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FrozenT [24]
3 years ago
9

Can you help me If y= -4 when x = 10, find y when x = 5

Mathematics
2 answers:
Artyom0805 [142]3 years ago
8 0
The correct answer is-2
Dafna1 [17]3 years ago
7 0

Answer:

-2

I think it's -2 because you're dividing x by 2 so you'd divide -4 by 2.

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The first digit of a string of 2002 digits is a 1. Any two-digit number formed by consecutive digits within this string is divis
Umnica [9.8K]

The largest possible last digit in the string of 2002 digits and number divisible by 19 or 31 is 9.

Given the first digit of a string of 2002 digits is 1 and the two digit number formed by consecutive digits within the string is divisible by 19 or 31.

We have to tell the last largest digit of such number.

Two digit numbers divisible by 19=19,38,57,76,95.

Two digit numbers divisible by 31=31,62,93,124

Number started with 1 =19

Last digit is 9

We have said that the number should be divisible by 19 or 31 not from both and started with 1.

Hence the largest possible last digit and number divisible by 19 or 31 in this string is 9.

Learn more about digits at brainly.com/question/26856218

#SPJ4

4 0
2 years ago
HELP!!! Use the properties of exponents to simplify the expression:
Dvinal [7]
There you go the answer key

6 0
3 years ago
Rewrite this expression using distributive property: 4a + 4b - 5
Triss [41]

Answer:

4(a + b) - 5

Step-by-step explanation:

Pull out the common factor

5 0
2 years ago
Which figure best represents a kite?
suter [353]

Answer:

4th one: A kite is the answer.

4 0
3 years ago
Set up but do not solve for the appropriate particular solution yp for the differential equation y′′+4y=5xcos(2x) using the Meth
taurus [48]

Answer:

So, solution of  the differential equation is

y(t)=-\frac{5x^2}{4}\cot 2x\cdot \cos  2x+c_1e^{-2it}+c_2e^{2it}\\

Step-by-step explanation:

We have the given differential equation: y′′+4y=5xcos(2x)

We use the Method of Undetermined Coefficients.

We first solve the homogeneous differential equation y′′+4y=0.

y''+4y=0\\\\r^2+4=0\\\\r=\pm2i\\\\

It is a homogeneous solution:

y_h(t)=c_1e^{-2i t}+c_2e^{2i t}

Now, we finding a particular solution.

y_p(t)=A5x\cos 2x\\\\y'_p(t)=A5\cos 2x-A10x\sin 2x\\\\y''_p(t)=-A20\sin 2x-A20x\cos 2x\\\\\\\implies y''+4y=5x\cos 2x\\\\-A20\sin 2x-A20x\cos 2x+4\cdot A5x\cos 2x=5x\cos 2x\\\\-A20\sin 2x=5x\cos 2x\\\\A=-\frac{x}{4} \cot 2x\\

we get

y_p(t)=A5\cos 2x\\\\y_p(t)=-\frac{5x^2}{4}\cot 2x\cdot \cos  2x\\\\\\y(t)=y_p(t)+y_h(t)\\\\y(t)=-\frac{5x^2}{4}\cot 2x\cdot \cos  2x+c_1e^{-2it}+c_2e^{2it}\\

So, solution of  the differential equation is

y(t)=-\frac{5x^2}{4}\cot 2x\cdot \cos  2x+c_1e^{-2it}+c_2e^{2it}\\

7 0
3 years ago
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