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goldfiish [28.3K]
3 years ago
8

1/4 | x + 2 | + 7 = 5

Mathematics
1 answer:
Katarina [22]3 years ago
7 0
Hope this is the correct answer! :)

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Evaluate the expression for x = 3, y=13 , and z = 5.
hoa [83]

Answer:

-437

Step-by-step explanation:

x=3,y=13,z=5

12x-3y4x-z

12(3)-3(13)(4)(3)-5

36-468-5

-437

5 0
3 years ago
You drink a beverage with 120 mg of caffeine. Each hour, the caffeine in your system decreases by about 12%. How long until you
uysha [10]

Answer:

  19.44 hours, about 19 hours 26 minutes

Step-by-step explanation:

The exponential equation that describes your caffeine level can be written as ...

  c(t) = 120·(1 -0.12)^t . . . . where t is in hours and c(t) is in mg

We want to find t for c(t) = 10, so ...

  10 = 120(0.88^t)

  10/120 = 0.88^t . . . . . . . divide by 120

  log(1/12) = t·log(0.88) . . . take logarithms

  t = log(1/12)/log(0.88) ≈ 19.4386

It will take about 19.44 hours, or 19 hours 26 minutes, for the caffeine level in your system to decrease to 10 mg.

3 0
3 years ago
You pick a card at random. 4 5 6 7 What is P(less than 7)? Write your answer as a fraction or whole number.
marta [7]

Answer:

3/4 is less than 7.

Step-by-step explanation:

There are 4 numbers but there are only 3 less than 7.

6 0
2 years ago
how can you use a graph of a linear relationship to predict an unknown value of y for a given value of x within the region of th
lbvjy [14]
If you know what the equation of the line IS(y=mx+b), then you can:

Substitute the x coordinate, slope and y-intercept into the line, then solve for y.
Hope this helped!
7 0
3 years ago
The sum of the squares of 3 consecutive positive integers is 116. What are the numbers?
elixir [45]
Let's say, the first number is "n"

well, the consecutive number of "n" is "n + 1"and the consecutive number of "n+1" is "n + 1" +1 or n + 2

so, the numbers are (n), (n+1) and (n+2), whatever "n" is

now, the sum of their squares is 116

\bf (n)^2+(n+1)^2+(n+2)^2=116

expand the binomials by FOIL or binomial theorem, and then simplify
3 0
3 years ago
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