Answer:
70/3 πcm
Step-by-step explanation:
hope this helps
Answer:
g(0.9) ≈ -2.6
g(1.1) ≈ 0.6
For 1.1 the estimation is a bit too high and for 0.9 it is too low.
Step-by-step explanation:
For values of x near 1 we can estimate g(x) with t(x) = g'(1) (x-1) + g(1). Note that g'(1) = 1²+15 = 16, and for values near one g'(x) is increasing because x² is increasing for positive values. This means that the tangent line t(x) will be above the graph of g, and the estimates we will make are a bit too big for values at the right of 1, like 1.1, and they will be too low for values at the left like 0.9.
For 0.9, we estimate
g(0.9) ≈ 16* (-0.1) -1 = -2.6
g(1.1) ≈ 16* 0.1 -1 = 0.6
the center is at the origin of a coordinate system and the foci are on the y-axis, then the foci are symmetric about the origin.
The hyperbola focus F1 is 46 feet above the vertex of the parabola and the hyperbola focus F2 is 6 ft above the parabola's vertex. Then the distance F1F2 is 46-6=40 ft.
In terms of hyperbola, F1F2=2c, c=20.
The vertex of the hyperba is 2 ft below focus F1, then in terms of hyperbola c-a=2 and a=c-2=18 ft.
Use formula c^2=a^2+b^2c
2
=a
2
+b
2
to find b:
\begin{gathered} (20)^2=(18)^2+b^2,\\ b^2=400-324=76 \end{gathered}
(20)
2
=(18)
2
+b
2
,
b
2
=400−324=76
.
The branches of hyperbola go in y-direction, so the equation of hyperbola is
\dfrac{y^2}{b^2}- \dfrac{x^2}{a^2}=1
b
2
y
2
−
a
2
x
2
=1 .
Substitute a and b:
\dfrac{y^2}{76}- \dfrac{x^2}{324}=1
76
y
2
−
324
x
2
=1 .
Step-by-step explanation:
3/4 ×1/-24 +5/5
3×1/4×-24+4/5
3×1/4×-24+4/5
3/-96+4/5
-3/96+4/5
-1/32+4/5
ans:123/160
