Ask question

Ask question

All categories

3 years ago

9

David has 56 baseball card I wish he sells three cars per week write an equation representing the number of cards n he has left

at the end of any given week W

1 answer:

aev [14]3 years ago

7 0

Answer:

56-3=N

Step-by-step explanation:

You might be interested in

Use inductive reasoning to describe the pattern. Then find the next two numbers in the pattern. 1, 4, 7, 10, . . .

Colt1911 [192]

The reasoning would be that the pattern is going up by 3 every time. 13, 16

5 0

3 years ago

Emma was given a system of equations to solve by graphing. Which statement correctly identifies Emma’s error?

Alexxandr [17]

Answer:

line one should have a y-intercept at (0,2).

Step-by-step explanation:

8 0

3 years ago

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

8 to the power of 15 over 8 to the power 13 simplified

Aleksandr-060686 [28]

$\dfrac{8^{15}}{8^{13}}=8^{15-13}=8^2=64\\\\Used:\\\\a^n:a^m=\dfrac{a^n}{a^m}=a^{n-m}$

4 0

3 years ago

How would you change it so that it is not misleading?

Eduardwww [97]

Answer:

c. It appears that Mike's grade point average for the second quarter is 2 times his first quarter average. Change the intervals on the horizontal scale so they are consistent

Step-by-step explanation:

It looks like mikes GPA shot up and doubled but in reality it barely moved.y changing the horizontal scale it would show that his GPA only slightly Improved

8 0

3 years ago

Read 2 more answers