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3 years ago

Help! Math Homework.

Mathematics

1 answer:

Alenkinab [10]3 years ago

8 0

To solve problem 1 what you need to do is figure out how many miles the girls have walked so far and then subtract that from the total distance they must complete.

3/4-(3/10+1/4)

Step 1:Change the denominators so that way they are all the same

3/4=15/20
3/10=6/20 15/20-(6/20+5/20)
1/4=5/20

Step 2: Begin solving the problem

6/20+5/20=11/20
15/20-11/20=4/20

Step 3: Simplify your answer

4/20=1/5

Step 4 (is optional): Write your answer in a complete statement

The three girls walk 1/5mile together to school

Answer: A

I hope this is correct and if not I apologize for giving false information.

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A store is having a sale on walnuts and chocolate chips. For 7 pounds of walnuts and 9 pounds of chocolate chips, the total cost

Jlenok [28]

Answer:

Walnuts cost $1.75, chocolate chips cost $2.75

Step-by-step explanation:

<h3>Let the costs be:</h3>

Walnuts - x
Chocolate chips - y

<h3>Set equations as per question</h3>

For 7 pounds of walnuts and 9 pounds of chocolate chips, the total cost is $37:

7x + 9y = 37

For 5 pounds of walnuts and 3 pounds of chocolate chips, the total cost is $17:

5x + 3y = 17

<h3>Solve the system by elimination</h3>

Multiply the second equation by 3 and subtract the first equation, solve for x:

3(5x + 3y) - (7x + 9y) = 3(17) - 37
15x + 9y - 7x - 9y = 51 - 37
8x = 14
x = 14/8
x = 1.75

Find the value of y:

7*1.75 + 9y = 37
12.25 + 9y = 37
9y = 37 - 12.25
9y = 24.75
y = 24.75/9
y = 2.75

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2 years ago

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Kryger [21]

Answer:

Step-by-step explanation:

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2 years ago

How many millilitres in a centimeter

marta [7]

Answer:1 millilitres

Step-by-step explanation:

One millilitres equals one centimeters

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3 years ago

13. The least common multiple of two non-zero integers a and b is the unique positive integer m such that (i) m is a common mult

Vlad [161]

Answer:

[a,b] divides n

Step-by-step explanation:

Let us denote the least common multiple of a and b [a,b]=m.

We want to prove that m divides n, where n is a multiple of a and b.

We suppose m does not divide n, then by the Division Theorem, there exists q and r integers such that:

(1) ... n=mq+r, where 0<r<m

As n is a multiple of a and b, there exists s and t integers such that:

sa=n and tb=n

Same thing happens to m as it is the least common multiple, there exists u and v such that:

ua=m and vb=m

So (1) has the following form:

n=mq+r ⇒ sa=uaq+r ⇒sa-uaq=r⇒(s-uq)a=r and

n=mq+r ⇒ tb=vbq+r ⇒ tb-vbq=r⇒ (t-vq)b=r

So r is a multiple of a and b, but r<m which is a contradiction as, m is the least common multiple of a and b. So this concludes the proof.

So this means that $\frac{ab}{m}$ is and integer.

As m= vb, then $\frac{m}{b}$ is an integer, lets say $\frac{m}{b}$ =v; and as m=ua, then $\frac{m}{a}$ =u.

So $\frac{ab}{m}$ v= $\frac{ab}{m}$ $\frac{m}{b}$ =a, so $\frac{ab}{m}$ divides a; on the other hand, $\frac{ab}{m}$ u= $\frac{ab}{m}$ $\frac{m}{a}$ =b, so $\frac{ab}{m}$ divides b. From this we can conclude that $\frac{ab}{m}$ is a common divisor of a and b.

4 0

3 years ago

Expanded form for 67.943

ratelena [41]

67.943

= 60.000 + 7.000 + 0.900 + 0.040 + 0.003

6 0

3 years ago