How do you find the area of this
How do you find the area of this - 1

Answers

Answer 1
Answer:

Answer:

156 cm^2

Step-by-step explanation:

area of a paralelogram is Base times hieght

13 times 12


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How to factorise this?!
4x-14xy

Answers

Find the highest common factor, which is 2X, therefore, this goes outside of the bracket. So inside goes 4x/2x= 2 & 14xy/2x= 7y2x(2-7y)When you multiply out of the brackets, everything should be equal
Take a factor of x out of the equation:
2x(2-7y)

What is the graph of 3x + 5y = –15?

Answers

Answer:

Graph is option A

Step-by-step explanation:

3x + 5y = –15

To graph the equation , we need to get y alone

3x + 5y = –15

Subtract 3x on both sides

5y = -3x–15

Divide whole equation by 5

y=-(3x)/(5) -3

To graph this equation , we need to make a table

Plug in some number for x  and find out y

x                y=-(3x)/(5) -3

0                -3

-5                0

Two points on the graph is (-5,0)  and (0,-3)

Graph is option A

*15 points* any help would be appreciated

Answers

Answer:

1

3

4

7

Step-by-step explanation:

this is what i think it might not be right but i tried my best ;3

can i please have brainly

Answer:

1, 4, 6 those are correct

Step-by-step explanation:

What is 1.2 % of 0.5?

Answers

The answer is 2.4 because you have to divide 1.2 by 0.5 and that gives you your answer 

For what values of r does the function y = erx satisfy the differential equation y'' + 5y' + 6y = 0?

Answers

Answer:

\huge\boxed{r=-2,-3}

Step-by-step explanation:

To solve for the values of r where the differentialequationy'' + 5y' + 6y = 0 is satisfied by the function y=e^(rx), we first need to find the first and second derivatives of y with respect to x, treating r as a constant.

\left[\frac{}{}y\frac{}{}\right]'=\left[\frac{}{}e^(rx)\frac{}{}\right]'

↓ applying the chain rule to the right side:   \displaystyle \left[\frac{}{}f(x)^a\frac{}{}\right]' = a \cdot f(x)^((a\, -\, 1)) \cdot f'(x) where f(x) = e^x and a = r

y'=r\cdot e^((rx \,-\, 1)) \cdot e^x

↓ simplifying using the exponentbaseproduct rule:   x^a \cdot x^b = x^((a \, +\, b))

y' = re^(\left[(rx \,-\, 1)\, +\, 1\right])

\boxed{y' = re^(rx)} \ \ \leftarrow \ \ \text{first derivative}

─────────────────────────────────

↓ taking the derivative of y with respect to x

y'' = \left[\frac{}{} re^(rx)\frac{}{}\right]'

takingout the constant (r) on the right side

y'' = r\left[\frac{}{} e^(rx)\frac{}{}\right]'

↓ simplifying by substituting in the first derivative

y'' = r \cdot y'

y'' = r \cdot re^(rx)

\boxed{y'' = r^2e^(rx)} \ \ \leftarrow \ \ \text{second derivative}

Now, we can plug these derivative expressions into the differential equation and solvefor r.

y'' + 5y' + 6y = 0

pluggingin the derivativeexpressions (think of y as the zeroth derivative of itself)

r^2e^(rx) + 5(re^(rx)) + 6(e^(rx)) = 0

factoringoute^(rx) from the left side

e^(rx)(r^2 + 5r + 6) = 0

factoring the second-degree polynomial factor

e^(rx)(r + 2)(r + 3) = 0

splitting into 3 equations using the zero product property: \text{If } ABC = 0,\text{ then } A=0\text{ or }B=0\text{ or }C=0.

First Equation

e^(rx)=0

↓ taking the natural log of both sides

rx = \ln(0)

\implies \text{un}\text{de}\text{fi}\text{ne}\text{d}

Second Equation

r+2=0

subtracting 2 from both sides

\huge\boxed{r=-2}

Third Equation

r+3=0

subtracting 3 from both sides

\huge\boxed{r=-3}

How to solve equation 2x/5+1=7?
How to solve equation x/3-4=10?

Answers

If you would like to solve the equation 2x / 5 + 1 = 7, you can do this using the following steps:

2x / 5 + 1 = 7
2x + 5 = 5 * 7
2x = 35 - 5
2x = 30
x = 15

If you would like to solve the equation x/3 - 4 = 10, you can do this using the following steps:

x/3 - 4 = 10
x/3 = 10 + 4
x/3 = 14
x = 14 * 3 = 42