HELP I CANT GET A BAD GRADEThe gestation period is the length of the time it takes a baby mammal to develop before it is born. The gestation period of a horse is, on average, 336 days. About how many weeks is that? There are 7 days in a week.

A.
50 weeks

B.
60 weeks

C.
40 weeks

D.
5 weeks

Answers

Answer 1

Answer:

Answer:

a. 50 weeks

Step-by-step explanation:

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Heres a the new picture I need help with ASAP

Answers

for #3 Nelson scored 29, because Craig scored 12 and Marla scored twice as craig, which was 24. After that we found out that she has 5 fewer than Nelson, witch would be plus 5.

i think the answers would be
1.) 4 1/2 * 4 2/3= 9/2 * 14/3 then i canceled out 9 and 3 and 2 and 14 and got 3*7= 21 pounds
2.) 5 5/8 * 1 5/9= 45/8 * 14/9 then i canceled out 45 and 9 and got 5/8 * 14= 19/8
3.) 12*2-5 and if you want it solved its 12*2= 24 -5= 19
4.) 66/8 = 8.25 thats how much yvette earns and then you would just do 68.8/8=8.60 thats how much lizbeth earns
5.) it would be 100 because 10*10 = 100 and 4*25=100 and no other smaller number could fit into it
6.) 4 5/8 + 3 3/4 - 11 1/8
37/8 +15/4= 37/8+30/8=67/8
67/8-11 1/8= 67/8-89/8= 22/8 or 2 3/4

How many solutions?
1
2
3
0

Answers

Answer:

The number of solutions of the system is 0

Step-by-step explanation:

we know that

When solving a system of equations by graphing, the solution of the system is equal to the intersection point both graphs.

In this problem the graphs do not intersect

therefore

The number of solutions of the system is 0

In the rhombus, m<1=18x, m<2-x+y, and m<3=30z. Find the value of x+y+z. The diagram is not drawn to scale.(1point)

Answers

x + y = 18x y = 17x18x = 30z x = 30z/1830z = 18x z = 18x/30 hope that helps

Cos3x=4cos^3x-3cosx prove

Answers

$\bf cos(\alpha + \beta)= cos(\alpha)cos(\beta)- sin(\alpha)sin(\beta)\n\n\n\textit{also recall that }cos(2\theta)=\begin{cases}cos^2(\theta)-sin^2(\theta)\n1-2sin^2(\theta)\n\boxed{2cos^2(\theta)-1}\end{cases}\n\n\nand\qquad sin^2(\theta)+cos^2(\theta)=1\implies sin^2(\theta)=1-cos^2(\theta)\n\n-------------------------------$

$\bf cos(3x)=4cos^3(x)-3cos(x)\n\n-------------------------------\n\ncos(3x)\implies cos(2x+x)\implies cos(2x)cos(x)-sin(2x)sin(x)\n\n\n\[2cos^2(x)-1]cos(x)-[2sin(x)cos(x)]sin(x)\n\n\n2cos^3(x)-cos(x)~~~~-~~~~2sin^2(x)cos(x)\n\n\n2cos^3(x)-cos(x)~~~~-~~~~2[1-cos^2(x)]cos(x)\n\n\n2cos^3(x)-cos(x)~~~~-~~~~[2cos(x)-2cos^3(x)]\n\n\n2cos^3(x)-cos(x)~~~~-~~~~2cos(x)+2cos^3(x)\n\n\n4cos^3(x)-3cos(x)$

A cup of coffee at 181 degrees is poured into a mug and left in a room at 66 degrees. After 6 minutes, the coffee is 139 degrees. Assume that the differential equation describing Newton's Law of Cooling is (in this case) dT/dt=k(T-66).1) What is the temperature of the coffee after 16 minutes?
2) After how many minutes will the coffee be 100 degrees?

Answers

The temperature of the coffee after 16 minutes is 134 degrees and after 1.3 minutes the coffee be 100 degrees.

What is Differential equation?

A differential equation is an equation that contains one or more functions with its derivatives.

A cup of coffee at 181 degrees is poured into a mug and left in a room at 66 degrees.

After 6 minutes, the coffee is 139 degrees.

Assume that the differential equation describing Newton's Law of Cooling is (in this case) dT/dt=k(T-66).

T=∫k(t-66)dt

=k(t²/2-66)+c

When t=0 and T=181

181=k(0-66)+c

181=-66k+c

when t=6, T=139

139=k(6²/2-66)+c

139=-48k+c

-42=-18k

Divide both sides by 18

k=7/3

139=-48×7/3+c

c=139+112=251

T=7/3(t-66)+251

The temperature of the coffee after 16 minutes

T=7/3(16-66)+251

T=7/3(-50)+251

T=134 degrees

After how many minutes will the coffee be 100 degrees

100=7/3(t-66)+251

100=7/3t-7/3(66)+251

100=7/3t-154+251

100=7/3t+97

100-97=7/3t

3=7/3t

9/7=t

1.3=t

Hence, the temperature of the coffee after 16 minutes is 134 degrees and after 1.3 minutes the coffee be 100 degrees.

To learn more on Differentiation click:

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Answer:

Step-by-step explanation:

$(dT)/(dt) =k(t-66)\nT=\int\ {k(t-66)} \, dt=K((t^2)/(2) -66)+c\nwhen t=0,T=181\n181=K(0-66)+c\n181=-66k+c\nwhen t=6,T=139\n139=k((6^2)/(2) -66)+c\n139=-48k+c\n181-139=-66k+48k\n-42=-18k\n7=3k\nk=(7)/(3) \n139=-48*(7)/(3) +c\nc=139+112=251\nT=(7)/(3) (t-66)+251\nnow complete the question$