Question

Hi. This is Acacia's little sister. I have 5 math problems I need help with. If you could help that would be great. It is beginner 8th grade work.

Answer 1

i can help

Answer 2

Sure :)

Answer 3

Write a variable expression for the word phrase.
"The product of 3 and a number decreased by 7"
a: 3n + 7
b: 7- 3n
c: 7n - 3
d: 3n - 7

Answer 4

i believe its d

Answer 5

So, when they say product of 3 and a number, the mean 3 times a number. Produce = Times. So we know its \(3\times n\), or \(3n\) for the first bit

Then it says it is decreased by \(7\), that means \(3n\) is decreased, or minus \(7\). That is equal to: \(3n-7\), which is \(D\)

Answer 6

When they say decreased, in past tense, then it means \(3n\) minus \(7\). Now, if it sayd 3n decreases 7, then it would be \(7-3n\)

Answer 7

Select the mathematical sentence that is true
a: 3 times (7 - 3) + 5 < 6 + 2 times 1
b: 3 times (7 - 3) + (< with line under it) 6 + 2 times 1
c: 3 times (7 - 3) + 5 = 6+ 2 times 1
d: 3 times (7 - 3) + 5 > 6 + 2 times 1

Answer 8

Simply solve each one, see if it is right. I will help with the first one
3 times (7 - 3) + 5 < 6 + 2 times 1
Turn that into equation form:
\[3\times(7-3)+5<6+2\times1\]First, remove the \(\times\) sign for the brackets
\[3\times(7-3)+5<6+2\times1\]\[3(7-3)+5<6+2\times1\]Then simply solve using BOMDAS, or BODMAS
\[3(7-3)+5<6+2\times1\]
Brackets first
\[3(4)+5<6+2\times1\]\[12+5<6+2\times1\]
Then the multiply sign
\[12+5<6+2\times1\]\[12+5<6+2\]
Add the numbers up
\[12+5<6+2\]\[17<6+2\]\[17<8\]
I do not believe \(17<8\), so that would be wrong. Just keep going through each one until you get one that makes sense.

Be careful for the \(\le\) sign, since it means LESS THAT or EQUAL TO, not just LESS THAT, like <

Answer 9

Find the solution set for the given inequality if the replacement set for x is the set {1, 2, 3, 4}.
4x - 3 (> with under line) 5
a: 1,2
b: 3,4
c: 2,3,4
d: 1,2,3,4

Answer 10

Just change \(x\) to 1, then 2, then 3 and lastly 4, and see if the equation \(4x-3\ge5\) makes sense. Let me help you with 1 and 2

\(4x-3\ge5\)
Sub 1 into \(x\)
\(4\times 1-3\ge5\)
\(4-3\ge5\)
\(1\ge5\)
1 is not Greater than or equal to 5, so it cannot be a posibility

\(4x-3\ge5\)
Sub 2into \(x\)
\(4\times 2-3\ge5\)
\(8-3\ge5\)
\(5\ge5\)
5 is Greater than or equal to 5, so it is a possibility

Answer 11

Thank you.

Answer 12

No problem @Kat_Caca_1996. If you thinked I helped, please consider pressing the best response button. Thank you :)