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3 years ago

TRUE OR FALSE:

1 answer:

7 0

It is possible to have an obtuse triangle that also contains a 35° angle.
Yes, this is true.
That's because an obtuse angle just means it's more than 90 but less than 180 degrees.

It is possible to have a right triangle that also contains a 110° angle.
No, this is false.
That's because a right triangle already has 90. That means that the two remaining angles must add up to 90 and not more or less.

It is possible to have an acute triangle that also contains a 70° angle.
Yes, this is true.
An acute triangle means that all of the angles are more than 0 but less than 90.

Hope this helps :)

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Charlie reads quickly. He reads 1 and 3 seventh pages every 2 third minutes. Charlie reads at a constant rate. How many pages do

andrey2020 [161]

The unit rate would be 1 3/7 ÷ 2/3 10/7 ÷ 2/3 10/7 · 3/2 separate by division, rearrange and multiply 15/7

Charlies reads 15/7, or 2 1/7, page per minute.

Hope this helps a bit.

7 0

3 years ago

Solve?? 3x + 10 > 4

Artemon [7]

Answer:

x>−2

Step-by-step explanation:

1 Subtract 1010 from both sides.

3x>4-103x>4−10

2 Simplify 4-104−10 to -6−6.

3x>-63x>−6

3 Divide both sides by 33.

x>-\frac{6}{3}x>−

4 Simplify \frac{6}{3}

to 22.

x>-2x>−2

8 0

3 years ago

Read 2 more answers

N - 4 = 3n + 6 Yikesss

raketka [301]

Answer: n=-5

Step-by-step explanation:

4 0

3 years ago

Read 2 more answers

Enter your answer

satela [25.4K]

Answer:So, since it is continuously, we need to think of PERT

A=P\times e^{rt}

So, we just plug our numbers in to solve for T

8000=4000e^{.07t}\\\ln(8000)=\ln(4000e^{.07t})\\\ln(8000) = \ln(4000) + \ln(e^{.07t})\\\ln(8000)=\ln(4000)+.07t\ln(e)\\\ln(8000)=\ln(4000)+.07t\\\frac{\ln(8000)}{\ln(4000)}=.07t\\\frac{\ln(8000)}{.07\ln(4000)}=t\\ 15.479=t

It will take 15.479 years

Step-by-step explanation:

8 0

2 years ago

A sample of tritium-3 decayed to 94.5% of its original amount after a year.

Zolol [24]

(a) If $y(t)$ is the mass after $t$ days and $y(0) = A$ then $y(t) = Ae^{kt}.$

$y(1) = Ae^k = 0.945A \implies e^k = 0.945 \implies k = \ln 0.945$

$\text{Then } Ae^{(\ln 0.945)t} = \frac{1}{2}A\ \Leftrightarrow\ \ln e^{(\ln 0.945)t} = \ln \frac{1}{2}\ \Leftrightarrow\ ( \ln 0.945) t = \ln \frac{1}{2} \Leftrightarrow \\ \\
t = - \frac{\ln 5}{\ln 0.945} \approx 12.25 \text{ years}$

(b)
$Ae^{(\ln 0.945) t} = 0.20 A \ \Leftrightarrow\ (\ln 0.945) t \ln \frac{1}{5}\ \implies \\ \\
t = - \frac{\ln 5}{\ln 0.945} \approx 28.45\text{ years}$

3 0

3 years ago