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3 years ago

6 eggs in 7 dауs what would be the answer?

Mathematics

2 answers:

ANTONII [103]3 years ago

4 0

Answer:

42 Eggs

Step-by-step explanation:

6×7=42

6 eggs every day for 7 days, gives us 42 eggs at the end of the week

Hope this Helps!

sp2606 [1]3 years ago

4 0

42 eggs. 6 ega multiplied by 7 days= 42 eggs

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Please only give me the check from numbers 1 to 5

Marizza181 [45]

Answer:

1) f=30

2) w=21

3) x=72

4) t=12

5) y=6

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2 years ago

an ordered pair where the x and y coordinate are the same lies in the 1st or 3rd quadrant. true, sometimes true or not true?. ex

Zarrin [17]

The quadrants are as follows:

The 1st quadrant has the points which have both x and y positive and the 3rd quadrant has the points which have both x and y negative. If the ordered pair and the same x and y value, if ons is positive, the other also is, and the same for negative.

So, at first we see that there are point where the x and y are the same and that are in the 1st or 3rd quadrant.

However, there is one special case:

When x and y are 0, that is, the ordered pair is (0, 0).

Since this point is the origin, it doesn't lie on any of the quadrants.

Thus, this affirmative is sometimes true. Every point but (0, 0) that have same x and y values are in the 1st or 3rd quadrant except for (0, 0).

7 0

1 year ago

Solve 5y'' + 3y' – 2y = 0, y(0) = 0, y'(0) = 2.8 y(t) = 0 Preview

mario62 [17]

Answer: The required solution is

$y(t)=-\dfrac{7}{3}e^{-t}+\dfrac{7}{3}e^{\frac{1}{5}t}.$

Step-by-step explanation: We are given to solve the following differential equation :

$5y^{\prime\prime}+3y^\prime-2y=0,~~~~~~~y(0)=0,~~y^\prime(0)=2.8~~~~~~~~~~~~~~~~~~~~~~~~(i)$

Let us consider that

$y=e^{mt}$ be an auxiliary solution of equation (i).

Then, we have

$y^prime=me^{mt},~~~~~y^{\prime\prime}=m^2e^{mt}.$

Substituting these values in equation (i), we get

$5m^2e^{mt}+3me^{mt}-2e^{mt}=0\\\\\Rightarrow (5m^2+3y-2)e^{mt}=0\\\\\Rightarrow 5m^2+3m-2=0~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~[\textup{since }e^{mt}\neq0]\\\\\Rightarrow 5m^2+5m-2m-2=0\\\\\Rightarrow 5m(m+1)-2(m+1)=0\\\\\Rightarrow (m+1)(5m-1)=0\\\\\Rightarrow m+1=0,~~~~~5m-1=0\\\\\Rightarrow m=-1,~\dfrac{1}{5}.$

So, the general solution of the given equation is

$y(t)=Ae^{-t}+Be^{\frac{1}{5}t}.$

Differentiating with respect to t, we get

$y^\prime(t)=-Ae^{-t}+\dfrac{B}{5}e^{\frac{1}{5}t}.$

According to the given conditions, we have

$y(0)=0\\\\\Rightarrow A+B=0\\\\\Rightarrow B=-A~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(ii)$

and

$y^\prime(0)=2.8\\\\\Rightarrow -A+\dfrac{B}{5}=2.8\\\\\Rightarrow -5A+B=14\\\\\Rightarrow -5A-A=14~~~~~~~~~~~~~~~~~~~~~~~~~~~[\textup{Uisng equation (ii)}]\\\\\Rightarrow -6A=14\\\\\Rightarrow A=-\dfrac{14}{6}\\\\\Rightarrow A=-\dfrac{7}{3}.$

From equation (ii), we get

$B=\dfrac{7}{3}.$

Thus, the required solution is

$y(t)=-\dfrac{7}{3}e^{-t}+\dfrac{7}{3}e^{\frac{1}{5}t}.$

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2 years ago

The Gotemba walking trail up Mount Fuji is about 9km long. Walkers need to return from the 18km walk by 8pm. Toshi estimates tha

stiv31 [10]

Answer:

Toshi must begin his walk at 11:00 AM in order that he can return by 8:00 PM.

Step-by-step explanation:

Since the Gotemba walking trail up Mount Fuji is about 9km long, and walkers need to return from the 18km walk by 8pm, if Toshi estimates that he can walk up the mountain at 1.5km / h on average, and down at twice that speed , these speeds taking into account meal breaks and rest times, to determine what is the latest time he can begin his walk so that he can return by 8pm the following calculation must be performed:

Climb: 1.5 km / h

Descent: 2 x 1.5 km / h = 3 km / h

Climb: 9 km / 1.5 km / h = 6 hours

Descent: 9km / 3 km / h = 3 hours

Total: 9 hours

8 PM = 20:00

20:00 - 09:00 = 11:00

Thus, Toshi must begin his walk at 11:00 AM in order that he can return by 8:00 PM.

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3 years ago

I just don't know where to start. I'm doing Algebra 2 stuff with sohcahtoa.

mezya [45]

Hello!

As you can see, 22 is the opposite angle. We are looking for the adjacent angle. Let's just call it x. Using SOH CAH TOA, we see that the tangent is the opposite divided by the adjacent side. Let's use that.

tan54≈1.38

This gives us the equation below.

1.38=22/x

Let's solve this equation for x.

1.38x=22

x=22/1.38

x=15.94

Therefore, the length of the shadow is about 15.94 meters.

I hope this helps!

7 0

2 years ago

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