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3 years ago

Simplify (x − 4)(x2 + 3x + 5)

Mathematics

2 answers:

uranmaximum [27]3 years ago

7 0

Answer:

The correct option is d. x³ - x²- 7x - 20

Step-by-step explanation:

(x − 4)(x² + 3x + 5)

To simplify the expression given, we will follow the steps below;

First, we will open the bracket

x(x² + 3x + 5) -4(x² + 3x + 5)

x³ + 3x² + 5x -4x² - 12x - 20

Then, we can now proceed to simplify the values by adding the values with the like coefficient

x³ + 3x² + 5x -4x² - 12x - 20

x³ + 3x² -4x²+ 5x - 12x - 20

x³ - x²- 7x - 20

Simplifying (x − 4)(x2 + 3x + 5) will yield x³ - x²- 7x - 20

Therefore the correct option is d. x³ - x²- 7x - 20

morpeh [17]3 years ago

3 0

The answer for your question is letter D

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Step-by-step explanation:

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Find the distance from the point to the line. (-1,-2,1);x=4+4t, y=3+t, z=6-t .The distance is ____ Typn exact answer, using radi

lawyer [7]

Answer:

The distance is 4.726

Step-by-step explanation:

we need to find the distance from the point to the line

Given:- point (-1,-2,1) and line ; x=4+4t, y=3+t, z=6-t .

used formula $d=\frac{|a\times b|}{|a|}$

Let point P be (-1,-2,1)

using value t=0 and t=1

The point Q (4 , 3, 6) and R ( 8, 4, 5)

Let a be the vector from Q to R : a = < 8 - 4, 4 - 3, 5 - 6 > = < 4, 1, -1 >

Let b be the vector from Q to P: b = < -1 - 4, -2 - 3, 1 - 6> = < -5, -5, -5 >

The cross product of a and b is:

$a \times b= \begin{vmatrix} i & j & k\\ 4 &1&-1\\-5 &-5&-5\\ \end{vmatrix}$

= -6i+15j-15k

The distance is : $d=\frac{\sqrt{(-6)^{2}+(15)^{2}+(-15)^{2}}}{\sqrt{(4)^{2}+(1)^{2}+(-1)^{2}}}$

$=\frac{\sqrt{36+225+225}}{\sqrt{16+1+1}}$

$d=\frac{\sqrt{486}}{\sqrt{18}}$

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3 years ago

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Natali5045456 [20]

Step-by-step explanation:

∠2 and ∠7 are alternate exterior angles so they are equal.

So you can create the equation

40x-48 = 20x+52 Subtract 20x from both sides. Add 48 to both sides.

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if you need m∠2 Substtute 5 for x in 20x+52

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