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3 years ago

Which expression shows the result of applying the distributive property to 9(2+5m)9(2+5m) ?

Mathematics

2 answers:

valina [46]3 years ago

4 0

aev [14]3 years ago

4 0

Answer:

18 + 45m

Step-by-step explanation:i took the test got it right

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5. Nathan can join a rock climbing club for $52 a month or a flat rate of

Trava [24]

Answer:

Solution: 9 or more months

Inequality: x > or = 9

Step-by-step explanation:

1. Divide 450 by 52 and you get 8.65...

2. Round up since you can't pay for half a month

4 0

3 years ago

A back contains 120 marbles some are red and the rest are black They are 19 red marbles for every black marble How many red marb

egoroff_w [7]

Answer:

6.3 ?

Step-by-step explanation:

4 0

3 years ago

Please answer ASAP ❤️Will earn 15 points (hope your answers are valid )

svetoff [14.1K]

Answer: The cost of one rose bush is $7 and the cost of one shrub is also $7

Step-by-step explanation:

The situtation can be represented by the systems of the equations.

10x + 4y = 98 x in this case is the cost of one rose bushes

9x + 9y = 126 y is the cost of one shrub.

Solve the system of equation using the elimination method.

10x +4y = 98

9x + 9y = 126 eliminate the y variable so you will have to multiply 9 on top and -4 down.

9(10x +4) = (98)(9)

-4(9x + 9y) = 126(-4)

You will now have the new two systems of equations

90x +36y = 882

-36x +-36y = -504 Now add the equations

0 + 54x = 378

54x = 378

x= 7

Now we know that the cost of one rose bush is 7 so we will plot it into one of the equations and solve for the cost of one shrub.

90(7) +36y=882

630 +36y = 882

-630 -630

36y = 252

y = 7

Check: 10(7) + 4(7)= 98

70 + 28 = 98

98= 98

so one rose bush is actually 7 dollars the same as 1 shrub.

5 0

3 years ago

Leah gets paid $450 a week +5% commission on her sales she hopes to earn at least $600 each week right in any quality to represe

ElenaW [278]

x = sales

450 + 0.05x = 600

5 0

3 years ago

in a AP the first term is 8,nth term is 33 and sum to first n terms is 123.Find n and common difference

allsm [11]

I believe there is no such AP...

Recursively, this sequence is supposed to be given by

$\begin{cases}a_1=8\\a_k=a_{k-1}+d&\text{for }k>1\end{cases}$

so that

$a_k=a_{k-1}+d=a_{k-2}+2d=\cdots=a_1+(k-1)d$

$a_n=a_1+(n-1)d$

$33=8+(n-1)d$

$21=(n-1)d$

$n$ has to be an integer, which means there are 4 possible cases.

Case 1: $n-1=1$ and $d=21$ . But

$\displaystyle\sum_{k=1}^2(8+21(k-1))=37\neq123$

Case 2: $n-1=21$ and $d=1$ . But

$\displaystyle\sum_{k=1}^{22}(8+1(k-1))=407\neq123$

Case 3: $n-1=3$ and $d=7$ . But

$\displaystyle\sum_{k=1}^4(8+7(k-1))=74\neq123$

Case 4: $n-1=7$ and $d=3$ . But

$\displaystyle\sum_{k=1}^8(8+3(k-1))=148\neq123$

8 0

3 years ago