Find the volume of a sphere if the surface area of the sphere is 64π cm²? Explain your

Mathematics

1 answer:

yarga [219]2 years ago

6 0

Answer:

the answer is 85.33π cm

Step-by-step explanation:

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Stella [2.4K]

Answer:

The first one is d: x=0 or x=6

the second one is the 3rd one

Step-by-step explanation:

8 0

3 years ago

The blueprint of a pool has a scale of 2 inches equals 7 feet. The scale drawing is shown below. the length is 20 and the width

AlladinOne [14]

Answer:

Part A:

Blueprint 10 inches ⇒ Actual 35 feet

Blueprint 20 inches ⇒ Actual 70 feet

Part B:

The cost of the cover is $857.5

Step-by-step explanation:

Part A:

∵ The drawing scale is 2 inches for 7 feet

∵ The length = 20 inches and the width = 10 inches

∴ The actual length = (20 × 7) ÷ 2 = 70 feet

∴ The actual width = (10 × 7) ÷ 2 = 35 feet

Part B:

∵ The cost per square foot = $0.35

∵ The area of the pool = 70 × 35 = 2450 feet ²

∴ The cost of the cover = 2450 × 0.35 = $857.5

8 0

3 years ago

Solve the equation for x∈Z <br><br>-x² +8x -14 ≥ 0

kenny6666 [7]

$ax^2+bx+c=0\\\Delta=b^2-4ac\\if\ \Delta < 0\ then\ no\ solutions\\if\ \Delta=0\ then\ one\ solution:x=\frac{-b}{2a}\\if\ \Delta > 0\ then\ two\ solutions:x=\frac{-b\pm\sqrt\Delta}{2a}\\================================\\\\-x^2+8x-14\geq0\ \ \ \ |multiply\ both\ sides\ by\ (-1)\ \{change\ \geq\ on\ \leq\}\\\\x^2-8x+14\leq0\\\\a=1;\ b=-8;\ c=14\\\\\Delta=(-8)^2-4\cdot1\cdot14=64-56=8 > 0\\\sqrt\Delta=\sqrt8=\sqrt{4\cdot2}=\sqrt4\cdot\sqrt2=2\sqrt2$

$x_1=\frac{-(-8)-2\sqrt2}{2\cdot1}=\frac{8-2\sqrt2}{2}=\frac{8}{2}-\frac{2\sqrt2}{2}=\boxed{4-\sqrt2}\\\\x_2=\frac{-(-8)+2\sqrt2}{2\cdot1}=\frac{8+2\sqrt2}{2}=\frac{8}{2}+\frac{2\sqrt2}{2}=\boxed{4+\sqrt2}\\\\============================$

$ax^2+bx+c=0\\\\if\ a > 0\ then\ the\ parabola\ o pen\ up\\if\ a < 0\ then\ the\ parabola\ o pen\ down\\========================\\\\a=1 > 0-therefore\ o pen\ up\ (look\ at\ the\ picture)\\\\===============================\\\\Answer:x\in\left$

$Solutions\ in\ \mathbb{Z}:\\\\\sqrt2\approx1.4\\\\threfore:4-\sqrt2\approx4-1.4=2.6\ and\ 4+\sqrt2\approx4+1.4=5.4\\\\look\ at\ the\ second\ picture:x=3\ or\ x=4\ or\ x=5\ (x\in\{3;\ 4;\ 5\})$