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2 years ago

Sharnae raised $52 less than lee during a school fundraiser. Lee raised $384. How much did sharnae raised?

Mathematics

1 answer:

neonofarm [45]2 years ago

6 0

Answer:

$352

Step-by-step explanation:

So subtract the amount Lee made from the $52 and you get your answer. i hope i help u

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The age of cars in the staff parking lot of a suburban college is uniformly distributed from six months (0.5 years) to 9.5 years

Nikolay [14]

5.5 is the answer stay safe

4 0

3 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

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

From equation (ii), we get

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

Thus, the required solution is

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

7 0

2 years ago

In 1999 Daniel was 146 cm tall. He grew to be 176 cm by the year 2006. What was Daniel's rate of growth over this period of his

andreyandreev [35.5K]

Answer:

he grows by 5 cm every year between 1999 and 2006

Step-by-step explanation:

This is a arithmetic progression problem with the formula;

T_n = a + (n - 1)d

We are told that In 1999 Daniel was 146 cm tall. He grew to be 176 cm by the year 2006.

Thus;

a = 146

d = 2006 - 1999 = 7

Thus;

176 = 146 + (7 - 1)d

176 - 146 = 6d

30 = 6d

d = 30/6

d = 5 cm

Thus, he grows by 5 cm every year between 1999 and 2006

3 0

2 years ago

The time, T (seconds) it takes for a pot of water to boil is inversely proportional to the cooker setting, H, applied to the pot

kherson [118]

Answer:

200

Step-by-step explanation:

T = x * (1/H)

x = T*H

x = 2800

so for H = 14

T = 2800 * (1/14) = 200

7 0

3 years ago

Forty-seven is less than or equal to five times three,added to nine

irina1246 [14]

Answer:

You need to do 5 times 3 for 15 a d then add 9 to get 24, so 47 is not less than or equal to 5 times 3 add 9

Step-by-step explanation:

8 0

2 years ago

Read 2 more answers