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Lunna [17]
3 years ago
6

Help please I cant get this right

Mathematics
1 answer:
ruslelena [56]3 years ago
8 0

if it were me I would start by inputting 3 into the equation and then trying to simplify it

\sqrt{3^2+6*3+9}

\sqrt{9+18+9}

\sqrt{36}

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Substitute (–5, 0.5) into x – 4y = –7 to get .
lesya [120]

Answer:

-7

Step-by-step explanation:

Coordinates are listed as (x,y) so x = -5 and y = .5

If you substitute the numbers in place of the letters, you get:

(-5) - 4(.5) = -7

-5 - 2 = -7

-7 = -7

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4 years ago
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trapecia [35]

Answer:

y = - 6x - 2

Step-by-step explanation:

The straight line equation is y = mx + c

First step is to bring the equation given to this form, where you make y the subject.

Now you can easily note down the gradient(m1). You will need it to find the gradient(m2) of the perpendicular line, because m1 × m2 = -1

After you have that substitute the value and the values of x and y given to find c .

Lastly, substitute the value of m2 and c into y = mx + c to find the line that is perpendicular .

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3 years ago
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SVETLANKA909090 [29]

Answer: (How much does your dog weigh?)

Step-by-step explanation: A statistical question is a question that has more than one possible answer. The only question that has more than one possible answer is "How much does your dog weigh?"

Statistical questions are more than one answer with numbers.

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3 years ago
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Readme [11.4K]

Answer:I might be wrong but im pretty sure the answer is 5.4

Step-by-step explanation:

8 0
4 years ago
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The equation of a line is y=3. Write an equation in slope intercept form of a line parallel to y=3 that passes through (0,6).
Dimas [21]

Answer:

y = 6

Step-by-step explanation:

We have to consider the equation of a line ( in slope - intercept form ) that is parallel to y = 3, and passes through the point ( 0, 6 ). Now y = 3 can be plotted as a horizontal line that passes through the point ( 0, 3 ). If we want a line that is parallel to this, it must be horizontal as well;

Let us consider the second point now. This line must pass through the point ( 0, 6 ). We can conclude that the line must be 1. Horizontal, and 2. Pass through point ( 0, 6 );

Equation - y = 6

7 0
3 years ago
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