Answer:
QPT is obtuse
SPQ is acute
TPR is right
Step-by-step explanation:
Took geometry two years ago
Answer:
52°i think
Step-by-step explanation:
148°-96°=52°
The function is
f(x) = (1/3)x² + 10x + 8
Write the function in standard form for a parabola.
f(x) = (1/3)[x² + 30x] + 8
= (1/3)[ (x+15)²- 225] + 8
= (1/3)(x+15)² -75 + 8
f(x) = (1/3)(x+15)² - 67
This is a parabola with vertex at (-15, -67).
The axis of symmetry is x = -15
The curve opens upward because the coefficient of x² is positive.
As x -> - ∞, f -> +∞.
As x -> +∞, f -> +∞
The domain is all real values of x (see the graph below).
Answer: The domain is (-∞, ∞)
Answer:
1 17/42.
Step-by-step explanation:
Find the common denominator, which is 42.
So change -4 1/6 to -4 7/42.
And change 5 4/7 to 5 24/52.
So add -4 7/42 + 5 24/52 and you'll get: 1 17/42.
(The decimal form of this is 1.404762)
The function you seek to minimize is
()=3‾√4(3)2+(13−4)2
f
(
x
)
=
3
4
(
x
3
)
2
+
(
13
−
x
4
)
2
Then
′()=3‾√18−13−8=(3‾√18+18)−138
f
′
(
x
)
=
3
x
18
−
13
−
x
8
=
(
3
18
+
1
8
)
x
−
13
8
Note that ″()>0
f
″
(
x
)
>
0
so that the critical point at ′()=0
f
′
(
x
)
=
0
will be a minimum. The critical point is at
=1179+43‾√≈7.345m
x
=
117
9
+
4
3
≈
7.345
m
So that the amount used for the square will be 13−
13
−
x
, or
13−=524+33‾√≈5.655m