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professor190 [17]

3 years ago

11

Bob sells gift cartons each December. They shipped a total of 3,300 cartons. If each carton sells for $28, how much money did Bo

b earn?

1 answer:

topjm [15]3 years ago

6 0

To solve this problem, all we need to do is multiply the amount charged for each carton by the amount of cartons that Bob shipped.

28 * 3300 = 92400

Bob earned $92,400 in sales from the gift cartons.
Hope that helped! =)

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Amy is using a drawing program to complete a construction with which she is almost finished. Which construction is she completin

Vedmedyk [2.9K]

Given options : Two intersecting circles are drawn with a radius in each marked. the image will be linked.

Given options : An equilateral triangle inscribed in a circle

A square inscribed in a circle

A regular pentagon inscribed in a circle

A regular hexagon inscribed in a circle.

<u>Note. When we join an intersection point of two circles and centers of the circles it would form an equilateral triangle that would be inscribe inside a common portion of both circles..</u>

Therefore, an equilateral triangle inscribed in a circle would be correct option.

She is completing an equilateral triangle inscribed in a circle.

4 0

3 years ago

Brennan has already run 16 miles on his own, and he expects to run 3 miles during each track practice. How many track practices

pentagon [3]

Answer:

It would take Brennan 4 track practice runs.

Step-by-step explanation:

He has already run 16 miles and needs to run 28 in total. If we are counting the 16 in the 28 miles, he would need to run only 12 more to reach his goal. And if he runs 3 miles a run, it would take him about 4 runs to finish his 28-mile goal.

(28-16=12 and 12/3=4)

8 0

3 years ago

Read 2 more answers

Find the range of the following piecewise function.

mash [69]

Answer:

$\huge\boxed{[-5;\ -2)\ \cup\ [3;\ 13)}$

Step-by-step explanation:

The following piecewise functions are linear functions. The graph of any of them is a line segment.

We just need to calculate the value of the function at each end specified in the brace.

$y=3x-2\ \text{if}\ -1\leq x$

Substitute x =-1 and x = 0:

$x=-1\\y=3(-1)-2=-3-2=-5\\\\x=0\\y=3(0)-2=0-2=-2$

Range of this piece is [-5; -2)

$y=2x+3\ \text{if}\ 0\leq x$

Substitute x =0and x = 5:

$x=0\\y=2(0)+3=0+3=3\\\\x=5\\y=2(5)+3=10+3=13$

Range of this piece is [3; 13)

Therefore the range of the following piecewise function is:

$[-5;\ -2)\ \cup\ [3;\ 13)$

Look at the picture.

6 0

3 years ago

1/3 of 9 is 3<br> How can you figure out these problems

gtnhenbr [62]

I don’t see the problems you want me to solve. Sorry

6 0

3 years ago

Drag the tiles to the correct boxes to complete the pairs not all tiles will be used match each quadratic graph to its respectiv

ch4aika [34]

Answer:

Part 1) The function of the First graph is $f(x)=(x-3)(x+1)$

Part 2) The function of the Second graph is $f(x)=-2(x-1)(x+3)$

Part 3) The function of the Third graph is $f(x)=0.5(x-6)(x+2)$

See the attached figure

Step-by-step explanation:

we know that

The quadratic equation in factored form is equal to

$f(x)=a(x-c)(x-d)$

where

a is the leading coefficient

c and d are the roots or zeros of the function

Part 1) First graph

we know that

The solutions or zeros of the first graph are

x=-1 and x=3

The parabola open up, so the leading coefficient a is positive

The function is equal to

$f(x)=a(x-3)(x+1)$

Find the value of the coefficient a

The vertex is equal to the point (1,-4)

substitute and solve for a

$-4=a(1-3)(1+1)$

$-4=a(-2)(2)$

$a=1$

therefore

The function is equal to

$f(x)=(x-3)(x+1)$

Part 2) Second graph

we know that

The solutions or zeros of the first graph are

x=-3 and x=1

The parabola open down, so the leading coefficient a is negative

The function is equal to

$f(x)=a(x-1)(x+3)$

Find the value of the coefficient a

The vertex is equal to the point (-1,8)

substitute and solve for a

$8=a(-1-1)(-1+3)$

$8=a(-2)(2)$

$a=-2$

therefore

The function is equal to

$f(x)=-2(x-1)(x+3)$

Part 3) Third graph

we know that

The solutions or zeros of the first graph are

x=-2 and x=6

The parabola open up, so the leading coefficient a is positive

The function is equal to

$f(x)=a(x-6)(x+2)$

Find the value of the coefficient a

The vertex is equal to the point (2,-8)

substitute and solve for a

$-8=a(2-6)(2+2)$

$-8=a(-4)(4)$

$a=0.5$

therefore

The function is equal to

$f(x)=0.5(x-6)(x+2)$

3 0

3 years ago