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3 years ago

- 3(4 - 9) Evaluate.

Mathematics

2 answers:

WARRIOR [948]3 years ago

8 0

-3(4-9)

Solve what is inside the parenthesis first:

4-9 = -5

-3 (-5)

(-) x (-) = (+)

snow_lady [41]3 years ago

4 0

Answer:

-15

Step-by-step explanation:

4-9=-5 -5×-3=15

negative times negative is positive

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Please help me, so I can help my son

Grace [21]

1:a would be 3. Subtract 9 & 6 to find 3. B= 7. Again 9-2 would help you find the answer. So then it would be 3+6=7+2.
2: 2+8 gets you 10. 10+5 gets you 15. 8+5 gets you 13. 2+13 gets you 15. So it’s 15=15.
3:7 needs to be changed into a fraction so, 7/1(1/7). Multiply across, you get 7/7, which gives you 1, so it’s 1=1.
4:I’m sorry I don’t know how to do this one
5:you multiply 3 and 6x, which gives you 18x. Then multiply 3 and 9y, and you get 27y. So then you have 18x+27y=18x+27y.
6: 15+0 gets you 15. So it’s 15=15.
Hope this helped. I’m not the best at explaining

4 0

4 years ago

Please help me again

WARRIOR [948]

X=46

Explanation:
You just have to make 3x-28 and x+64 equal to each other and x would be 46.

m∠1=44

Explanation:
You just have to subtract the 46 that u solved earlier from 90 degree so it’s 44

7 0

3 years ago

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.