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lapo4ka [179]
3 years ago
9

Write an expression for 20 times the sum of 4 and 2

Mathematics
2 answers:
marta [7]3 years ago
8 0

Answer:

20x6

Step-by-step explanation:

20x (x stands for multiply/times)

6(sum/add 4+2)

20x6

Hope this helps.

Lady bird [3.3K]3 years ago
3 0

Answer:

20 x (4+2)

Step-by-step explanation:

20 x (4+2)

20 x 6

120

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Answer: what ???? lol

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3 years ago
What is the value of n in the equation 2.3 × 10 9power = (1 × 10 3power)(2.3 × 10n)
Amanda [17]
Hello,

2.3 x 10^9 = ( 1 x 10^3) ( 2.3 x 10^n) 

Solution:

Take the logarithm of both sides of the equation to remove the variable from the exponent. 

n=6


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7 0
3 years ago
Please this urgent! I need your help in these questions, it would be very appreciated.
Paraphin [41]

Answer:

3 is continous. 4 is discrete

Step-by-step explanation:

3 houses 2 terms, therefore continuing it. 4 houses 1, stopping the graph.

I might be dumb

3 0
3 years ago
Let f be defined as shown on the graph. Which statement is true?
satela [25.4K]
The inverse of f does NOT exist. The reason why is because the function fails the horizontal line test. Recall that the horizontal line test is a test where you try to see if you can pass a single horizontal line through more than one point on the function curve. If you can get the horizontal line to pass through more than one point, then it fails the test. It's very similar to the vertical line test.
4 0
3 years ago
(1 point) A fish tank initially contains 15 liters of pure water. Brine of constant, but unknown, concentration of salt is flowi
Drupady [299]

Answer:

a. \dfrac{dx}{dt} = 6\frac{2}{3} \cdot c

b. x(t) = 6\frac{2}{3} \cdot c \cdot t

c. c  = \dfrac{3}{8}  \ g/L

Step-by-step explanation:

a. The volume of water initially in the fish tank = 15 liters

The volume of brine added per minute = 5 liters per minute

The rate at which the mixture is drained = 5 liters per minute

The amount of salt in the fish tank after t minutes = x

Where the volume of water with x grams of salt = 15 liters

dx =  (5·c - 5·c/3)×dt = 20/3·c = 6\frac{2}{3} \cdot c \cdot dt

\dfrac{dx}{dt} = 6\frac{2}{3} \cdot c

b. The amount of salt, x after t minutes is given by the relation

\dfrac{dx}{dt} = 6\frac{2}{3} \cdot c

dx = 6\frac{2}{3} \cdot c \cdot dt

x(t) = \int\limits \, dx  = \int\limits \left ( 6\frac{2}{3} \cdot c \right) \cdot dt

x(t) = 6\frac{2}{3} \cdot c \cdot t

c. Given that in 10 minutes, the amount of salt in the tank = 25 grams, and the volume is 15 liters, we have;

x(10) = 25 \ grams(15 \ in \ liters) = 6\frac{2}{3} \times c \times 10

6\frac{2}{3} \times c  =\dfrac{25 \ grams }{10}

c  =\dfrac{25 \ g/L }{10 \times 6\frac{2}{3} }  = \dfrac{25 \ g/L}{10 \times \dfrac{20}{3} } =\dfrac{3}{200} \times 25 \ g/L= \dfrac{75}{200}  \ g/L = \dfrac{3}{8}  \ g/L

c  = \dfrac{3}{8}  \ g/L

4 0
3 years ago
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