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Elan Coil [88]
3 years ago
7

Discord Link in pic. If you know what this is you should cos yesterday remember.

Mathematics
2 answers:
marin [14]3 years ago
6 0

e p i c

im here before this gets reported :P

VLD [36.1K]3 years ago
3 0

Answer:

Step-by-step explanation:

Its symbols from ancient Egypt welded in to the rock that the gods had walked on

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What is the measurement of angle B
Liono4ka [1.6K]
Angles B and D are congruent so:

(360-112-74)/2=87°
6 0
2 years ago
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I'd appreciate it if anyone could help! :)<br><br> Which graph represents the function?
Monica [59]

Answer: Upper right corner

=========================================================

How I got that answer:

The line y = x goes through (0,0) and (1,1). You would normally extend this line out as far as you can in both directions. However, the inequality x < -1 says you can only graph this if x is less than -1. We will not graph any part of the graph that is beyond x = -1 to the right. So we only have a small piece of it. The left piece of y = x. There is an open hole at the endpoint.

Similarly, y = -x is only graphed if x >= -1. We have a closed endpoint here. This graph goes through (0,0) and (1,-1). We erase the portion that is to the left of x = -1.

Doing all this leads to the upper right corner choice as our answer. The bottom right corner is close to the answer, but the open and closed endpoints are in the wrong spots.

3 0
3 years ago
NEED ANSWER ASAP REALLY IMPORTANT i will take any of the answers
dedylja [7]

1. \: MC  \: is \:  tangent \:  to \:  (B)\Rightarrow \widehat{BCM}=90° \\ 2. \: MC  \: and \: MZ \: are \:  tangent  \: to  \: (B)\Rightarrow MC = MZ\Leftrightarrow 5x - 9 = x + 7\Leftrightarrow x = 4 \\ 3. \: BM =  \sqrt{ {BC}^{2} + {MC}^{2}  }  =  \sqrt{ {5}^{2} +  {12}^{2}  }  = 13 \\ \Rightarrow EM = BM - BE = BM - BC = 13 - 5 = 8 \\ 4. \:  \tan(60°) = \frac{MC}{BC}  =  \frac{13 \sqrt{3}  }{ BC}\Leftrightarrow BC =  \frac{13 \sqrt{3} }{\tan(60°)} =   \frac{13 \sqrt{3} }{ \sqrt{3} }  = 13 \\ 5. \: MC =  \sqrt{ {BM}^{2}  -  {BC}^{2} }  =  \sqrt{ {20}^{2}  -  {12}^{2} }  = 16 \\ 6. \:BZ =  \sqrt{ {BM}^{2}  -  {MZ}^{2} } =  \sqrt{ {25}^{2}  -  {20}^{2} }  = 15 \\ \Rightarrow XE=BX+BE=BZ+BZ=15+15=30

7 0
3 years ago
The half-life of cesium-137 is 30 years. Suppose we have a 180-mg sample. (a) Find the mass that remains after t years. (b) How
DedPeter [7]

Answer:

a) Q(t) = 180e^{-0.023t}

b) 11.4mg of cesium-137 remains after 120 years.

c) 225.8 years.

Step-by-step explanation:

The following equation is used to calculate the amount of cesium-137:

Q(t) = Q(0)e^{-rt}

In which Q(t) is the amount after t years, Q(0) is the initial amount, and r is the rate at which the amount decreses.

(a) Find the mass that remains after t years.

The half-life of cesium-137 is 30 years.

This means that Q(30) = 0.5Q(0). We apply this information to the equation to find the value of r.

Q(t) = Q(0)e^{-rt}

0.5Q(0) = Q(0)e^{-30r}

e^{-30r} = 0.5

Applying ln to both sides of the equality.

\ln{e^{-30r}} = \ln{0.5}

-30r = \ln{0.5}

r = \frac{\ln{0.5}}{-30}

r = 0.023

So

Q(t) = Q(0)e^{-0.023t}

180-mg sample, so Q(0) = 180

Q(t) = 180e^{-0.023t}

(b) How much of the sample remains after 120 years?

This is Q(120).

Q(t) = 180e^{-0.023t}

Q(120) = 180e^{-0.023*120}

Q(120) = 11.4

11.4mg of cesium-137 remains after 120 years.

(c) After how long will only 1 mg remain?

This is t when Q(t) = 1. So

Q(t) = 180e^{-0.023t}

1 = 180e^{-0.023t}

e^{-0.023t} = \frac{1}{180}

e^{-0.023t} = 0.00556

Applying ln to both sides

\ln{e^{-0.023t}} = \ln{0.00556}

-0.023t = \ln{0.00556}

t = \frac{\ln{0.00556}}{-0.023}

t = 225.8

225.8 years.

8 0
3 years ago
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Solve for x 3x+8=35 can u plz plz plz plz plz
tamaranim1 [39]

Answer: x=9

Step-by-step explanation:

3x+8=35

Subtract 35-8=27

3x=27

27/3=9

x=9

7 0
2 years ago
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