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3 years ago

It's the same possible answers for both questions

Mathematics

1 answer:

Anettt [7]3 years ago

5 0

Answer:

q10, vertically opp angles

q11, corresponding angles

Step-by-step explanation:

Topic: angle properties

If you like to venture further, feel free to check out my insta (learntionary). It would be best if you could give it a follow. I'll be constantly posting math tips and notes! Thanks!

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What are the coordinates of the center of the ellipse shown below?

photoshop1234 [79]

Answer:

Option D (7, -3)

Step-by-step explanation:

We know that the general equation of an ellipse has the form:

$\frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1$

Where the point (h, k) are the coordinates of the center of the ellipse

In this case the equation of the ellipse is:

$\frac{(x-7)^2}{4} + \frac{(y+3)^2}{16} = 1$

Then

$h=7\\\\k = -3$

So The coordinates of the center of the ellipse are (7, -3)

8 0

3 years ago

Read 2 more answers

Jesse takes his dog and cat for their annual vet visit.Jesse's dog weighs 23 pounds.The vet tells him his cat's weight is 5/8 as

Kobotan [32]

If I’m not mistaken the cats weight is 14.4

5 0

4 years ago

PLEASE HELP WITH THIS IF I DONR GET IT RIGHT I WILL FAIL I WILL GIVE BRAINLIEST AND CORRECT ANSWE ONLY

Akimi4 [234]

Answer:

A= 21 m

Step-by-step explanation:

7 0

3 years ago

1496 mm2 to square centimetres.

Kazeer [188]

Answer:

14.96

Step-by-step explanation:

1496mm *mm = 1496 * 1/10 cm * 1/10cm =1496/100 cm2

3 0

3 years ago

Read 2 more answers

I need help with this problem from the calculus portion on my ACT prep guide

LenaWriter [7]

Given a series, the ratio test implies finding the following limit:

$\lim _{n\to\infty}\lvert\frac{a_{n+1}}{a_n}\rvert=r$

If r<1 then the series converges, if r>1 the series diverges and if r=1 the test is inconclusive and we can't assure if the series converges or diverges. So let's see the terms in this limit:

$\begin{gathered} a_n=\frac{2^n}{n5^{n+1}} \\ a_{n+1}=\frac{2^{n+1}}{(n+1)5^{n+2}} \end{gathered}$

Then the limit is:

$\lim _{n\to\infty}\lvert\frac{a_{n+1}}{a_n}\rvert=\lim _{n\to\infty}\lvert\frac{n5^{n+1}}{2^n}\cdot\frac{2^{n+1}}{\mleft(n+1\mright)5^{n+2}}\rvert=\lim _{n\to\infty}\lvert\frac{2^{n+1}}{2^n}\cdot\frac{n}{n+1}\cdot\frac{5^{n+1}}{5^{n+2}}\rvert$

We can simplify the expressions inside the absolute value:

$\begin{gathered} \lim _{n\to\infty}\lvert\frac{2^{n+1}}{2^n}\cdot\frac{n}{n+1}\cdot\frac{5^{n+1}}{5^{n+2}}\rvert=\lim _{n\to\infty}\lvert\frac{2^n\cdot2}{2^n}\cdot\frac{n}{n+1}\cdot\frac{5^n\cdot5}{5^n\cdot5\cdot5}\rvert \\ \lim _{n\to\infty}\lvert\frac{2^n\cdot2}{2^n}\cdot\frac{n}{n+1}\cdot\frac{5^n\cdot5}{5^n\cdot5\cdot5}\rvert=\lim _{n\to\infty}\lvert2\cdot\frac{n}{n+1}\cdot\frac{1}{5}\rvert \\ \lim _{n\to\infty}\lvert2\cdot\frac{n}{n+1}\cdot\frac{1}{5}\rvert=\lim _{n\to\infty}\lvert\frac{2}{5}\cdot\frac{n}{n+1}\rvert \end{gathered}$

Since none of the terms inside the absolute value can be negative we can write this with out it:

$\lim _{n\to\infty}\lvert\frac{2}{5}\cdot\frac{n}{n+1}\rvert=\lim _{n\to\infty}\frac{2}{5}\cdot\frac{n}{n+1}$

Now let's re-writte n/(n+1):

$\frac{n}{n+1}=\frac{n}{n\cdot(1+\frac{1}{n})}=\frac{1}{1+\frac{1}{n}}$

Then the limit we have to find is:

$\lim _{n\to\infty}\frac{2}{5}\cdot\frac{n}{n+1}=\lim _{n\to\infty}\frac{2}{5}\cdot\frac{1}{1+\frac{1}{n}}$

Note that the limit of 1/n when n tends to infinite is 0 so we get:

$\lim _{n\to\infty}\frac{2}{5}\cdot\frac{1}{1+\frac{1}{n}}=\frac{2}{5}\cdot\frac{1}{1+0}=\frac{2}{5}=0.4$

So from the test ratio r=0.4 and the series converges. Then the answer is the second option.

8 0

1 year ago

It's the same possible answers for both questions​

It's the same possible answers for both questions