Answer:
17.3 cm
Step-by-step explanation:
20² = 10² + x²
400 = 100 + x²
300 = x²
x = 17.3 cm
392.4 in2 should be correct
example of a net...
Answer:
1356cm3
Step-by-step explanation:
when the stone is dropped in to the water the new height is 17.4cm(14+3.4)
find the volume when the height is 14cm and when the height is 17.4cm
24*14*17.4=4716
24*14*14=3360
then subtract the two volumes
4716-3360=1356cm3

Solve the following using Substitution method
2x – 5y = -13
3x + 4y = 15


- To solve a pair of equations using substitution, first solve one of the equations for one of the variables. Then substitute the result for that variable in the other equation.

- Choose one of the equations and solve it for x by isolating x on the left-hand side of the equal sign. I'm choosing the 1st equation for now.

- Add 5y to both sides of the equation.


- Multiply
times 5y - 13.

- Substitute
for x in the other equation, 3x + 4y = 15.

- Multiply 3 times
.

- Add
to 4y.

- Add
to both sides of the equation.

- Divide both sides of the equation by 23/2, which is the same as multiplying both sides by the reciprocal of the fraction.

- Substitute 3 for y in
. Because the resulting equation contains only one variable, you can solve for x directly.


- Add
to
by finding a common denominator and adding the numerators. Then reduce the fraction to its lowest terms if possible.

- The system is now solved. The value of x & y will be 1 & 3 respectively.

Answer:
60 students passed, and 75 appeared in examination.
Step-by-step explanation:
Let's say s is the total number of students and p is the number of students who passed.
80% of the students passed, so:
0.8 s = p
If there were 10 less passers, and 15 less students (5 less failures), then the ratio of passers to failures would be 5/1.
(p − 10) / (s − p − 5) = 5 / 1
Simplify the second equation:
p − 10 = 5 (s − p − 5)
p − 10 = 5s − 5p − 25
6p = 5s − 15
Substitute the first equation.
6 (0.8 s) = 5s − 15
4.8 s = 5s − 15
0.2 s = 15
s = 75
p = 0.8 s
p = 60
60 students passed, and 75 appeared in examination.