficient and variable, so we'll use "x" to find the last term.

$\bf \stackrel{\textit{middle term}}{2(x)(?)}=\stackrel{\textit{middle term}}{x}\implies ?=\cfrac{x}{2x}\implies ?=\cfrac{1}{2}$

so, there's our fellow, however, let's recall that all we're doing is borrowing from our very good friend Mr Zero, 0, so if we add (1/2)², we also have to subtract (1/2)²

$\bf \left( x^2 -x +\left[ \cfrac{1}{2} \right]^2-\left[ \cfrac{1}{2} \right]^2 \right)=6\implies \left( x^2 -x +\left[ \cfrac{1}{2} \right]^2 \right)-\left[ \cfrac{1}{2} \right]^2=6 \\\\\\ \left(x-\cfrac{1}{2} \right)^2=6+\cfrac{1}{4}\implies \left(x-\cfrac{1}{2} \right)^2=\cfrac{25}{4}\implies x-\cfrac{1}{2}=\sqrt{\cfrac{25}{4}} \\\\\\ x-\cfrac{1}{2}=\cfrac{\sqrt{25}}{\sqrt{4}}\implies x-\cfrac{1}{2}=\cfrac{5}{2}\implies x=\cfrac{5}{2}+\cfrac{1}{2}\implies x=\cfrac{6}{2}\implies \boxed{x=3}$

6 0

3 years ago