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3 years ago

Which statements are true regarding the sides and angles of the triangle? Select three options.

Mathematics

1 answer:

ss7ja [257]3 years ago

3 0

Answer:

What are the options?

Step-by-step explanation:

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Helppp meee please this one confuses me

slava [35]

$= (3x + 2)(x + 1) \\ = 3x(x + 1) + 2(x + 1) \\ = {3x}^{2} + 3x + 2x + 2 \\ = {3x}^{2} + 5x + 2$

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2 years ago

Need help finding out something, refer to picture

Alja [10]

Answer:

He made a mistake in the second line.

Step-by-step explanation:

He factored out x out of 6-2x and 5+4x, which is not possible because there is not an x in both 6 and 5.

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3 years ago

Read 2 more answers

Please answer question b.

ser-zykov [4K]

It is 10.l; Divide 1000 by 93 1/2 and you get 10.6951872... which is a really long decimal, but keep in mind they are asking for the GREATEST POSSIBLE NUMBER that Jodi can cut out, which means that they are looking for a whole number; how much possible?

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3 years ago

Can someone help with this!

Sveta_85 [38]

In the given image, $a_{23}$ is 5. The correct option is the first option 5

<h3>Matrix notation</h3>

From the question, we are to determine which entry in the given matrix represents $a_{23}$

NOTE: For any entry denoted as $a_{mn}$ , it represents the entry in the <em>m</em>-th row and <em>n</em>-th column.

Thus,

$a_{23}$ represents the entry in the 2nd row and 3rd column.

In the given image, the entry in the 2nd row and 3rd column is 5.

Hence, in the given image, $a_{23}$ is 5. The correct option is the first option 5

Learn more on Matrix notation here: brainly.com/question/2382978

#SPJ1

5 0

2 years ago

If y=e5t is a solution to the differential equation

a_sh-v [17]

Answer:

k = 30, $y(t) = C_1e^{5t}+C_2e^{6t}$

Step-by-step explanation:

Since $y=e^{5t}$ is a solution, then it must satisfy the differential equation. So, we calculate the derivatives and replace the value in the equation. We have that

$\frac{d^2y}{dt^2} = 25 e^{5t},\frac{dy}{dt} = 5e^{5t}$

Then, replacing the derivatives in the equation we have:

$25e^{5t}-11(5)e^{5t}+ke^{5t}=0 e^{5t}(25-55+k) =0$

Since $e^{5t}$ is a positive function, we have that

$25-55+k = 0 \rightarrow k = 30$ .

Now, consider a general solution $y(t) = Ae^{rt}, A \in \mathbb{R}$ , then, by calculating the derivatives and replacing them in the equation, we get

$Ae^{rt}(r^2-11r+30)=0$

We already know that r=5 is a solution of the equation, then we can divide the polynomial by the factor (r-5) to the get the other solution. If we do so, we get that (r-6)=0. So the other solution is r=6.

Therefore, the general solution is

$y(t) = C_1e^{5t}+C_2e^{6t}$

8 0

3 years ago