Answer:
$50
Step-by-step explanation:
.40C = $20
C = $20/.40 = $50
bro i literally had the same question about this
you in honors or something? sry i only got A
I might have B later on or something
Answer:
A.
41.125π ≈ 129
149.875π ≈ 471
Step-by-step explanation:
Surface area of smaller ganza:
3.5/2 = 1.75 x 1.75 = 3.0625 x 3.14 = 9.61625 x 2 = 19.2325
1.75 x 10 = 17.5 x 3.14 = 54.95 x 2 = 109.9 + 19.2 = 129 (Rounded)
Surface area of larger ganza:
5.5/2=2.75 x 2.75 = 7.5625 x 3.14 = 23.74625 x 2 = 47.4925
2.75 x 24.5 = 67.375 x 3.14 = 211.5575 x 2 = 423.115 + 47.4925 = 471 (Rounded)
π
129/3.14 = 41.125
471/3.14 = 149.875
x=3 and y=−2
Step-by-step explanation:
Solve8x+5y=14for x:
8x+5y=14
8x+5y+−5y=14+−5y(Add -5y to both sides)
8x=−5y+14
8x8=−5y+148(Divide both sides by 8)
x=−58y+74
Step: Substitute−58y+74forxin−4x−5y=−2:
−4x−5y=−2
−4(−58y+74)−5y=−2
−52y−7=−2(Simplify both sides of the equation)
−52y−7+7=−2+7(Add 7 to both sides)
−52y=5
−52y−52=5−52(Divide both sides by (-5)/2)
y=−2
Step: Substitute−2foryinx=−58y+74:
x=−58y+74
x=−58(−2)+74
x=3(Simplify both sides of the equation)
Answer:
x=3 and y=−2
For the mix you need 60 Ib of nuts and raisins, and 60 lb of mix will sell for $300.
Let <em>x</em> be the weight of nuts and <em>y</em> the weight of raisins needed for the mix.
Total weight = <em>x</em>+<em>y</em> = 60, so <em>y</em> = 60 - <em>x</em>
Total value = 6<em>x</em> + 3<em>y</em> = 300. From the above, we can substitute the <em>y</em> for 60 - x:
Total value = 6<em>x</em> + 3(60 - <em>x</em>) = 300, which we can simplify to:
6<em>x</em> + 180 - 3<em>x</em> = 300, and simplify again to:
3<em>x</em> = 120, and again to:
<em>x</em> = 40. We can now calculate<em> y</em> from the first equation y = 60 - <em>x</em>, so:
y = 20.
40 lb of nuts and 20 lb of raisins. Check:
40 lb x $6/lb = $240, and 20 lb x $3/lb = $60, $240 + $60 = $300.
Answer:
Step-by-step explanation:
A cone is a three-dimensional closed figure that has a circular base connected to a vertex (or apex) point outside the plane of the base. Similar Cross Sections. (parallel to base) • The one and only base of the cone is a circle (or other curved figure).
