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3 years ago

Find the product ( 6x+4)^2

Mathematics

2 answers:

topjm [15]3 years ago

7 0

(6x + 4) * (6x + 4)
= 36x^2+ 24x + 24x + 16
= 36x^2 + 48x + 16

Sophie [7]3 years ago

3 0

The answer should be 22. hope that helps.

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What is the value of x in the equation <img src="https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B5%7Dx%20-%5Cfrac%7B2%7D%7B3%7D" id="Te

SVEN [57.7K]

Answer:

x= 200

Step-by-step explanation:

The given equation is:

1/5 x - 2/3 y = 30

The value of y is 15

Put the value of y in the given equation:

1/5 x - 2/3(15) = 30

1/5 x- 10 = 30

Take the L.C.M of the denominator

x-50/5 = 30

Move the denominator to the R.H.S

x-50 = 30*5

x-50 = 150

Combine the like terms

x= 150+50

x = 200

The value of x is 200...

5 0

3 years ago

Read 2 more answers

|18| what is the absolute value?

Tju [1.3M]

The answer is 18, the answer is always just the number in the brackets( hope this helps)

7 0

3 years ago

Zachary invested $5,300 in an account paying an interest rate of 9 % compounded

son4ous [18]

Answer:625

625 in delta math

Step-by-step explanation:

3 0

3 years ago

Can someone please help me with these? I will give you the brainliest answer and also give extra points! I really need help :(

Ierofanga [76]

Answer:

1. No solution

2. Infinite many solutions

3. One solution

4. No solution

5. No solution

6. One solution

7. No solution

8. One solution

9. Infinite many solutions

10. Infinite many solutions

Step-by-step explanation:

6 0

2 years ago

Given: y" - 2y' = 6t + 5e^2t. Find the correct form to use for y_p if the equation is solved using Undetermined coefficients. Do

const2013 [10]

Answer:

$y_p=A+Bt+Ce^{2t}$

Step-by-step explanation:

Given: $y'' - 2y' = 6t + 5e^{2t}$ .

we need to find the correct form for $y_p$ if the equation is solve using undetermined coefficients.

A first order differential equation $\frac{\mathrm{d} y}{\mathrm{d} x}=f\left ( x,y \right )$ is said to be homogeneous if $f(tx,ty)=f(x,y)$ for all t.

Consider homogeneous equation $y''-2y'=0$

Let $y=e^{rt}$ be the solution .

We get $(r^2-2r)e^{rt}=0$

Since $e^{rt}\neq 0$ , $r^2-2r=0$ .

So, we get solution as $y_c=c_1+c_2e^{2t}$

As constant term and $e^{2t}$ are already in the R.H.S of equation

$y" - 2y' = 6t + 5e^{2t}$ , we can take $y_p$ as $y_p=A+Bt+Ce^{2t}$

6 0

3 years ago