Question 1 (1 point)

Y=1/3x+2 Is this a linear equation or nonlinear

Vanyuwa [196]

Answer:

C

Step-by-step explanation:

i took the test

7 0

3 years ago

Read 2 more answers

An artificial lake is in the shape of a rectangle and has an area of 9/20 square mile. The width of the lake is 1/5 the length o

Zina [86]

X=width of the lake
y=length of the lake.
we set out this system of equiations
x*y=9/20
x=y/5
we solve by sustitution method
(y/5)*y=9/20
y²/5=9/20
y²=(5*9)/ 20
y²=2.25
y=√2.25=1.5

x=y/5
x=1.5/5=0.3

Solution:
width of the lake=0.3 miles
lenght of the lake=1.5 miles

6 0

4 years ago

Help me please !<br> 4,5 and 6

andreev551 [17]

9514 1404 393

Answer:

4a. ∠V≅∠Y

4b. TU ≅ WX

5. No; no applicable postulate

6. see below

Step-by-step explanation:

a. When you use the ASA postulate, you are claiming you have shown two angles and the side between them to be congruent. Here, you're given side TV and angle T are congruent to their counterparts, sides WY and angle W. The angle at the other end of segment TV is angle V. Its counterpart is the other end of segment WY from angle W. In order to use ASA, we must show ...

∠V≅∠Y

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b. When you use the SAS postulate, you are claiming you have shown two sides and the angle between them are congruent. The angle T is between sides TV and TU. The angle congruent to that, ∠W, is between sides WY and WX. Then the missing congruence that must be shown is ...

TU ≅ WX

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The marked congruences are for two sides and a non-included angle. There is no SSA postulate for proving congruence. (In fact, there are two different possible triangles that have the given dimensions. This can be seen in the fact that the given angle is opposite the shortest of the given sides.)

"No, we cannot prove they are congruent because none of the five postulates or theorems can be used."

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The first statement/reason is always the list of "given" statements.

1. ∠A≅∠D, AC≅DC . . . . given

2. . . . . vertical angles are congruent

3. . . . . ASA postulate

4. . . . . CPCTC

8 0

3 years ago

Whats the answer to that question?

gavmur [86]

Answer:

-0.9090... can be written as $\frac{10}{11}$ .

Explanation:

Any <em>repeating </em>decimal can be written as a fraction by dividing the section of the pattern to be repeated <em>by </em>9's.

We can start by listing out

0.909090... = 9/10 + 0/100 + 9/1000 + 0/10000 + 9/100000 + 0/1000000 + ...

Now. we let this series be equal to x, that is

$x$ = 9/10 + 0/100 + 9/1000 + 0/10000 + 9/100000 + 0/1000000 + ...

Now, we'll multiply both sides by 100 .

$100x$ = 90 + 0 + 9/10 + 0/100 + 9/1000 + 0/10000 + ...

Then, subtract the 1st equation from the second like so:

$100x$ = 90 + 0 + 9/10 + 0/100 + 9/1000 + 0/10000 + 9/100000 + 0/1000000 + ...

$-x$ = - 9/10 - 0/100 - 9/1000 - 0/10000 - 9/100000 - 0/1000000 - ...

And we end up with this:

$99x=90$

Finally, we divide both sides by 99 in order to isolate x and get the fraction we're looking for.

$x=\frac{90}{99}$

Which can be reduced and simplified to

$x=\frac{10}{11}$

Hope this helps!

4 0

3 years ago

Cynthia invests some money in a bank which pays 5% compound interest per year. She wants it to be worth over £8000 at the end of

Agata [3.3K]

$\bf \qquad \textit{Compound Interest Earned Amount}\\\\A=P\left(1+\frac{r}{n}\right)^{nt}\quad \begin{cases}A=\textit{accumulated amount}\to &8000\\P=\textit{original amount deposited}\\r=rate\to 5\%\to \frac{5}{100}\to &0.05\\n=\begin{array}{llll}\textit{times it compounds per year}\\\textit{is only once, so}\end{array}\to &1\\t=years\to &3\end{cases}$

$\bf 8000=P\left(1+\frac{0.05}{1}\right)^{1\cdot 3}\implies 8000=P(1.05)^3\implies \cfrac{8000}{1.05^3}=P\\\\\\6910.70\approx P\implies \stackrel{\textit{rounded up}}{6911 = P}\qquad therefore\qquad 6911~~\leqslant ~~P$

6 0

3 years ago