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3 years ago

14

See has at most $150 to spend on restocking dolls and trains at her toy store. Dolls cost $7.50, and trains cost $5.00. Dee need

s no more than 10 trains, and at least 8 dolls. What is the system of inequality to represent this situation?

1 answer:

Burka [1]3 years ago

8 0

Answer:

Answer would be 30

Step-by-step explanation:

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An avid traveler is investigating whether travel websites differ in their pricing. She chooses a random sample of 32 hotel rooms

Tatiana [17]

The answer is D beachside trust

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2 years ago

X+3y=5<br> -2x-6y=-10<br><br> Solve using elimination

RSB [31]

Slimination
multiply first equaiton by 2
2x+6y=10
add to second equiaotn
2x-2x+6y-6y=10-10
0=0
this is true so therfor they are the same equaiton

so solve
x+3y=5
3y=-x+5
y=-1/3x+5/3

subsituter values for x and get valuse for y

if x=2 then y=1

3 0

3 years ago

2. How do you<br> multiply and divide<br> rational<br> expressions?

tatuchka [14]

Answer:

1.Factor both the numerator and the denominator.

2.Write as one fraction.

3: Simplify the rational expression.

4: Multiply any remaining factors in the numerator and/or denominator.

Step-by-step explanation:

7 0

3 years ago

Select all the expressions that are equivalent to 2•2•2•2•2•2

aliina [53]

Answer:

2^6, 64

Step-by-step explanation: The simplified version is two to the fifth power and the answer is 64. Those are the two I know

8 0

3 years ago

Read 2 more answers

A village fete has a children’s running race each year, run in heats of up to ten children. For each heat the first three contes

bekas [8.4K]

Answer: 1) 1/3,654 2) 3/406 3) 72,684,900,288,000 4) 120

<u>Step-by-step explanation:</u>

1) First and Second and Third

$\dfrac{3\ total\ prizes}{29\ total\ people}\times \dfrac{2\ remaining\ prizes}{28\ remaining\ people}\times \dfrac{1\ remaining\ prize}{27\ remaining\ people}=\dfrac{6}{21,924}\\\\\\.\qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \quad =\large\boxed{\dfrac{1}{3,654}}$

2) First and Second

$\dfrac{3\ total\ prizes}{29\ total\ people}\times \dfrac{2\ remaining\ prizes}{28\ remaining\ people}=\dfrac{6}{812}=\large\boxed{\dfrac{13}{406}}$

$3)\quad \dfrac{29!}{(29-10)!}=\large\boxed{72,684,900,288,000}$

$4)\quad _{10}C_3=\dfrac{10!}{3!(10-3)!}=\large\boxed{120}$

8 0

3 years ago