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SpyIntel [72]
3 years ago
12

A)Chris and Alex Were Building a sand Castle. They Decided to make it into a cone shape. After some time, the diameter of the th

e sand shaped cone was 12in and the height was 7in. What is the exact volume of the sand cone?
B) the kids kept adding sand, ten minutes later the volume of the sand cone had increased by 30%. what is the exact volume of the sand cone now?
Mathematics
1 answer:
Nesterboy [21]3 years ago
3 0

Answer:

A.) 84π | B.) 546π/5

Step-by-step explanation:

A.)

V(Cone) = πr²(h/3) where r = radius and h = height.

r = d/2 where d = diameter.

r = 12/2 = 6

V(Cone) = π(6)²(7/3) = π(36)(7/3) = 84π

B.)

84π · 0.3 = 126π/5

84π + 126π/5 = 546π/5  

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What are the solutions?<br>8x^2-12x-23=-5​
erastova [34]

Answer:

x = 3/4 + (3 sqrt(5))/4 or x = 3/4 - (3 sqrt(5))/4

Step-by-step explanation:

Solve for x over the real numbers:

8 x^2 - 12 x - 23 = -5

Divide both sides by 8:

x^2 - (3 x)/2 - 23/8 = -5/8

Add 23/8 to both sides:

x^2 - (3 x)/2 = 9/4

Add 9/16 to both sides:

x^2 - (3 x)/2 + 9/16 = 45/16

Write the left hand side as a square:

(x - 3/4)^2 = 45/16

Take the square root of both sides:

x - 3/4 = (3 sqrt(5))/4 or x - 3/4 = -(3 sqrt(5))/4

Add 3/4 to both sides:

x = 3/4 + (3 sqrt(5))/4 or x - 3/4 = -(3 sqrt(5))/4

Add 3/4 to both sides:

Answer:  x = 3/4 + (3 sqrt(5))/4 or x = 3/4 - (3 sqrt(5))/4

4 0
3 years ago
Verify that the given point is on the curve and find the lines that are (a) tangent and (b) normal to the curve at the given poi
lara [203]

For each curve, plug in the given point (x,y) and check if the equality holds. For example:

(I) (2, 3) does lie on x^2+xy-y^2=1 since 2^2 + 2*3 - 3^2 = 4 + 6 - 9 = 1.

For part (a), compute the derivative \frac{\mathrm dy}{\mathrm dx}, and evaluate it for the given point (x,y). This is the slope of the tangent line at the point. For example:

(I) The derivative is

x^2+xy-y^2=1\overset{\frac{\mathrm d}{\mathrm dx}}{\implies}2x+x\dfrac{\mathrm dy}{\mathrm dx}+y-2y\dfrac{\mathrm dy}{\mathrm dx}=0\implies\dfrac{\mathrm dy}{\mathrm dx}=\dfrac{2x+y}{2y-x}

so the slope of the tangent at (2, 3) is

\dfrac{\mathrm dy}{\mathrm dx}(2,3)=\dfrac74

and its equation is then

y-3=\dfrac74(x-2)\implies y=\dfrac74x-\dfrac12

For part (b), recall that normal lines are perpendicular to tangent lines, so their slopes are negative reciprocals of the slopes of the tangents, -\frac1{\frac{\mathrm dy}{\mathrm dx}}. For example:

(I) The tangent has slope 7/4, so the normal has slope -4/7. Then the normal line has equation

y-3=-\dfrac47(x-2)\implies y=-\dfrac47x+\dfrac{29}7

3 0
3 years ago
First, rewrite 5/9 and 11/21 so they have a common denominator. Then, use to order 5/9 and 11/21.
spayn [35]

5/9 and 11/21

LCM of the denominators is 63

changing their denominators to all have a common denominator of 63:

5(7)/63 = 35/63

11(3)/63 = 33/63

from the above;

5/9 > 11/21.

4 0
3 years ago
A research company surveys people in a community about a new recycling program. The company expects 6,000 people to respond favo
strojnjashka [21]

Answer:

5850

Step-by-step explanation:

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4 0
3 years ago
The following question involves a standard deck of 52 playing cards. In such a deck of cards there are four suits of 13 cards ea
olasank [31]

Answer:

(a)No. The probability of drawing a specific second card depends on the identity of the first card.

(b)4/663

(c) 4/663

(d) 8/663

Step-by-step explanation:

(a)The events are not independent because we are drawing cards without replacement and the probability of drawing a specific second card depends on the identity of the first card.

(b) P(ace on 1st card and jack on 2nd).

P$(Ace on 1st card) =\dfrac{4}{52}\\ P$(Jack on 2nd card)=\dfrac{4}{51}\\\\$Therefore:\\P(ace on 1st card and jack on 2nd) =\dfrac{4}{52}\times \dfrac{4}{51}\\=\dfrac{4}{663}

(c)P(jack on 1st card and ace on 2nd)

P$(Jack on 1st card) =\dfrac{4}{52}\\ P$(Ace on 2nd card)=\dfrac{4}{51}\\\\$Therefore:\\P(jack on 1st card and ace on 2nd) =\dfrac{4}{52}\times \dfrac{4}{51}\\=\dfrac{4}{663}

(d)Probability of drawing an ace and a jack in either order.

We can either draw an ace first, jack second or jack first, ace second.

Therefore:

P(drawing an ace and a jack in either order) =P(AJ)+(JA)

From parts (b) and (c) above:

P$(jack on 1st card and ace on 2nd) =\dfrac{4}{663}\\P$(ace on 1st card and jack on 2nd) =\dfrac{4}{663}\\$Therefore:\\P(drawing an ace and a jack in either order)=\dfrac{4}{663}+\dfrac{4}{663}\\=\dfrac{8}{663}

6 0
3 years ago
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