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3 years ago

9

A used motorcycle is on sale for $3,600. Erik makes an offer equal to 3/4 of this price. How much does Erik offer for the motorc

ycle?

2 answers:

lesantik [10]3 years ago

4 0

3/4 of 3600..." of " means multiply
3/4 * 3600 =
10800/4 =
2700 <==

Elza [17]3 years ago

4 0

$3600x.75= $2700 Or leave it as a fraction

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If x+1 and x-1 are the factors of the polynomial ax^3 + x^2 - 2x +b,find the values of a and b

Nadusha1986 [10]

The polynomial remainder theorem says that dividing a polynomial $p(x)$ by $x-c$ leaves a remainder of $p(c)=0$ if $x-c$ is a factor of $p(x)$ . In this case, check $c=-1$ and $c=1$ .

$a(-1)^3+(-1)^2-2(-1)+b=-a+b+3=0$

$a(1)^3+(1)^2-2(1)+b=a+b-1=0$

From the first equation,

$-a+b+3=0\implies a=b+3$

and substituting into the second gives

$(b+3)+b-1=0\implies2b+2=0\implies b=-1\implies a=2$

3 0

3 years ago

Explain how to use the vertex and the value of “A” to determine the range of an absolute value function. PLEASE HELP!!

Thepotemich [5.8K]

Answer:

First, a absolute value function is something like:

y = f(x) = IxI

remember how this work:

if x ≥ 0, IxI = x

if x ≤ 0, IxI = -x

Notice that I0I = 0.

And the range of this function is all the possible values of y.

For example for the parent function IxI, the range will be all the positive reals and the zero.

First, if A is the value of the vertex of the absolute function, then we know that A is the maximum or the minimum value of the function.

Now, if the arms of the graph open up, then we know that A is the minimum of the function, and the range will be:

y ≥ A

Or all the real values equal to or larger than A.

if the arms of the graph open downwards, then A is the maximum of the function, and we have that the range is:

y ≤ A

Or "All the real values equal to or smaller than A"

5 0

3 years ago

41=12d-7 can someone help me?

OverLord2011 [107]

Answer:

d = 4

Step-by-step explanation:

41 = 12d - 7

48 = 12d

d = 4

6 0

4 years ago

Read 2 more answers

Charlie wants to join a golf club. Club A offers $80 for a new membership fee plus $45 per month. Club B offers $110 for a new m

xeze [42]

Ummm...i think it is 265 because i did it the easy way i added

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4 years ago

Read 2 more answers

Identify the interval on which the quadratic function is positive.

Alenkasestr [34]

Answer:

$\textsf{1. \quad Solution: $1 < x < 4$,\quad Interval notation: $(1, 4)$}$

$\textsf{2. \quad Solution: $-2 < x < 4$,\quad Interval notation: $(-2, 4)$}$

Step-by-step explanation:

<h3><u>Question 1</u></h3>

The intervals on which a <u>quadratic function</u> is positive are those intervals where the function is above the x-axis, i.e. where y > 0.

The zeros of the <u>quadratic function</u> are the points at which the parabola crosses the x-axis.

As the given <u>quadratic function</u> has a negative leading coefficient, the parabola opens downwards. Therefore, the interval on which y > 0 is between the zeros.

To find the zeros of the given <u>quadratic function</u>, substitute y = 0 and factor:

$\begin{aligned}y&= 0\\\implies -7x^2+35x-28& = 0\\-7(x^2-5x+4)& = 0\\x^2-5x+4& = 0\\x^2-x-4x+4& = 0\\x(x-1)-4(x-1)&= 0\\(x-1)(x-4)& = 0\end{aligned}$

Apply the <u>zero-product property</u>:

$\implies x-1=0 \implies x=1$

$\implies x-4=0 \implies x=4$

Therefore, the interval on which the function is positive is:

Solution: 1 < x < 4
Interval notation: (1, 4)

<h3><u>Question 2</u></h3>

The intervals on which a <u>quadratic function</u> is negative are those intervals where the function is below the x-axis, i.e. where y < 0.

The zeros of the <u>quadratic function</u> are the points at which the parabola crosses the x-axis.

As the given <u>quadratic function</u> has a positive leading coefficient, the parabola opens upwards. Therefore, the interval on which y < 0 is between the zeros.

To find the zeros of the given <u>quadratic function</u>, substitute y = 0 and factor:

$\begin{aligned}y&= 0\\\implies 2x^2-4x-16& = 0\\2(x^2-2x-8)& = 0\\x^2-2x-8& = 0\\x^2-4x+2-8& = 0\\x(x-4)+2(x-4)&= 0\\(x+2)(x-4)& = 0\end{aligned}$

Apply the <u>zero-product property</u>:

$\implies x+2=0 \implies x=-2$

$\implies x-4=0 \implies x=4$

Therefore, the interval on which the function is negative is:

Solution: -2 < x < 4
Interval notation: (-2, 4)

3 0

1 year ago