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-BARSIC- [3]
3 years ago
13

(x+3)(2x-1) whats the product im sooo confused

Mathematics
2 answers:
Elodia [21]3 years ago
8 0

Answer: 2x

​2

​​ +5x−3

Step-by-step explanation:

2x

​2

​​ −x+6x−3

2x

​2

​​ +(−x+6x)−3

trapecia [35]3 years ago
4 0

Answer:

2 {x}^{2}  + 5x - 3

Step-by-step explanation:

1) multiply the the parentheses

2) calculate the products

3)collect like terms

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Answer:

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Y' = (-1, -1)

Z' = (4, -8)

General rule: (x,y) = (y, -x)

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8 0
3 years ago
Group of answer choices<br><br> Phone A<br><br> Phone B<br><br> Phone C<br><br> Phone D
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Answer:

Phone B

Step-by-step explanation:

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3 years ago
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8. On a trip, a student drove 40 miles per hour for 2 hours and then drove 30 miles per hour for 3
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Write an appropriate direct variation equation if y = 45 when x = 5.
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3 years ago
Which of the following is a solution to 2cos2x − cos x − 1 = 0?
Pavel [41]

Answer:

Option A is correct.

Solution for the given equation is, x = 0^{\circ}

Step-by-step explanation:

Given that : 2\cos^2x -\cos x -1 =0

Let \cos x =y

then our equation become;

2y^2-y-1= 0           .....[1]

A quadratic equation is of the form:

ax^2+bx+c =0.....[2] where a, b and c are coefficient and the solution is given by;

x = \frac{-b\pm \sqrt{b^2-4ac}}{2a}

Comparing equation [1] and [2] we get;

a = 2 b = -1 and c =-1

then;

y = \frac{-(-1)\pm \sqrt{(-1)^2-4(2)(-1)}}{2(2)}

Simplify:

y = \frac{ 1 \pm \sqrt{1+8}}{4}

or

y = \frac{ 1 \pm \sqrt{9}}{4}

y = \frac{ 1 \pm 3}{4}

or

y = \frac{1+3}{4} and y = \frac{1 -3}{4}

Simplify:

y = 1 and y = -\frac{1}{2}

Substitute y = cos x we have;

\cos x = 1

⇒x = 0^{\circ}

and

\cos x = -\frac{1}{2}

⇒ x = 120^{\circ} \text{and} x = 240^{\circ}

The solution set:  \{0^{\circ}, 120^{\circ} , 240^{\circ}\}

Therefore, the solution for the given equation  2\cos^2x -\cos x -1 =0 is, 0^{\circ}





8 0
3 years ago
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