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3 years ago

Which points are located in Quadrant l? Check all that apply. (0, 0) (1, 1) (7, 3) (5, 0) (8, 0) (1, 10)

2 answers:

5 0

When both the x and the y in coordinates are positive and not equal to zero, the points are in Quadrant I. So, the correct answers are (1,1), (7,3), and (1,10). Hope this helps!

Helga [31]3 years ago

4 0

Answer:

1,1 7,3 and 1,10

Step-by-step explanation:

I just took the quiz and got it correct! Yay

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gizmo_the_mogwai [7]

Answer:

x = 9

Step-by-step explanation:

-5(1 + 1x) = -50

-5 + -5x = -50

-5 + -5x + 5 = -5x

-50 + 5 = -45

-5x = -45

-5x / -5 = x

-45 / -5 = 9

x = 9

5 0

3 years ago

Is -2.7 an integer??????????

ale4655 [162]

Answer:

ya you can take it both like integer and decimal

cause - the negative mark make it negative the . make it decimal

3 0

3 years ago

A line passes through the points R(-1,-5) and S(-3,-1). Write the equation of the line through the point A(4,0) that is parallel

IRISSAK [1]

Sorry I was moving around taking the picture and my handwriting is sloppy. But I hope you understand how I got the answer

4 0

3 years ago

Enter the simplified form of the complex fraction in the box. Assume no denominator equals zero.

Olenka [21]

Answer:

$\frac{(9-x)}{3x}$

Step-by-step explanation:

we have

$\frac{\frac{2}{x-3}-\frac{3}{x}}{\frac{3}{x-3}}$

step 1

Solve the numerator of the quotient

$\frac{2}{x-3}-\frac{3}{x}=\frac{2x-3(x-3)}{(x-3)x}=\frac{2x-3x+9}{(x-3)x}=\frac{9-x}{(x-3)x}$

step 2

substitute in the original expression

$\frac{\frac{9-x}{(x-3)x}}{\frac{3}{x-3}}$

$\frac{(x-3)(9-x)}{3x(x-3)}$

step 3

simplify

$\frac{(9-x)}{3x}$

5 0

3 years ago

A right triangle has a leg of 11 and a hypotenuse of 16. Find the missing leg (in simplest radical form)

igomit [66]

Answer:

11^2 + b^2 = 16^2

121 + b^2 = 256

b^2 = 135

b = 3 X q(15)

Step-by-step explanation:

8 0

3 years ago