Answer:
the answer is option B. angle S.
when naming an angle we place the vertex of the angle in the middle. here the angle is RST. But that option is unavailable. very often when there are no other angles interfering with the parent angle, we represent it using one letter that is the mid letter, the vertex. here in this case it is S.
Correct question :
If the perimeters of each shape are equal, which equation can be used to find the value of x? A triangle with base x + 2, height x, and side length x + 4. A rectangle with length of x + 3 and width of one-half x. (x + 4) + x + (x + 2) = one-half x + (x + 3) (x + 2) + x + (x + 4) = 2 (one-half x) + 2 (x + 3) 2 (x) + 2 (x + 2) = 2 (one-half x) + 2 (x + 3) x + (x + 2) + (x + 4) = 2 (x + 3 and one-half)
Answer: (x + 2) + x + (x + 4) = 2 (one-half x) + 2 (x + 3)
Step-by-step explanation:
Given the following :
A triangle with base x + 2, height x, and side length x + 4 - - - -
b = x + 2 ; a = x ; c = x + 4
Perimeter (P) of a triangle :
P = a + b + c
P =( x + 2) + x + (x + 4) - - - (1)
A rectangle with length of x + 3 and width of one-half x
l = x + 3 ; w = 1/2 x
Perimeter of a rectangle (P) = 2(l+w)
P = 2(x+3) + 2(1/2x)
If perimeter of each same are the same ; then;
(1) = (2)
(x + 2) + x + (x + 4) = 2(x+3) + 2(1/2x)
Answer: E. 120
The number of seniors were there in high school X at the beginning of the year = 120
Step-by-step explanation:
Given : At the beginning of the year, the ratio of juniors to seniors in high school X was 3 to 4.
Let the number of juniors be 3x and the number of seniors be 4x.
Since , during the year, 10 juniors and twice as many seniors transferred to another high school, while no new students joined high school X.
i.e. Number of juniors at the end of the year = 3x-10
Number of seniors at the end of the yea = 4x-2(10)=4x-20
At the end of the year, the ratio of juniors to seniors was 4 to 5.
The number of seniors were there in high school X at the beginning of the year = 4(30)=120
Hence, the correct answer is E. 120 .
