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3 years ago

Don’t know the answer.

Mathematics

1 answer:

kvasek [131]3 years ago

6 0

They are both isosceles triangles

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What is the least common denominator of 1/4, 3/7, 1/11

inna [77]

Alright. One way you can do this is to list multiples of each denominator:

1/4:
4, 8, 12, 16, 20, 24, 28, 32, 36...... and so on

3/7:
7, 14, 21, 28, 35, 42, 49, 57, 63......

1/11;
11, 22, 33, 44, 55, 66, 77, 88......

Keep going until you see the least common denominator!

Now an easier way to do this is by multiplying all of the numbers in the denominator then get the factors:

4*7*11 = 308

And since those are not divisible by a whole number (besides 4 but 308 won't be divisible by 2*7*11)

Therefore 308 is the least common denominator.

6 0

3 years ago

I made 12 dog treats using 1 cup of peanut butter. If I want to make 36 dog treats, how much peanut butter will I need? Write a

MaRussiya [10]

Answer:

You will need 3 cups peanut butter to make 36 dogs treats.

Peanut butter Cup : The dog treats made = 1: 12

Step-by-step explanation:

12 dog treats = 1 cup

So, by UNITARY METHOD

1 dog treat = (1/12) cup

So, 36 dog treats = 36 x (1/12) cup = 3 cups

So, you will need 3 cups peanut butter to make 36 dogs treats.

If 1 cup = 12 Dog Treats

then 1 x 3 cup = 12 x 3 dog treats

or, 3 Cups = 36 dog treats

The ratio of the Peanut butter cup : The dog treats made = 1: 12

6 0

3 years ago

Where is the center of CD given C(-6, 12) and D (13, 20)?

Law Incorporation [45]

(3.5, 16)

Midpoint Formula is
((x1+x2)/2, (y1+y2)/2)

3 0

3 years ago

Can someone check whether its correct or no? this is supposed to be the steps in integration by parts

Gwar [14]

Answer:

$\displaystyle - \int \dfrac{\sin(2x)}{e^{2x}}\: \text{d}x=\dfrac{\sin(2x)}{4e^{2x}}+\dfrac{\cos(2x)}{4e^{2x}}+\text{C}$

Step-by-step explanation:

$\boxed{\begin{minipage}{5 cm}\underline{Integration by parts} \\\\$\displaystyle \int u \dfrac{\text{d}v}{\text{d}x}\:\text{d}x=uv-\int v\: \dfrac{\text{d}u}{\text{d}x}\:\text{d}x$ \\ \end{minipage}}$

Given integral:

$\displaystyle -\int \dfrac{\sin(2x)}{e^{2x}}\:\text{d}x$

$\textsf{Rewrite }\dfrac{1}{e^{2x}} \textsf{ as }e^{-2x} \textsf{ and bring the negative inside the integral}:$

$\implies \displaystyle \int -e^{-2x}\sin(2x)\:\text{d}x$

Using <u>integration by parts</u>:

$\textsf{Let }\:u=\sin (2x) \implies \dfrac{\text{d}u}{\text{d}x}=2 \cos (2x)$

$\textsf{Let }\:\dfrac{\text{d}v}{\text{d}x}=-e^{-2x} \implies v=\dfrac{1}{2}e^{-2x}$

Therefore:

$\begin{aligned}\implies \displaystyle -\int e^{-2x}\sin(2x)\:\text{d}x & =\dfrac{1}{2}e^{-2x}\sin (2x)- \int \dfrac{1}{2}e^{-2x} \cdot 2 \cos (2x)\:\text{d}x\\\\& =\dfrac{1}{2}e^{-2x}\sin (2x)- \int e^{-2x} \cos (2x)\:\text{d}x\end{aligned}$

$\displaystyle \textsf{For }\:-\int e^{-2x} \cos (2x)\:\text{d}x \quad \textsf{integrate by parts}:$

$\textsf{Let }\:u=\cos(2x) \implies \dfrac{\text{d}u}{\text{d}x}=-2 \sin(2x)$

$\textsf{Let }\:\dfrac{\text{d}v}{\text{d}x}=-e^{-2x} \implies v=\dfrac{1}{2}e^{-2x}$

$\begin{aligned}\implies \displaystyle -\int e^{-2x}\cos(2x)\:\text{d}x & =\dfrac{1}{2}e^{-2x}\cos(2x)- \int \dfrac{1}{2}e^{-2x} \cdot -2 \sin(2x)\:\text{d}x\\\\& =\dfrac{1}{2}e^{-2x}\cos(2x)+ \int e^{-2x} \sin(2x)\:\text{d}x\end{aligned}$

Therefore:

$\implies \displaystyle -\int e^{-2x}\sin(2x)\:\text{d}x =\dfrac{1}{2}e^{-2x}\sin (2x) +\dfrac{1}{2}e^{-2x}\cos(2x)+ \int e^{-2x} \sin(2x)\:\text{d}x$