Answer:
The area in which Taft contributed equal or surpass Roosevelt's was in conservation.....
Step-by-step explanation:
U 2 can help me by marking as brainliest........
Sphere Surface Area = <span> 4 • <span>π <span>• r²
For it to equal 16 PI, then radius must equal 2
4*PI*2*2 = 16 PI
</span></span></span>
Sphere Volume = <span> 4/3 • <span>π <span>• r³
</span></span></span>
Sphere Volume = <span> 4/3 • <span>π <span>• 2^3
</span></span></span>
Sphere Volume = <span> 4/3 *PI * 8
</span>
Sphere Volume = <span> 32 / 3 PI
</span>
Sphere Volume = <span> 10.666 PI cubic feet AND I think that is answer B
which SHOULD read 10 (2/3) PI ft^3
</span>
Step-by-step explanation:
1 Remove parentheses.
8{y}^{2}\times -3{x}^{2}{y}^{2}\times \frac{2}{3}x{y}^{4}
8y
2
×−3x
2
y
2
×
3
2
xy
4
2 Use this rule: \frac{a}{b} \times \frac{c}{d}=\frac{ac}{bd}
b
a
×
d
c
=
bd
ac
.
\frac{8{y}^{2}\times -3{x}^{2}{y}^{2}\times 2x{y}^{4}}{3}
3
8y
2
×−3x
2
y
2
×2xy
4
3 Take out the constants.
\frac{(8\times -3\times 2){y}^{2}{y}^{2}{y}^{4}{x}^{2}x}{3}
3
(8×−3×2)y
2
y
2
y
4
x
2
x
4 Simplify 8\times -38×−3 to -24−24.
\frac{(-24\times 2){y}^{2}{y}^{2}{y}^{4}{x}^{2}x}{3}
3
(−24×2)y
2
y
2
y
4
x
2
x
5 Simplify -24\times 2−24×2 to -48−48.
\frac{-48{y}^{2}{y}^{2}{y}^{4}{x}^{2}x}{3}
3
−48y
2
y
2
y
4
x
2
x
6 Use Product Rule: {x}^{a}{x}^{b}={x}^{a+b}x
a
x
b
=x
a+b
.
\frac{-48{y}^{2+2+4}{x}^{2+1}}{3}
3
−48y
2+2+4
x
2+1
7 Simplify 2+22+2 to 44.
\frac{-48{y}^{4+4}{x}^{2+1}}{3}
3
−48y
4+4
x
2+1
8 Simplify 4+44+4 to 88.
\frac{-48{y}^{8}{x}^{2+1}}{3}
3
−48y
8
x
2+1
9 Simplify 2+12+1 to 33.
\frac{-48{y}^{8}{x}^{3}}{3}
3
−48y
8
x
3
10 Move the negative sign to the left.
-\frac{48{y}^{8}{x}^{3}}{3}
−
3
48y
8
x
3
11 Simplify \frac{48{y}^{8}{x}^{3}}{3}
3
48y
8
x
3
to 16{y}^{8}{x}^{3}16y
8
x
3
.
-16{y}^{8}{x}^{3}
−16y
8
x
3
Done
It would be 0.4:) hope this helps
