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Deffense [45]
3 years ago
13

Find the height of the cylinder whose radius is 7cm and tsa is 96cm2

Mathematics
1 answer:
Advocard [28]3 years ago
7 0

Answer:

0.62

Step-by-step explanation:

Height = Volume / πr²

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Jacob consumes an energy drink that contains caffeine. After consuming the energy drink, the amount of caffeine in Jacob's body
sasho [114]

Answer: 0.87400mg of caffeine.

Step-by-step explanation:

You have

N(t)=N0(e^−rt)(1)

as a general Exponential decay equation where N0 is the amount at t=0, N(t) is the amount remaining at time t and r is the exponential decay constant. You're specifically given that after 10 hours, the decay factor is 0.2601, i.e.,

N(10)/N(0)=N0(e^−10r)/N0(e^0)= e^−10r=0.2601 . .(2)

Taking the last 2 parts of (2) to the power of 0.1t gives

e^−rt=0.2601^.1t . .(3)

This means that

N(t)=N0(e^−rt)=N0(0.2601^.1t). .(4)

Also,

N(2.56)N(1.56)=N0(0.2601.1(2.56))N0(0.2601.1(1.56))=0.2601.1(2.56−1.56)=0.2601^.1

= 0.87400mg of caffeine.

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3 years ago
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32.both of their ideas are correct whether it is to replace x or y first it doesn't matter only one can be replaced to find the value of the other.

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Triangle ABC has vertices A(-5, -2), B(7, -5), and C(3, 1). Find the coordinates of the intersection of the three altitudes
Darina [25.2K]

Answer:

Orthocentre (intersection of altitudes) is at (37/10, 19/5)

Step-by-step explanation:

Given three vertices of a triangle

A(-5, -2)

B(7, -5)

C(3, 1)

Solution A by geometry

Slope AB = (yb-ya) / (xb-xa) = (-5-(-2)) / (7-(-5)) = -3/12 = -1/4

Slope of line normal to AB, nab = -1/(-1/4) = 4

Altitude of AB = line through C normal to AB

(y-yc) = nab(x-xc)

y-1 = (4)(x-3)

y = 4x-11           .........................(1)

Slope BC = (yc-yb) / (xc-yb) = (1-(-5) / (3-7)= 6 / (-4) = -3/2

Slope of line normal to BC, nbc = -1 / (-3/2) = 2/3

Altitude of BC

(y-ya) = nbc(x-xa)

y-(-2) = (2/3)(x-(-5)

y = 2x/3 + 10/3 - 2

y = (2/3)(x+2)    ........................(2)

Orthocentre is at the intersection of (1) & (2)

Equate right-hand sides

4x-11 = (2/3)(x+2)

Cross multiply and simplify

12x-33 = 2x+4

10x = 37

x = 37/10  ...................(3)

substitute (3) in (2)

y = (2/3)(37/10+2)

y=(2/3)(57/10)

y = 19/5  ......................(4)

Therefore the orthocentre is at (37/10, 19/5)

Alternative Solution B using vectors

Let the position vectors of the vertices represented by

a = <-5, -2>

b = <7, -5>

c = <3, 1>

and the position vector of the orthocentre, to be found

d = <x,y>

the line perpendicular to BC through A

(a-d).(b-c) = 0                          "." is the dot product

expanding

<-5-x,-2-y>.<4,-6> = 0

simplifying

6y-4x-8 = 0 ...................(5)

Similarly, line perpendicular to CA through B

<b-d>.<c-a> = 0

<7-x,-5-y>.<8,3> = 0

Expand and simplify

-3y-8x+41 = 0 ..............(6)

Solve for x, (5) + 2(6)

-20x + 74 = 0

x = 37/10  .............(7)

Substitute (7) in (6)

-3y - 8(37/10) + 41 =0

3y = 114/10

y = 19/5  .............(8)

So orthocentre is at (37/10, 19/5)  as in part A.

8 0
3 years ago
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