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worty [1.4K]
3 years ago
11

Find the equivalent expressions to 12a+6b

Mathematics
1 answer:
anzhelika [568]3 years ago
6 0

Answer:

6a+6a+6b

Step-by-step explanation:

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3 intersecting lines are shown. A horizontal line contains points R, A, T, S. Another line intersects at point A and also contai
svlad2 [7]

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C or A I think it's correct

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3 years ago
The highest service temperature on any of the Solar System planets is found on Venus partly because it's nearest to the Sun and
SpyIntel [72]

Answer:

842 Fahrenheit, 723.15 Kelvin

Step-by-step explanation:

(450C * 9/5) + 32 = 810 + 32 = 842 Fahrenheit

450C + 273.15 = 723.15 Kelvin

7 0
3 years ago
(2x^2-4x+8)+(3x^2+15x-12)
Neko [114]

Answer:

5x2+11x−4

Step-by-step explanation:

=2x2+−4x+8+3x2+15x+−12

Combine Like Terms:

=2x2+−4x+8+3x2+15x+−12

=(2x2+3x2)+(−4x+15x)+(8+−12)

=5x2+11x+−4

have a great weekend <3

6 0
3 years ago
Read 2 more answers
The point slope form of the equation of the line that passes through -5, -1 and 10, -7 is Y +7=
MrMuchimi

Answer:

e standard form of a line is in the form Ax + By = C where A is a positive integer, and B, and C are integers. The standard form of a line is just another way of writing the equation of a line.

Step-by-step explanation:

6 0
4 years ago
A lake currently has a depth of 30 meters. As sediment builds up in the lake, its depth decreases by 2% per year. This situation
lubasha [3.4K]

Answer:

This situations represents the depth of the lake after t years.

The rate of decay is equal to 0.02 a year.

So the depth of the lake each year is 0.98 times the depth in the previous year.

It will take 5.77 years for the depth of the lake to reach 26.7 meters.

Step-by-step explanation:

Exponential equation of decay:

The exponential equation for the decay of an amount is given by:

D(t) = D(0)(1-r)^t

In which D(0) is the initial amount and r is the decay rate, as a decimal.

A lake currently has a depth of 30 meters. As sediment builds up in the lake, its depth decreases by 2% per year.

This means that D(0) = 30, r = 0.02

So

D(t) = D(0)(1-r)^t

D(t) = 30(1-0.02)^t

D(t) = 30(0.98)^t

This situation represents

The depth of the lake after t years.

The rate of growth or decay, r, is equal to

The rate of decay is equal to 0.02 a year.

So the depth of the lake each year is times the depth in the previous year.

1 - 0.02 = 0.9

So the depth of the lake each year is 0.98 times the depth in the previous year.

It will take between years for the depth of the lake to reach 26.7 meters.

This is t for which D(t) = 26.7. So

D(t) = 30(0.98)^t

26.7 = 30(0.98)^t

(0.98)^t = \frac{26.7}{30}

\log{(0.98)^t} = \log{\frac{26.7}{30}}

t\log{(0.98)} = \log{\frac{26.7}{30}}

t = \frac{\log{\frac{26.7}{30}}}{\log{0.98}}

t = 5.77

It will take 5.77 years for the depth of the lake to reach 26.7 meters.

4 0
3 years ago
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