6. One variable only so pretty straightforward.
length-x+4
width-x
x+x+4=80
2x=76
x=38
x+4=42
answer: length 42cm and width 38cm
7. Another money problem!
n-# of nickels
q-# of quarters
n=3+q
0.05n+0.25q=2.85
Substitution works like a charm!
0.05(3+q)+0.25q=2.85
0.15+0.05q+0.25q=2.85
0.3q=2.7
q=9
n=3+q
n=3+9
n=12
answer: 9 nickels and 12 quarters
8. One variable situation again.
Ann's money-2b+9
Betty's money-b
b+2b+9=60
3b=51
b=17
2b+9=43
answer: Ann has $43 and Betty has $17.
9. # of red m&m's-x+1
# of blue m&m's-x
x+1+x=13
x=6
x+1=7
answer: 6 blue and 7 red m&m's
10. a-number of adult tickets
s-number of student tickets
a+s=785 ----> a=785-s
5a+2s=3280
5(785-s)+2s=3280
-3s=-645
s=215
a+s=785
a+215=785
a=570
answer: 215 children tickets and 570 adult tickets
If Sa=2πrh+2π
v=π
then the surface area is π
and volume is
(rh-2h)/2r.
Given Sa=2πrh+2π
=π.
We have to find surface area and volume from the given expression.
Volume is basically amount of substance a container can hold in its capacity.
First we will find the value of v from the expression. Because they are in equal to each other, we can easily find the value of v.
2πrh+2πv=πh
Keeping the term containing v at left side and take all other to right side.
2πv=π-2πrh
v=(πh-2πrh)/2π
v=π/2π-2πrh/2π
v=h/2-h/r
v=h(r-2)/2r
Put the value of v in Sa=2πrh+2π
Sa=2πrh+2π*h(r-2)/2r
=2πrh+2πrh(r-2)/2
=2πrh+πrh(r-2)
=2πrh+πh-2πrh
=πh
Hence surface area is πh and volume is h(r-2)/2.
Learn more about surface area at brainly.com/question/16519513
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the answer is D and i know cause i just did that and thats the answer
For simple integer factors as this one has, you want to find two values for the quadratic in the form ax^2+bx+c. Let the two values be j and k. These two values must satisfy two conditions.
jk=ac=10 and j+k=b=11, so j and k must be 1 and 10.
Now replace bx with jx and kx...
2x^2+x+10x+5 now factor 1st and 2nd pair of terms.
x(2x+1)+5(2x+1)
(x+5)(2x+1)
