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3 years ago

How many whole sandwiches can be made with 3 pounds of turkey if each sandwich has 5 ounces of turkey on it? 4 6 9 10

2 answers:

6 0

The answer is 9. there are 16 ounces in a pound so if you times that by three it is 48 ounces. 9 sandwiches equals 45 ounces

alekssr [168]3 years ago

4 0

Answer:

9 whole sandwiches can be made

Step-by-step explanation:

As we know 1 pound = 16 ounces

Therefore, 3 pounds = 16 × 3 = 48 ounces

There are 48 ounces (3 pounds) of turkey.

In each sandwich turkey is used = 5 ounces

48 ounces turkey is used = 48 ÷ 5 = 9.6 ≈ 9

9 whole sandwiches can be made with 3 pounds of turkey.

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Combine like terms 4a2-3b2+ 4b + 10a2

jolli1 [7]

Answer:

14a² - 3b² + 4b

Step-by-step explanation:

Like terms are the terms that have similar stuff on it, like same variables, no variables, and stuff like that.

4a2-3b2+ 4b + 10a2

4a² and 10a² is similar. 4a² + 10a² = 14a²

Now, the equation is 14a² - 3b² + 4b. There is no more like terms left, so that is the answer.

Hope this helps!!

-Ketifa

4 0

2 years 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

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