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kirill [66]
3 years ago
13

1 The measure of Z1 =

Mathematics
1 answer:
ollegr [7]3 years ago
3 0

Answer:

∠1 = 90°

Step-by-step explanation:

The angle inscribed in a semicircle is always 90°

∠1 = 90°

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Graph g(x)=-2x-3<br><br> ........
nekit [7.7K]

Answer:

Well, the x intercept will be -1.5 and the y-intercept will be -3. It will be a negative slope with a slope of -4/2

5 0
3 years ago
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Zoe prepares some lemonade for the party.
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eyeey

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3 years ago
−19.89 −3.9 find the quotient for me please
Sunny_sXe [5.5K]

Answer:

5.1

Step-by-step explanation:

I'm guessing you wanted me to divide these to find the quotient.

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4 years ago
What is the slope of the line 5x+y=15?
Reptile [31]

Answer:

-5

Step-by-step explanation:

Write in y = mx + b form

y = -5x + 15

m = slope

Answer: -5

7 0
4 years ago
Please help me! Write the slope intercept form of an equation for the line that passes through (0,3) and is perpendicular to 9x-
lesya692 [45]

Answer:

y = -4/9x + 3

Step-by-step explanation:

Keep in mind:

- the equation for slope-intercept form is y = mx + b.

- "y" means output, "x" means input, "m" means slope, and "b" means y intercept

The first step in this equation is to turn 9x - 4y = -8 into slope intercept form.

Step 1: subtract 9x from both sides to begin to isolate y.

9x - 4y = -8

-4y = -9x - 8

Step 2: divide both sides by -4 to further isolate y.

y = 9/4x + 2

Now, we have the equation for the perpendicular line. Remember, the slope of a line is the opposite reciprocal of the slope of the line perpendicular to it.

This means that the slope of the line we are trying to find the equation for is -4/9.

Now that we have a slope and a point the line passes through, (0, 3), we can find the equation. Plug the coordinates of the point and the slope into the equation for slope-intercept form.

3 = -4/9(0) + b

To find the equation, solve for b.

Step 1: Simplify -4/9(0)

3 = 0 + b

Step 2: Simplify 0 + b

3 = b

Now that we know b and m, we have our equation.

The final answer is y = -4/9x + 3.

6 0
4 years ago
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