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3 years ago

I WILL GIVE BRAINLIEST!!

Mathematics

1 answer:

irinina [24]3 years ago

6 0

Answer:

The are 8 dimes and 12 nickels

Step-by-step explanation:

1.We set n+d=20 because 20 is the total number of dimes and nickels in the jar.We have $0.05 n because for ever nickel,we earn $0.05.Adding this to the value of dimes,represented similarly by $0.1,then we set this to the amount of money we have in total,which is $1.40.

2.Subtract d from both sides of the first equation to get it by itself.Substitute-d for n in the second equation.The divide both sides by 0.5 to obtain the answer.

3.Substitute 8 in for d to obtain the value or n.

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In the figure, m ║ n and p is a transversal. Which of the following are alternate interior angles?

Lyrx [107]

The answer is c. Because there inside and opposite

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3 years ago

FUNDRAISING A school is raising money by selling calendars for $20 each. Mrs. Hawkins promised a party to whichever of her Engli

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Rj, this is the solution to question 19:

• 1st period class sold 60 calendars in total

• 2nd period class sold 123 calendars in total

We calculate the average per week, this way:

• 1st period class sold 60/4 = 15 calendars per week

• 2nd period class sold 123/4 = 30.75 calendars per week

In consequence the multiplication equation that represents this situation is:

15w + 30.75w = 183, where w is the number of weeks Mrs. Hawkins' classes sold calendars.

This is the solution to question 20:

We already know the average number of calendars that 1st and 2nd period classes sold per week. Now, let's calculate the average for 3nd 4th peiod classes, this way:

• 3rd period class sold 89 calendars in total

• 4th period class sold 126 calendars in total

Therefore, the average is:

• 3rd period class sold 89/4 = 22.25 calendars per week

• 4th period class sold 126/4 = 31.5 calendars per week

Summarizing, we have:

15 + 30.75 + 22.25 + 31.5 = 99.5 calendars was the average number sold by all her classes in a week.

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7 0

1 year ago

A manufacturer cuts squares from the corners of a rectangular plece of sheet metal that measures 2 inches by 7 inches (see Figur

Vladimir79 [104]

Answer:

The value of x that maximizes the volume enclosed by this box is 0.46 inches

The maximum volume is 3.02 cubic inches

Step-by-step explanation:

see the attached figure to better understand the problem

we know that

The volume of the open-topped box is equal to

$V=LWH$

where

$L=(7-2x)\ in\\W=(2-2x)\ in\\H=x\ in$

substitute

$V=(7-2x)(2-2x)x$

Convert to expanded form

$V=(7-2x)(2-2x)x\\V=(14-14x-4x+4x^{2})x\\V=14x-14x^2-4x^2+4x^{3}\\V=4x^{3}-18x^{2} +14x$

using a graphing tool

Graph the cubic equation

Remember that

The domain for x is the interval -----> (0,1)

Because

If x>1

then

the width is negative (W=2-2x)

The maximum is the point (0.46,3.02)

see the attached figure

therefore

The value of x that maximizes the volume enclosed by this box is 0.46 inches

The maximum volume is 3.02 cubic inches

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3 years ago

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The Answer to this question is:

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3 years ago

Read 2 more answers

Please come and help really quick please thank you

professor190 [17]

Answer:

1: 6

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Step-by-step explanation:

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3 years ago