Please help really confused on all

Mathematics

1 answer:

ollegr [7]3 years ago

5 0

$\left \{ {{x - y = 1.50} \atop {40x + 40y = 940}} \right. 
 \\ \\
*x - y = 1.50$

x = 1.50 + y

*40x + 40y = 940
40 (1.50 + y) + 40y = 940
60 + 40y +40y = 940
60 + 80y = 940
80y = 940 - 60
80y = 880
y = 880/80
y = 11

*x - y = 1.50
x - 11 = 1.50
x = 1.50 + 11
x = 12.5

x=DAVID
Y=PETER

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dalvyx [7]

Answer:

$\boxed{\sf \ \ \ m = -\dfrac{tn}{t-s} \ \ \ }$

Step-by-step explanation:

Hello,

let s assume that m+n is different from 0

we have this equation and we need to find m as a function of t, s, and n

$t=\dfrac{ms}{m+n}$

<=>

$(m+n)*t=ms\\\\ tm+tn=sm\\ (t-s)m = -tn\\ m = -\dfrac{tn}{t-s}$

for t-s different from 0, so t different from s

hope this helps

4 0

3 years ago

HELP ME PLEASE!! ASAP!!

Levart [38]

The total area is given by:
A1 = (300) * (500)
A1 = 150000 feet ^ 2
The area occupied by 5 people is:
A2 = (5) * (5)
A2 = 25 feet ^ 2
Then, we can make the following rule of three:
5 --------> 25
x --------> 150000
Clearing x we have:
x = (150000/25) * (5)
x = 30000
Answer:
D. 30,000 people

3 0

4 years ago

Please help me with these. God bless you

Digiron [165]

Counting all of the lines, there is About 17

6 0

2 years ago

Read 2 more answers

How many pounds of cashew nuts worth 75 cents a pound must be mixed with 10 pounds of pecans worth $1.50 a pound to make a mixtu

Trava [24]

Let c represent the weight of cashews and p the weight of pecans.

Then c + 10 = total weight of the nut mixture.

An equation for the value of the mixture follows:

$1.50(10 lb) + $0.75c = (c+10)($1.00)

Solve this equation for c: 15 + .75c = c + 10. Subtract .75c from both sides:

15 = 1c - 0.75c + 10. Then 5=0.25c, and c = 5/0.25, or 20.

Need 20 lb of cashews.

Check: the pecans weigh 10 lb and are worth $1.50 per lb, so the total value of the pecans is $15. The total value of the cashews is (20 lb)($0.75/lb), or $15. Does (20 lb + 10 lb)($1/lb) = $15 + $15? Yes. So c= 20 lb is correct.

3 0

3 years ago

One container is filled with mixture that is 25% acid. A second container is filled with a mixture that is 55% acid. The second

algol13

Answer:

Therefore the third container contains $\frac{735}{17}\%$ = 43.23 % acid.