Can someone simplify (3x)^4

Which of the following would be the criterion for establishing similarity in the two triangles?

Luba_88 [7]

Answer:

<h3>A) Not enough information or not similar.</h3>

Step-by-step explanation:

We are given a triangle with a segment that connects two points of the two sides of given triangle.

There are different triangle similarity theorems are available there as given in options.

Side Side Side (SSS) triangle similarity theorem.

Side Angle Side (SAS) triangle similarity theorem.

Angle Angle (AA) triangle similarity theorem.

But in the shown figure, we are not provided much information about it's sides and angles.

Therefore, correct option is :

<h3>A) Not enough information or not similar.</h3>

5 0

3 years ago

Factor 9x2+24x+16<br> Enter your answer in the boxes.<br><br> 9x2+24x+16= ( )2

Vitek1552 [10]

Answer:

123

Step-by-step explanation:

if the perimeter of a rectrangular field of length 25 m is 86 m find its breadthif the perimeter of a rectrangular field of length 25 m is 86 m find its breadthif the perimeter of a rectrangular field of length 25 m is 86 m find its breadthif the perimeter of a rectrangular field of length 25 m is 86 m find its breadthif the perimeter of a rectrangular field of length 25 m is 86 m find its breadthif the perimeter of a rectrangular field of length 25 m is 86 m find its breadthif the perimeter of a rectrangular field of length 25 m is 86 m find its breadthif the perimeter of a rectrangular field of length 25 m is 86 m find its breadthif the perimeter of a rectrangular field of length 25 m is 86 m find its breadthif the perimeter of a rectrangular field of length 25 m is 86 m find its breadthif the perimeter of a rectrangular field of length 25 m is 86 m find its breadthif the perimeter of a rectrangular field of length 25 m is 86 m find its breadthif the perimeter of a rectrangular field of length 25 m is 86 m find its breadthif the perimeter of a rectrangular field of length 25 m is 86 m find its breadthif the perimeter of a rectrangular field of length 25 m is 86 m find its breadthif the perimeter of a rectrangular field of length 25 m is 86 m find its breadthif the perimeter of a rectrangular field of length 25 m is 86 m find its breadthif the perimeter of a rectrangular field of length 25 m is 86 m find its breadthif the perimeter of a rectrangular field of length 25 m is 86 m find its breadthif the perimeter of a rectrangular field of length 25 m is 86 m find its breadth

7 0

3 years ago

julio has 6 pounds of flour to make 5 cakes. if the cakes are equal in size how much flour will be used in each cake?

NeTakaya

Answer:

6×5=30

Step-by-step explanation:

6 0

3 years ago

Read 2 more answers

A community garden center host a plant giveaway every spring to help community members start there gardens. Last year the give a

OlgaM077 [116]

Answer:

<h2>195 plants (verified) ✅</h2>

Step-by-step explanation:

We will use a proportion to solve this.

x/y = x/y

let x = the number of families in each case

let y = the number of plants in each case

Since we don't know the second y, we will leave that as y and solve for it.

50/150 = 65/y

Cross-multiply

50y = 9750

Divide

195 = y

Now, we can check this by substituting the value of y back into the equation and seeing if both sides equal.

50 (195) = 9750

9750 = 9750 ✅

5 0

3 years ago

Solve this differential equation using power series and indicial roots about (0,0):

Ivanshal [37]

Let $y=\displaystyle\sum_{k\ge0}a_kx^k$ , so that

$y'=\displaystyle\sum_{k\ge1}ka_kx^{k-1}$
$y''=\displaystyle\sum_{k\ge2}k(k-1)a_kx^{k-2}$

Substituting into the ODE gives

$\displaystyle3x\sum_{k\ge2}k(k-1)a_kx^{k-2}+6\sum_{k\ge1}ka_kx^{k-1}+\sum_{k\ge0}a_kx^k=0$
$\displaystyle3\sum_{k\ge2}k(k-1)a_kx^{k-1}+6\sum_{k\ge1}ka_kx^{k-1}+\sum_{k\ge0}a_kx^k=0$

The first series starts with a linear term, while the other two start with a constant. Extract the first term from each of the latter two series:

$\displaystyle6\sum_{k\ge1}ka_kx^{k-1}=6\sum_{k\ge2}ka_kx^{k-1}+6a_1$
$\displaystyle\sum_{k\ge0}a_kx^k=\sum_{k\ge1}a_kx^k+a_0$

Finally, to get the series to start at the same index, shift the index of the first two series by replacing $k$ with $k+1$ . Then the ODE becomes

$\displaystyle3\sum_{k\ge1}k(k+1)a_{k+1}x^k+6\sum_{k\ge1}(k+1)a_{k+1}x^k+\sum_{k\ge1}a_kx^k+6a_1+a_0=0$

which can be consolidated to get

$\displaystyle\sum_{k\ge1}\bigg[(3k(k+1)+6(k+1))a_{k+1}+a_k\bigg]x^k+6a_1+a_0=0$
$\displaystyle\sum_{k\ge1}\bigg[3(k+1)(k+2)a_{k+1}+a_k\bigg]x^k+6a_1+a_0=0$

You're fixing the solution so that it contains the origin, which means

$y(0)=\displaystyle\sum_{k\ge0}a_kx^k=a_0=0$

which in turn means $a_1=0$ . With the given recurrence, it follows that $a_k=0$ for all $k\ge2$ , so the solution would be $y=0$ . This is to be expected, since $x=0$ is clearly a singular point for the ODE.

8 0

4 years ago