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3 years ago

Can someone help me solve this?

Mathematics

1 answer:

julsineya [31]3 years ago

8 0

Answer:

im sorry i cant see that

Step-by-step explanation:

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❗️HELP ASAP❗️

tangare [24]

I think the answer is G.

I think this because of the fact families (like my elementary teacher used to say)
It’s like about multiplication of fact families so make it where it would have fact families with those numbers as the sums.
I hope this helps! Be safe out there.

3 0

3 years ago

Which of the following irrational numbers will result in a rational number when squared? Put a check mark next to all correct an

Anastaziya [24]

Answer:

The square root of 7 and the square root of 2

Step-by-step explanation:

5 0

3 years ago

Solve y = x^2 +11 <br> for x.

lisabon 2012 [21]

Answer:

x = ± $\sqrt{y-11}$

Step-by-step explanation:

Given

y = x² + 11 ( subtract 11 from both sides )

y - 11 = x² ( take the square root of both sides )

± $\sqrt{y-11}$ = $\sqrt{x^2}$ , hence

x = ± $\sqrt{y-11}$

7 0

4 years ago

Read 2 more answers

Write 2x+y=5 in slope intercept form

mylen [45]

2x+y=5
To put this in slope intercept form, you simply have to get y by itself on the left side of the equation.
2x+y=5
2x+y-2x=5-2x
y=-2x+5
Now, I had to switch the right side around like that, because slope intercept form is y=mx+b
Because you're subtracting 2x from 5, it becomes 5 added to -2x when it's put in the correct form.
y=-2x+5

6 0

3 years ago

Read 2 more answers

Determine whether the given function is a solution to the given differential equation.

lesantik [10]

Given :

A function , x = 2cos t -3sin t .....equation 1.

A differential equation , x'' + x = 0 .....equation 2.

To Find :

Whether the given function is a solution to the given differential equation.

Solution :

First derivative of x :

$x'=\dfrac{d(2cos t - 3sin t)}{dt}\\\\x'=\dfrac{d(2cost)}{dt}-\dfrac{(3sint)}{dt}\\\\x'=-2sint-3cost$

Now , second derivative :

$x''=\dfrac{d(-2sint-3cost)}{dt}\\\\x''=-\dfrac{d(2sint)}{dt}-\dfrac{d(3cost)}{dt}\\\\x''=-2cost+3sint$

( Note : derivative of sin t is cos t and cos t is -sin t )

Putting value of x'' and x in equation 2 , we get :

=(-2cos t + 3sin t ) + ( 2cos t -3sin t )

= 0

So , x'' and x satisfy equation 2.

Therefore , x function is a solution of given differential equation .

Hence , this is the required solution .

8 0

3 years ago