So... the radiator has 15 liters of 70% antifreeze.. but needs an 80% antifreeze
well, so, you need to drain some and put some with higher percentage, seems to be, you will end up at the same 15 liters, possible the radiator's capacity, of 80% antifreeze
so, the same amount going out, of 70% is the same amount going in, of 100% antifreeze
now.. let's use the decimal format for the percents, or 70% is 70/100 or 0.7 and so on
![\bf \begin{array}{lccclll} &amount&concentration& \begin{array}{llll} concentration\\ amount \end{array}\\ &-----&-------&-------\\ \textit{70\% sol'n, out}&x&0.7&0.7x\\ \textit{100\% sol'n, in}&x&1.00&x\\ \textit{current 70\% sol'n}&15&0.7&10.5\\ -----&-----&-------&-------\\ mixture&15&0.8&12 \end{array}](https://tex.z-dn.net/?f=%5Cbf%20%5Cbegin%7Barray%7D%7Blccclll%7D%0A%26amount%26concentration%26%0A%5Cbegin%7Barray%7D%7Bllll%7D%0Aconcentration%5C%5C%0Aamount%0A%5Cend%7Barray%7D%5C%5C%0A%26-----%26-------%26-------%5C%5C%0A%5Ctextit%7B70%5C%25%20sol%27n%2C%20out%7D%26x%260.7%260.7x%5C%5C%0A%5Ctextit%7B100%5C%25%20sol%27n%2C%20in%7D%26x%261.00%26x%5C%5C%0A%5Ctextit%7Bcurrent%2070%5C%25%20sol%27n%7D%2615%260.7%2610.5%5C%5C%0A-----%26-----%26-------%26-------%5C%5C%0Amixture%2615%260.8%2612%0A%5Cend%7Barray%7D)
so.. let's subtract, from the current solution, 0.7x and add 1x or x, our antifreeze concentration amount, should be 12 though
10.5 - 0.7x + x = 12
solve for "x"
Answer:
1. A number decreased by seven.
2. A number increased by nine.
3. The product of six and a number, divided by eight.
4. The product of three and a number increased by five is equal to twenty.
5. The product of two and a number increased by eight is equal to twenty-three.
Step-by-step explanation:
1. x-7
A number decreased by seven.
2. y+9
A number increased by nine.
3. 6m/8
The product of six and a number, divided by eight.
4. 3y+5 =20
The product of three and a number increased by five is equal to twenty.
5. 2(x+8)= 23
The product of two and a number increased by eight is equal to twenty-three.
Answer:
Andre's height: y < x < 164
Step-by-step explanation:
We will first of all convert each of the statements to mathematical inequalities, before we combine them.
Let Andre's height be = x
Lin's height be = y
The maths teacher's height = 164
Andre is a little taller than Lin:
x > y
Andre is shorter than their maths teacher, who is 164 centimetres
x<164
Combining the two inequalities, we have Andre's height as
y < x < 164
This means that Lin is shorter than Andre who is in turn shorter than the Maths Teacher.
Answer:
181 mi^2
Step-by-step explanation:
Dimensions of large rectangle, on the left, are 6 mi by 8 m; area is 48 m^2;
Dimensions of small rectangle, in the middle of the diagram, are 4 mi by 5 mi, or 20 mi^2; area is 20 mi^2.
Dimensions of circle: r = 12 mi; area of quarter circle is (1/4)(3.14)(12 mi)^2, or 113 mi^2.
Total area is 113 mi^2 + 20 mi^2 + 48 mi^2, or 181 mi^2