find the perimeter of a triangle with sides 15 inches, 15 inches, and 21 inches length
To find the perimeter of a triangle we add all the sides of the triangle
The length of the sides of the triangle are given as 15 inches, 15 inches, and 21 inches
Perimeter of a triangle =
15 inches + 15 inches + 21 inches = 51 inches
So 51 inches is the perimeter
Which of these is equal to cos 47°?
a. cos 43°
b. sin 43°
c. tan 43°
d. cos 133°
B.....
cos (47°)= sin (90°-47°)=sin (43°) [complementary angles]
Answer B
Answer: The dimensions of the frame:.
W= 2√10 or about 6.3245 inches
L= 4√10 or about 12.6491 inches
Step-by-step explanation:
This would be so much easier if the area of the box was 72. Then the dimensions of the frame would be 6 × 12. So just a little bit larger and more complicated.
Here the equation is w(2w)=80
2w^2 = 80 (2w is easier to work with than L/2).
w^2=40 (^2 means squared)
√w^2 = √40 square root of both sides: Factor 40. Take out√4
W = √4*10
W= 2√10
L= 2w = 4√10
6.3245 × 12.6491 = 80
Answer:
Let's define the variables:
A = price of one adult ticket.
S = price of one student ticket.
We know that:
"On the first day of ticket sales the school sold 1 adult ticket and 6 student tickets for a total of $69."
1*A + 6*S = $69
"The school took in $150 on the second day by selling 7 adult tickets and student tickets"
7*A + 7*S = $150
Then we have a system of equations:
A + 6*S = $69
7*A + 7*S = $150.
To solve this, we should start by isolating one variable in one of the equations, let's isolate A in the first equation:
A = $69 - 6*S
Now let's replace this in the other equation:
7*($69 - 6*S) + 7*S = $150
Now we can solve this for S.
$483 - 42*S + 7*S = $150
$483 - 35*S = $150
$483 - $150 = 35*S
$333 = 35*S
$333/35 = S
$9.51 = S
That we could round to $9.50
That is the price of one student ticket.
