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3 years ago

(please break it down for me to understand):)

Mathematics

1 answer:

Ganezh [65]3 years ago

5 0

Answer:

1/50 times 2/99 = 2/4950

divide numerator and denominator by 2 and the answer you should get is 1/2475 and in decimal form it equals 0.00040404040404040

Step-by-step explanation:

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PLEASE HELP ME ANSWER ALL OF THESE QUESTIONS!!!!

daser333 [38]

Remember: a^2 + b^2 = c^2

64 + 49 = 113

√113 = $\sqrt{113}$

100 - 25 = 75

square root of 75 = 5 $\sqrt{3}$

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3 years ago

Determine which of the following statements is true concerning the values described in column #1 and column #2. Column #1 Column

Gelneren [198K]

Hi,

1)
$y=-2x^2-4x+12=-2(x^2+2x+1)+2+12=-2(x+1)^2+14\\\\
Vertex=(-1,14) \ x-coordinate=-1\\

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2)
$y=x^2-4x+3=x^2-4x+4-1=(x-2)^2-1\\\\
Vertex=(2,-1) \ x-coordinate=2\\


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3 0

3 years ago

Read 2 more answers

Find the solution(s) to (x-3)^2=49. Check all that apply

Natasha_Volkova [10]

Answer:

B. x = -4

E. x = 10

Step-by-step explanation:

$(x-3)^2=49 \\ \\ x - 3 = \pm \sqrt{49} \\ \\ x - 3 = \pm 7 \\ \\ x = 3 \pm 7 \\ \\ x = 3 + 7, \: \: x = 3 - 7, \\ \\ x = 10, \: \: x = - 4$

8 0

3 years ago

A car uses 1 2/5 gallons of gasoline to travel 50 3/4 miles. How far can the car travel, in miles, on 1 gallon of gasoline?

Leya [2.2K]

Answer:

30 9/20

Step-by-step explanation:

(50 3/4)-((2/5) x (50 3/4)

=30.45

=30.45/1

=30.45/1 x 100/100

=3045/100

=3045÷5/100÷5

=30 9/20

30.45= 30 9/20

Hope this helps have a good day.

8 0

3 years ago

Read 2 more answers

Use the variation of parameters method to solve the DR y" + y' - 2y = 1

postnew [5]

Answer:

$y(t)\ =\ C_1e^{-2t}+C_2e^t-t\dfrac{e^{-2t}}{3}-\dfrac{1}{3}$

Step-by-step explanation:

As given in question, we have to find the solution of differential equation

$y"+y'-2y=1$

by using the variation in parameter method.

From the above equation, the characteristics equation can be given by

$D^2+D-2\ =\ 0$

$=>D=\ \dfrac{-1+\sqrt{1^2+4\times 2\times 1}}{2\times 1}\ or\ \dfrac{-1-\sqrt{1^2+4\times 2\times 1}}{2\times 1}$

$=>\ D=\ -2\ or\ 1$

Since, the roots of characteristics equation are real and distinct, so the complementary function of the differential equation can be by

$y_c(t)\ =\ C_1e^{-2t}+C_2e^t$

Let's assume that

$y_1(t)=e^{-2t}$ $y_2(t)=e^t$

$=>\ y'_1(t)=-2e^{-2t}$ $y'_2(t)=e^t$

and g(t)=1

Now, the Wronskian can be given by

$W=y_1(t).y'_2(t)-y'_1(t).y_2(t)$

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

$=e^{-t}+2e^{-t}$

$=3e^{-t}$

Now, the particular solution can be given by

$y_p(t)\ =\ -y_1(t)\int{\dfrac{y_2(t).g(t)}{W}dt}+y_2(t)\int{\dfrac{y_1(t).g(t)}{W}dt}$

$=\ -e^{-2t}\int{\dfrac{e^t.1}{3.e^{-t}}dt}+e^{t}\int{\dfrac{e^{-2t}.1}{3.e^{-t}}dt}$

$=\ -e^{-2t}\int{\dfrac{1}{3}dt}+\dfrac{e^t}{3}\int{e^{-t}dt}$

$=\dfrac{-e^{-2t}}{3}.t-\dfrac{1}{3}$

$=-t\dfrac{e^{-2t}}{3}-\dfrac{1}{3}$

Now, the complete solution of the given differential equation can be given by

$y(t)\ =\ y_c(t)+y_p(t)$

$=C_1e^{-2t}+C_2e^t-t\dfrac{e^{-2t}}{3}-\dfrac{1}{3}$

5 0

3 years ago