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ki77a [65]
3 years ago
9

HELP ASAP WILL MARK BRAINLIEST ANSWER!!!!!!

Mathematics
1 answer:
Alekssandra [29.7K]3 years ago
7 0
The answer is this math is pointless and after the test you will never use it again
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There are 14 people on the city bus. At first stop, 4 people get off and 2 get on. At the second stop, 6 people get off. If ther
Dovator [93]

Answer:

6

Step-by-step explanation:

14-4=10+2=12-6=6+6=12

7 0
3 years ago
A ____ is a pair of equal ratios.
kobusy [5.1K]
A Equivalent Ratio Is A Pair Of Equal Ratios.
3 0
4 years ago
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What is the slope of a line with the eqation of 4×+5y=5
evablogger [386]

Answer:

- 4/5x

Step-by-step explanation:

So, this is a simple thing as long as you convert it into slope-int form. So, I am guessing from the question that the equation is

4x + 5y = 5

From here, you have to isolate y and leave all of the other variable by themselves. So, first you would subtract 4x from both sides and that leaves you with

5y = 5 -4x

Then, to isolate 5, you would divide 5 by both sides to get

y = 1- 4/5x

Now, you would put this is slope intercept form: y = mx + b to get

y = - 4/5 x + 1

Now, you can see that the slope is - 4/5x

Hope I helped!!!!!

6 0
3 years ago
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Write an expression (4x-2) • 6 (2x+7) in the standard form of a quadratic expression,
Butoxors [25]

Answer:

108

Step-by-step explanation:

7 0
4 years ago
Find a homogeneous linear differential equation with constant coefficients whose general solution is given by
frez [133]

Answer:

y" + 2y' + 2y = 0

Step-by-step explanation:

Given

y=c_1e^{-x}cosx+c_2e^{-x}sinx

Required

Determine a homogeneous linear differential equation

Rewrite the expression as:

y=c_1e^{\alpha x}cos(\beta x)+c_2e^{\alpha x}sin(\beta x)

Where

\alpha = -1 and \beta = 1

For a homogeneous linear differential equation, the repeated value m is given as:

m = \alpha \± \beta i

Substitute values for \alpha and \beta

m = -1 \± 1*i

m = -1 \± i

Add 1 to both sides

m +1= 1 -1 \± i

m +1= \± i

Square both sides

(m +1)^2= (\± i)^2

m^2 + m + m + 1 = i^2

m^2 + 2m + 1 = i^2

In complex numbers:

i^2 = -1

So, the expression becomes:

m^2 + 2m + 1 = -1

Add 1 to both sides

m^2 + 2m + 1 +1= -1+1

m^2 + 2m + 2= 0

This corresponds to the homogeneous linear differential equation

y" + 2y' + 2y = 0

6 0
3 years ago
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