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3 years ago

Brad has 3 pairs of pants, 7 tee shirts, and 2 pairs of sneakers. How many different outfit combinations can he wear

Mathematics

2 answers:

irakobra [83]3 years ago

8 0

Answer:

Step-by-step explanation:

MArishka [77]3 years ago

6 0

HE CAN WEAR 42 DIFFERENT COMBOS

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Which situation is an example of an observational study?

jarptica [38.1K]

Answer: this is how i feel about the answers A 80% b 0% c 90% d 0%

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2 years ago

If someone can help me out :D!

Nataliya [291]

Hello

Answer:

28.378

Explanation:

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2 years ago

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Find the average rate of change of the function between the given values of x. y = 9+ 5x + 0.5x2 between x = 2 and x = 4.

svet-max [94.6K]

Answer:

The average rate of change is 8.

Step-by-step explanation:

The formula to calculate the average rate of change of a function F(x) is:

$\frac{F(b)-F(a)}{b-a}$

In this case, F(x) = $0.5x^{2} +5x+9$

a=2 and b=4

You have to evaluate x=2 (which is a in the formula) and x=4 (which is b in the formula) in the function.

In order to obtain F(b) and F(a) you have to replace x=4 and x=2 in the given function:

F(b) = $(0.5)4^{2} + 5(4) +9= 37$

F(a) = $(0.5)2^{2} + 5(2)+9=21$

$\frac{F(b)-F(a)}{b-a} = \frac{37-21}{4-2}=\frac{16}{2} = 8$

The answer is 8.

3 0

2 years ago

You must use the LCD 1/7 + 3/14 *Convert Into into like fractions and add them

Tresset [83]

The LCD of 1/17 and 3/14 is 14.

1/7 = 2/14

3/14 = 3/14

2 + 3 = 5

The denominators are the same, so it will remain the same.

= 5/14 <===

5/14 is your answer.

I hope this helps! :)

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3 years ago

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Solve y'' + 10y' + 25y = 0, y(0) = -2, y'(0) = 11 y(t) = Preview

svetlana [45]

Answer: The required solution is

$y=(-2+t)e^{-5t}.$

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

$y^{\prime\prime}+10y^\prime+25y=0,~~~~~~~y(0)=-2,~~y^\prime(0)=11~~~~~~~~~~~~~~~~~~~~~~~~(i)$

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

$m^2e^{mt}+10me^{mt}+25e^{mt}=0\\\\\Rightarrow (m^2+10y+25)e^{mt}=0\\\\\Rightarrow m^2+10m+25=0~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~[\textup{since }e^{mt}\neq0]\\\\\Rightarrow m^2+2\times m\times5+5^2=0\\\\\Rightarrow (m+5)^2=0\\\\\Rightarrow m=-5,-5.$

So, the general solution of the given equation is

$y(t)=(A+Bt)e^{-5t}.$

Differentiating with respect to t, we get

$y^\prime(t)=-5e^{-5t}(A+Bt)+Be^{-5t}.$

According to the given conditions, we have

$y(0)=-2\\\\\Rightarrow A=-2$

and

$y^\prime(0)=11\\\\\Rightarrow -5(A+B\times0)+B=11\\\\\Rightarrow -5A+B=11\\\\\Rightarrow (-5)\times(-2)+B=11\\\\\Rightarrow 10+B=11\\\\\Rightarrow B=11-10\\\\\Rightarrow B=1.$

Thus, the required solution is

$y(t)=(-2+1\times t)e^{-5t}\\\\\Rightarrow y(t)=(-2+t)e^{-5t}.$

6 0

2 years ago