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3 years ago

Find the product: 9x(x3-4x2-3x)

Mathematics

2 answers:

nekit [7.7K]3 years ago

8 0

Answer:

9x^4 - 36x^3 - 27x^2

Step-by-step explanation:

9x(x^3 - 4x^2 - 3x)

Multiply 9x to each of the terms in the parentheses:

9x^4 - 36x^3 - 27x^2

Andre45 [30]3 years ago

7 0

Answer:

9x^4-36x^3-27x^2

Step-by-step explanation:

assuming those are exponents

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3 years ago

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What is -3x-32=-2(5-4x)

san4es73 [151]

Answer:

x=−2

Step-by-step explanation:

Step 1: Simplify both sides of the equation.

−3x−32=−2(5−4x)

−3x+−32=(−2)(5)+(−2)(−4x)(Distribute)

−3x+−32=−10+8x

−3x−32=8x−10

Step 2: Subtract 8x from both sides.

−3x−32−8x=8x−10−8x

−11x−32=−10

Step 3: Add 32 to both sides.

−11x−32+32=−10+32

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Step 4: Divide both sides by -11.

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Determine whether the statement is always, sometimes, or never true. Complete the explanation. In parallelogram MNPQ, the diagon

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This is never true. In a parallelogram, the diagonals will always bisect each other. Thus, each of these segments would always have to be equal.

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3 years ago

Would appreciate the help !

aleksandr82 [10.1K]

This is one pathway to prove the identity.

Part 1

$\frac{\sin(\theta)}{1-\cos(\theta)}-\frac{1}{\tan(\theta)} = \frac{1}{\sin(\theta)}\\\\\frac{\sin(\theta)}{1-\cos(\theta)}-\cot(\theta) = \frac{1}{\sin(\theta)}\\\\\frac{\sin(\theta)}{1-\cos(\theta)}-\frac{\cos(\theta)}{\sin(\theta)} = \frac{1}{\sin(\theta)}\\\\\frac{\sin(\theta)*\sin(\theta)}{\sin(\theta)(1-\cos(\theta))}-\frac{\cos(\theta)(1-\cos(\theta))}{\sin(\theta)(1-\cos(\theta))} = \frac{1}{\sin(\theta)}\\\\$

Part 2

$\frac{\sin^2(\theta)}{\sin(\theta)(1-\cos(\theta))}-\frac{\cos(\theta)-\cos^2(\theta)}{\sin(\theta)(1-\cos(\theta))} = \frac{1}{\sin(\theta)}\\\\\frac{\sin^2(\theta)-(\cos(\theta)-\cos^2(\theta))}{\sin(\theta)(1-\cos(\theta))} = \frac{1}{\sin(\theta)}\\\\\frac{\sin^2(\theta)-\cos(\theta)+\cos^2(\theta)}{\sin(\theta)(1-\cos(\theta))} = \frac{1}{\sin(\theta)}\\\\$

Part 3

$\frac{\sin^2(\theta)+\cos^2(\theta)-\cos(\theta)}{\sin(\theta)(1-\cos(\theta))} = \frac{1}{\sin(\theta)}\\\\\frac{1-\cos(\theta)}{\sin(\theta)(1-\cos(\theta))} = \frac{1}{\sin(\theta)}\\\\\frac{1}{\sin(\theta)} = \frac{1}{\sin(\theta)} \ \ {\checkmark}\\\\$

As the steps above show, the goal is to get both sides be the same identical expression. You should only work with one side to transform it into the other. In this case, the left side transforms while the right side stays fixed the entire time. The general rule is that you should convert the more complicated expression into a simpler form.

We use other previously established or proven trig identities to work through the steps. For example, I used the pythagorean identity $\sin^2(\theta)+\cos^2(\theta) = 1$ in the second to last step. I broke the steps into three parts to hopefully make it more manageable.

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3 years ago

Need some help, kinda stuck

ElenaW [278]

Answer:

7/4

Step-by-step explanation:

$\displaystyle\frac{2 + \sqrt{ - 3} }{2} ( \frac{2 - \sqrt{ - 3} }{2} )~~~~~~$

Evaluate.

Solution:

Rewrite it as,

$\displaystyle\frac{2 + \sqrt{ - 1 } \sqrt{ 3} }{2} ( \frac{2 - \sqrt{ - 1} \sqrt{3} }{2} )$
$\displaystyle\frac{2 + i\sqrt{ 3} }{2} ( \frac{2 - i\sqrt{ 3} }{2} )~~~~~~$