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2 years ago

HELP PLEASE ONE QUESTION I DONT GET IT 6th GRADE MATH

Mathematics

2 answers:

grigory [225]2 years ago

8 0

Answer:

-2.65

Step-by-step explanation:

-1.2 - 1.45 = -2.65

you could just simplify fraction into decimal and use calculator

kati45 [8]2 years ago

8 0

Answer:

-2.65

Step-by-step explanation:

We can change the fraction to a decimal

29/20 *5/5 = 145/100 = 1.45

-1.2 - 1.45

-2.65

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HELP<br> I also need the answer to this

ser-zykov [4K]

Hey!

To divide fractions, you have to keep, change, flip.

Keep 3 1/2(7/2)

Change ÷ to ×

Flip 2 1/4(9/4) = 4/9

Now multiply the improper fractions

$\frac{7}{2} \times \frac{4}{9} = \frac{28}{18} = 1 \frac{10}{18} = 1 \frac{5}{9}$

<em>The 1st choice matches the answer. The answer is the 1st choice.</em>

Good luck and hope this helps! :)

5 0

2 years ago

QUESTION 1

Alexeev081 [22]

Answer:

1. 9

2. 10

Step-by-step explanation:

idk if those first 2 are right

6 0

2 years ago

Identify the transformation of the function f(x)=|x| by observing the equation of the function g(x)=|x+3|

Alla [95]

Answer:

a horizontal translation by 3 units left

Step-by-step explanation:

f(x)= |x|

we are given with absolute function f(x)

g(x) = |x+3|

To get g(x) from f(x) , 3 is added with x

If any number is added with x then the graph of the function move to the left

Here 3 is added with x, so the graph of f(x) moves 3 units left to get g(x)

So there will be a horizontal translation by 3 units

6 0

2 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

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