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adell [148]
3 years ago
14

Dan counted all the coins in his bank, and he had 72 quarters. can he exchange the quarters for an even amount of dollar bills?

how do you know?
Mathematics
1 answer:
Vladimir [108]3 years ago
8 0
There are 4 quarters in a dollar. If you divide 72 by 4 and you do not get a remainder then he can.
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Fill in the blank to make this statement true: (0,0) is a solution to _________ equation(s) in the form y=kx. no one some all
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Answer:

All

Explanation:

When an equation is in the form of y=kx, it displays direct variation, meaning that the y-intercept is 0. For example, take an equation in slope-intercept form, it has an equation of y=mx+b. Since there is no "b", in the form y=kx, that means that the line must pass through the origin, or (0,0)

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A restaurant bill is $75. You want to leave about a 20% tip. Which is a 20% tip?
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75*.20= 15

The tip is $15
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Write in scientific notation 0.00042 urgent !!!!
Black_prince [1.1K]

42 \cdot 10^{-4}

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Help pls I'll give 25 points
RideAnS [48]

Answer:

\text{C. }60

Step-by-step explanation:

<u>Question from image</u>:

"The digits 1, 2, 3, 4, and 5 can be formed to arrange 120 different numbers. How many of these numbers will have the digits 1 and 2 in increasing order? For example, 14352 and 51234 are two such numbers."

Let's start by taking a look with the number 1. There are four possible places 1 could be, because there needs to be space for the 2 after it.

Checkmarks mark where the 1 can be, the <em>x</em> marks where it cannot be.

\underline{\checkmark}\:\underline{\checkmark}\:\underline{\checkmark}\:\underline{\checkmark}\:\underline{X}

Let's start with the first position:

\underline{\checkmark}\:\underline{\#}\:\underline{\#}\:\underline{\#}\:\underline{\#}

There are four places the 2 can be. For each of these four places, we can arrange the remaining 3 digits in 3!=6 ways. Therefore, there are 4\cdot 6=\boxed{24} possible numbers when 1 is the first digit of the number.

Continue this process with the remaining possible positions for 1.

Second position:

\underline{X}\underline{\checkmark}\:\underline{\#}\:\underline{\#}\:\underline{\#}

There are three places the 2 can be, since the 2 must be behind the 1. For each of these three places, the remaining 3 digits can be arranged in 3!=6 ways. Therefore, there are 3\cdot 6=\boxed{18} possible numbers when 1 is the second digit of the number.

The pattern continues. Next there will be 2 places to place the 2. For each of these, there are 3!=6 ways to rearrange the remaining 3 digits for a total of 2\cdot 6=\boxed{12} possible numbers when 1 is the third digit of the number.

Lastly, when 1 is the fourth digit of the number, there is only 1 place the 2 can be. For this one place, there are still 3!=6 ways to rearrange the remaining three numbers. Therefore, there are 1\cdot 6=\boxed{6} possible numbers when 1 is the fourth digit of the number.

Thus, there are 24+18+12+6=\boxed{60} numbers that will have the digits 1 and 2 in increasing order, from a set of 120 five-digit numbers created by the digits 1 through 5, where no digit may be repeated.

5 0
3 years ago
(HELP ASAP!)<br>Which postulate or theorem proves △HJZ∼△WJR ?
Tomtit [17]

Answer:

SAS Similarity Theorem

Step-by-step explanation:

we know that

<u>SAS Similarity Theorem</u>: States that if two sides in one triangle are proportional to two sides in another triangle and the included angle in both are congruent, then the two triangles are similar

In this problem

∠HJZ=∠RJW ----> by vertical angles

and

WJ/HJ=JR/JZ

substitute the values

8/4=6/3

2=2 ----> is true

so

Two sides of triangle HJZ are proportional with the two corresponding sides of triangle WJR and the included angle is congruent

therefore

Triangles are similar by SAS Similarity Theorem

8 0
4 years ago
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