Answer:
8 $10 notes
Step-by-step explanation:
I started by working it backwards. $93-$3 equals $90. So we have three of our 13 notes. We need ten more. Well the most $10 notes we can have is 8. if we had 8 $10 notes that would give us $80. $90-$80=10. We can take ten and divide it by five so we have 2 $5 notes.
So in total he would have 8 $10 notes, 2 $5 notes, and $3 one dollar notes which adds up to 13 notes
The answer is whole
Hope this helps
Answer:
1. ![2r+1.6b=10](https://tex.z-dn.net/?f=2r%2B1.6b%3D10)
2. ![r=5-0.8b](https://tex.z-dn.net/?f=r%3D5-0.8b)
3. ![b=6.25-1.25r](https://tex.z-dn.net/?f=b%3D6.25-1.25r)
Step-by-step explanation:
<h3> The complete exercise is: "Brown rice costs $2 per pound, and beans cost $1.60 per pound. Lin has $10 to spend on these items to make a large meal of beans and rice for a potluck dinner. Let b be the number of pounds of beans Lin buys and r be the number of pounds of rice she buys when she spends all her money on this meal.</h3><h3>1. Write an equation relating the two variables.</h3><h3>2. Rearrange the equation so "b" is the independent variable.</h3><h3>3. Rearrange the equation so "r" is the independent variable."</h3><h3 />
1. According to the information exercise given in the exercise, the cost per pounds of brown rice is $2. Since "r" represents the number of pounds of rice Lin buys, the total cost of the brown rice she buys can be represented with this expression:
![2r](https://tex.z-dn.net/?f=2r)
The beans cost $1.60 per pound. Since "b" represents the number of pounds of beans Lin buys, the total cost of beans she buys can be represented as:
![1.6b](https://tex.z-dn.net/?f=1.6b)
Knowing that she spends $10, you can write the following equation:
![2r+1.6b=10](https://tex.z-dn.net/?f=2r%2B1.6b%3D10)
2. In order to rearrange the equation so "b" is the independent variable, you need to solve for "r":
![2r+1.6b=10\\\\2r=10-1.6b\\\\r=\frac{10-1.6b}{2} \\\\r=5-0.8b](https://tex.z-dn.net/?f=2r%2B1.6b%3D10%5C%5C%5C%5C2r%3D10-1.6b%5C%5C%5C%5Cr%3D%5Cfrac%7B10-1.6b%7D%7B2%7D%20%5C%5C%5C%5Cr%3D5-0.8b)
3. To rearrange the equation so "r" is the independent variable, you must solve for "b". You get:
![2r+1.6b=10\\\\1.6b=10-2r\\\\b=\frac{10-2r}{1.6}\\\\b=6.25-1.25r](https://tex.z-dn.net/?f=2r%2B1.6b%3D10%5C%5C%5C%5C1.6b%3D10-2r%5C%5C%5C%5Cb%3D%5Cfrac%7B10-2r%7D%7B1.6%7D%5C%5C%5C%5Cb%3D6.25-1.25r)
Answer:
![p= \dfrac{1}{2}](https://tex.z-dn.net/?f=p%3D%20%5Cdfrac%7B1%7D%7B2%7D)
Step-by-step explanation:
Given equation:
![p^3=\dfrac{1}{8}](https://tex.z-dn.net/?f=p%5E3%3D%5Cdfrac%7B1%7D%7B8%7D)
Cube root both sides:
![\implies \sqrt[3]{p^3}= \sqrt[3]{\dfrac{1}{8}}](https://tex.z-dn.net/?f=%5Cimplies%20%5Csqrt%5B3%5D%7Bp%5E3%7D%3D%20%5Csqrt%5B3%5D%7B%5Cdfrac%7B1%7D%7B8%7D%7D)
![\implies p= \sqrt[3]{\dfrac{1}{8}}](https://tex.z-dn.net/?f=%5Cimplies%20p%3D%20%5Csqrt%5B3%5D%7B%5Cdfrac%7B1%7D%7B8%7D%7D)
![\textsf{Apply exponent rule} \quad \sqrt[n]{a}=a^{\frac{1}{n}}:](https://tex.z-dn.net/?f=%5Ctextsf%7BApply%20exponent%20rule%7D%20%5Cquad%20%5Csqrt%5Bn%5D%7Ba%7D%3Da%5E%7B%5Cfrac%7B1%7D%7Bn%7D%7D%3A)
![\implies p= \left(\dfrac{1}{8}\right)^{\frac{1}{3}}](https://tex.z-dn.net/?f=%5Cimplies%20p%3D%20%5Cleft%28%5Cdfrac%7B1%7D%7B8%7D%5Cright%29%5E%7B%5Cfrac%7B1%7D%7B3%7D%7D)
![\textsf{Apply exponent rule} \quad \left(\dfrac{a}{b}\right)^c=\dfrac{a^c}{b^c}:](https://tex.z-dn.net/?f=%5Ctextsf%7BApply%20exponent%20rule%7D%20%5Cquad%20%5Cleft%28%5Cdfrac%7Ba%7D%7Bb%7D%5Cright%29%5Ec%3D%5Cdfrac%7Ba%5Ec%7D%7Bb%5Ec%7D%3A)
![\implies p= \dfrac{1^{\frac{1}{3}}}{8^{\frac{1}{3}}}](https://tex.z-dn.net/?f=%5Cimplies%20p%3D%20%5Cdfrac%7B1%5E%7B%5Cfrac%7B1%7D%7B3%7D%7D%7D%7B8%5E%7B%5Cfrac%7B1%7D%7B3%7D%7D%7D)
![\textsf{Apply exponent rule} \quad 1^a=1:](https://tex.z-dn.net/?f=%5Ctextsf%7BApply%20exponent%20rule%7D%20%5Cquad%201%5Ea%3D1%3A)
![\implies p= \dfrac{1}{8^{\frac{1}{3}}}](https://tex.z-dn.net/?f=%5Cimplies%20p%3D%20%5Cdfrac%7B1%7D%7B8%5E%7B%5Cfrac%7B1%7D%7B3%7D%7D%7D)
Rewrite 8 as 2³:
![\implies p= \dfrac{1}{(2^3)^{\frac{1}{3}}}](https://tex.z-dn.net/?f=%5Cimplies%20p%3D%20%5Cdfrac%7B1%7D%7B%282%5E3%29%5E%7B%5Cfrac%7B1%7D%7B3%7D%7D%7D)
![\textsf{Apply exponent rule} \quad (a^b)^c=a^{bc}:](https://tex.z-dn.net/?f=%5Ctextsf%7BApply%20exponent%20rule%7D%20%5Cquad%20%28a%5Eb%29%5Ec%3Da%5E%7Bbc%7D%3A)
![\implies p= \dfrac{1}{2^{(3 \cdot \frac{1}{3})}}](https://tex.z-dn.net/?f=%5Cimplies%20p%3D%20%5Cdfrac%7B1%7D%7B2%5E%7B%283%20%5Ccdot%20%5Cfrac%7B1%7D%7B3%7D%29%7D%7D)
Simplify:
![\implies p= \dfrac{1}{2^{\frac{3}{3}}}](https://tex.z-dn.net/?f=%5Cimplies%20p%3D%20%5Cdfrac%7B1%7D%7B2%5E%7B%5Cfrac%7B3%7D%7B3%7D%7D%7D)
![\implies p= \dfrac{1}{2^{1}}](https://tex.z-dn.net/?f=%5Cimplies%20p%3D%20%5Cdfrac%7B1%7D%7B2%5E%7B1%7D%7D)
![\implies p= \dfrac{1}{2}](https://tex.z-dn.net/?f=%5Cimplies%20p%3D%20%5Cdfrac%7B1%7D%7B2%7D)
Answer:
h = 6
Step-by-step explanation:
Given he area of the banner expressed as;
A = ℎ(2ℎ−2)
h is the height of the banner
A is the area = 60
Substitute
60 = ℎ(2ℎ−2)
60 = 2h² - 2h
30 = h² - h
h²-h-30 = 0
Factorize;
h²-6h+5h-30 = 0
h(h-6)+5(h-6) = 0
(h-6)(h+5) = 0
h - 6 = 0 and h+5 = 0
h = 6 and -5
Since the height cannot be negative;
h = 6
Hence he height of the banner is 6