Use place value to find the product of 60 x 4 =

Please please please can someone help me :)

disa [49]

Answer:

Step-by-step explanation:

now the equation looks like

3-4Cos(x)=y

hmmm i'm wondering if you can figure it out from here??? it's the same as the other two.. plug in that $\sqrt{2}$ /2 .. can you? :)

8 0

2 years ago

What is the waist to hip ratio? Calculate the WHR of a male with a waist measurement of 40 inches and hip measurement 2 inches l

aev [14]

Waist to Hip Ratio (WHR) = Waist Measurement / Hip Measurement

Waist Measurement = 40 inches
Hip Measurement = 2 inches less = (40 -2) = 38 inches.

Waist to Hip Ratio (WHR) = 40/38 = 1.0526

8 0

3 years ago

The Pacific halibut fishery has been modeled by the differential equation dy dt = ky 1 − y K where y(t) is the biomass (the tota

Dafna1 [17]

Answer:

a. The biomass weighs 2.30 * 10^7 kg after a year

b. It'll take 2.56 years to get to 4*10^7kg

Step-by-step explanation:

a.

k = 0.78,K = 6E7 kg

Given

dy/dt = ky(1- y/K)

Make ky dt the subject of formula

ky dt = dy/(1-y/K) --- make k dt the subject of formula

k dt = dy/(y(1-y/K))

k dt = K dy / y(K-y)

k dt = ((1/y) + (1/(K-y)))dy ---- integrate both sides

kt + c = ln(y/(K-y))

Ce^(kt) = y/(K-y)

Substitute the values of k and K

Ce^(0.78t) = y/(6*10^7 - y) ----- (1)

Given that y(0) = 2 * 10^7kg

(1) becomes

Ce^(0.78*0) = (2 * 10^7)/(6*10^7 - 2*10^7)

Ce° = (2*10^7)/(4*10^7

C = 2/7

Substitute 2/7 for C in (1)

2/7e^0.78t = y/(6*10^7 - y) ---(2)

We're to find the biomass a year later

So, t = 1

2/7e^0.78 = y/(6*10^7 - y)

0.62 = y/(6*10^7 - y)

y = 0.62(6*10^7 - y)

y = 0.62*6*10^7 - 0.62y

y + 0.62y = 0.62*6*10^7

1.62y = 0.62*6*10^7

1.62y = 3.72 * 10^7

y = 2.30 * 10^7kg.

Hence, the biomass weighs 2.30 * 10^7 kg after a year

b.

Here, we're to calculate the time it'll take the biomass to get to 4*10^7 kg

Substitute 4*10^7 for y in (2)

2/7e^0.78t = 4*10^7/(6*10^7 - 4*10^7)

2/7e^0.78t = 4*10^7/2*10^7

2/7e^0.78t = 2

e^0.78t = (2*7)/2

e^0.78t = 2

t = 2 * 1/0.78

t = 2.56 years

Hence, it'll take 2.56 years to get to 4*10^7kg

8 0

3 years ago

How do I simplify the square root: 7 times the square root of 81

DENIUS [597]

$7\cdot\sqrt{81}=7\cdot\sqrt{9\cdot9}=7\cdot9=\boxed{63}$

7 0

3 years ago

Please help me out guys. The photo will show you what to do. 50 points cause I need it done fast

Nonamiya [84]

Answer:

a) starting height: 5.5 ft

b) hang time: 5.562 seconds

c) maximum height: 126.5 ft

d) time to maximum height: 2.75 seconds

Step-by-step explanation:

a) The starting height is the height at t=0.

h(0) = -16·0 +88·0 +5.5

h(0) = 5.5

The starting height is 5.5 feet.

__

b) The ball is in the air between t=0 and the non-zero time when h(t) = 0. We can find the latter by solving ...

-16t^2 + 8t +5.5 = 0

t^2 -(11/2)t = 5.5/16 . . . . . subtract 5.5, then divide by -16

t^2 -(11/2)t +(11/4)^2 = (5.5/16) +(11/4)^2 . . . . complete the square

(t -11/4)^2 = 126.5/16 . . . . . . . . . . . . . . . . . . . . call this [eq1] for later use

t -11/4 = √7.90625

t = 2.75 +√7.90625 ≈ 5.562

The ball will be in the air about 5.562 seconds.

__

c) If we multiply [eq1] above by -16 and add the constant on the right, we get the vertex form of the height equation:

h(t) = -16(t -11/4) +126.5

The vertex at (2.75, 126.5) tells us ...

The maximum height of the ball is 126.5 feet.

__

d) That same vertex point tells us ...

The maximum height will be reached at t = 2.75 seconds.

_____

If you really need answers fast, a graphing calculator can give them to you in very short order (less than a minute).

6 0

3 years ago