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zhuklara [117]
3 years ago
9

Solve each equation for the given variable. Use inverse operations to solve each equation.

Mathematics
1 answer:
andre [41]3 years ago
4 0
X = 15

Inverse operations would be + 5 to the -5 which cancels that out, & adding 5 to the 15 which gives you 15
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The equation for line r can be written as y=-1/4x+3. Line s which is parallel to lone r includes the point (-4,3). What is the e
MaRussiya [10]

Answer:

y=-\frac{1}{4}x+2

Step-by-step explanation:

Hi there!

<u>What we need to know:</u>

  • Linear equations are typically organized in slope-intercept form:y=mx+b where m is the slope and b is the y-intercept (the value of y when x is 0)
  • Parallel lines always have the same slope

<u>1) Determine the slope of line S using line R (m)</u>

y=-\frac{1}{4} x+3

We can identify clearly that the slope of the line is -\frac{1}{4}, as it is in the place of m. Because parallel lines always have the same slope, the slope of line S would also be -\frac{1}{4}. Plug this into y=mx+b:

y=-\frac{1}{4}x+b

<u>2) Determine the y-intercept of line S (b)</u>

y=-\frac{1}{4}x+b

Plug in the given point (-4,3) and solve for b

3=-\frac{1}{4}(-4)+b\\3=1+b

Subtract 1 from both sides to isolate b

3-1=1+b-1\\2=b

Therefore, the y-intercept is 2. Plug this back into y=-\frac{1}{4}x+b:

y=-\frac{1}{4}x+2

I hope this helps!

6 0
3 years ago
Alexander Litvinenko was poisoned with 10 micrograms of the radioactive substance Polonium-210. Since radioactive decay follows
koban [17]

Answer:

The amount of Polonium-210 left in his body after 72 days is 6.937 μg.

Step-by-step explanation:

The decay rate of Polonium-210 is the following:

N(t) = N_{0}e^{-\lambda t}     (1)

Where:

N(t) is the quantity of Po-210 at time t =?

N₀ is the initial quantity of Po-210 = 10 μg

λ is the decay constant  

t is the time = 72 d  

The decay rate is 0.502%, hence the quantity that still remains in Alexander is 99.498%.    

First, we need to find the decay constant:

\lambda = \frac{ln(2)}{t_{1/2}}    (2)

Where t(1/2) is the half-life of Po-210 = 138.376 days

By entering equation (2) into (1) we have:

N(t) = N_{0}e^{-\frac{ln(2)}{t_{1/2}}*t}} = 10* \frac{99.498}{100}*e^{-\frac{ln(2)}{138.376}*72} = 6.937 \mu g    

Therefore, the amount of Polonium-210 left in his body after 72 days is 6.937 μg.  

I hope it helps you!  

8 0
3 years ago
Explain how you can solve 40 x 60 By breaking apart the factors?
vlada-n [284]
Use the distributive property
8 0
3 years ago
If absolute value is always positive, then is |-45| &lt; |-26.5|?
allochka39001 [22]

Answer: The value is always positive with absolute value, so that is incorrect. Just take away all negative symbols and compare the numbers.

7 0
3 years ago
PLEASE PLEASE HELPPPP
Anastaziya [24]
You could expect about 1,248 students to play at least 4 hours every week.

Our tables have 100 students involved. All but 4 of the students are in the category of at least 4 hours. That is 96%.

Therefore, we simply need to multiply 0.96 by 1300 to get the estimated amount.

0.96 x 1300 = 1248
7 0
3 years ago
Read 2 more answers
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