we know that
Perimeter of a triangle is equal to
P=a+b+c
where
a,b and c are the length sides of the triangle
<u>Find the perimeter of the triangles</u>
<u>Triangle N 1</u>
P=(x-2)+(x)+(3x+1)-------> P=5x-1
<u>Triangle N 2</u>
P=(2x-5)+(x+4)+(6x-7)------> P=9x-8
equate the perimeters
5x-1=9x-8--------> 9x-5x=-1+8-------> 4x=7
x=7/4------> x=1.75
therefore
the answer is
x=1.75
Answer:
f(x) = 4.35 +3.95·sin(πx/12)
Step-by-step explanation:
For problems of this sort, a sine function is used that is of the form ...
f(x) = A + Bsin(2πx/P)
where A is the average or middle value of the oscillation, B is the one-sided amplitude, P is the period in the same units as x.
It is rare that a tide function has a period (P) of 24 hours, but we'll use that value since the problem statement requires it. The value of A is the middle value of the oscillation, 4.35 ft in this problem. The value of B is the amplitude, given as 8.3 ft -4.35 ft = 3.95 ft. Putting these values into the form gives ...
f(x) = 4.35 + 3.95·sin(2πx/24)
The argument of the sine function can be simplified to πx/12, as in the Answer, above.
Answer:
The answer to 75 - ( 8 + 45 ÷ 3 ) × 2= 29
Answer:
Uche's pumping rate is <u>300 mL/s</u>.
Step-by-step explanation:
Given:
Uche pumps gasoline at a rate of 18 L/min.
Now, to find Uche's pumping rate in mL/s.
The rate at which Uche pumps gasoline = 18 L/min.
So, to get the pumping rate in mL/s we use convert L/min to mL/s by using conversion factor:
<u>1 L/min = 16.6667 mL/s.</u>
18 L/min = 16.6667 × 18 mL/s.
18L/min = 300 mL/s.
Therefore, Uche's pumping rate is 300 mL/s.
Answer: 14 right and 8 up, or (14, 8)
Step-by-step explanation:
To find the distance, you can subtract or add numbers.
So from (-6, 2) to 0, I have to move right 6 spaces to get (0, 2).
Then, I must move 8 to the right to get (8, 2).
From here, I continue to move up 8 since we want to get to (8,10). So in all, you should move 14 spaces to the right and 8 up (14, 8).
