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3 years ago

PLZZ I NEED THIS FAST!

Mathematics

2 answers:

Travka [436]3 years ago

7 0

For those just looking for the answer: D

The answer is D- 14 × (2 + 5), as 14 x 2 is 28 and 14 x 5 is 70. 2 and 5 are prime numbers with no common factors, so they can't be further simplified.

Burka [1]3 years ago

5 0

So remember that GCFs are greatest common factors, or the greatest number and/or variable that divides 2 or more terms nicely (aka no decimals).

With this, you at least know that a factor of these two numbers is 7, which turns this expression to 7(4 + 10). However, what's inside the parentheses can still be factored out with a GCF of 2. So you can further factor the expression as 2 * 7(2 + 5), which can be simplified to 14(2 + 5), or D. Since 2 and 5 do not have GCFs greater than 1, that is your answer.

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Enter your answer and show all the steps that you use to solve this problem in the space provided. A.Solve a–9=20 B.Solve b–9&gt

Goryan [66]

Answer:

A) The value of a is 29.

B) The value of b is greater than 29.

C) In both part A and part B we have used a common property which is addition property and that we have add 9 on both side of equation in both parts.

D) The value of a in part A is equal to 29 whereas in part B the value of b is greater than 29.

Step-by-step explanation:

Solving for Part A.

Given,

$a-9=20$

We have to solve for a.

$a-9=20$

By using addition property of equality, we will add both side by 9;

$a-9+9=20+9\\a=29$

Hence the value of a is 29.

Solving for Part B.

Given,

$b-9>20$

We have to solve for b.

$b-9>20$

By using addition property of inequality, we will add both side by 9;

$b-9+9>20+9\\b>29$

Hence the value of b is greater than 29.

Solving for Part C.

In both part A and part B we have used a common property which is addition property and that we have add 9 on both side of equation in both parts.

Solving for Part D.

The value of a in part A is equal to 29 whereas in part B the value of b is greater than 29.

5 0

3 years ago

Precal help if you could could you answer both. Me finishing this helps me graduate.

oee [108]

Exercise 1:

The easiest way to compute powers of complex numbers is to write them in the form

$z = \rho e^{i\theta} = \rho(\cos(\theta)+i\sin(\theta))$

In this form, you have

$z^n = \rho^n e^{in\theta} = \rho^n(\cos(n\theta)+i\sin(n\theta))$

The magnitude of the number is given by

$z=a+bi \implies \rho = \sqrt{a^2+b^2}$

So, we have

$z=-1+\sqrt{3}i \implies \rho = \sqrt{1+3}=2$

As for the angle, we have

$z=a+bi \implies \theta = \text{atan2}\dfrac{b}{a}$

So, we have

$z=-1+\sqrt{3}i \implies \theta = \text{atan2}\left(\dfrac{-\sqrt{3}}{-1}\right) = -\dfrac{2\pi}{3}$

Finally,

$z=-1-\sqrt{3}i = 2\left(\cos\left( -\dfrac{2\pi}{3}\right) + i\sin\left( -\dfrac{2\pi}{3}\right)\right) \implies z^6 = 2^6\left(\cos\left( -4\pi\right) + i\sin\left(-4\pi\right)\right) = 64$