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3 years ago

9

Y=2x+3 in ordered pairs

1 answer:

Iteru [2.4K]3 years ago

4 0

Answer:

(0,3) , (1,5) , (2,7)

Step-by-step explanation:

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Use the graph of the function. Determine Over what interval(s) the function is positive or negative.

san4es73 [151]

Answer:

I can't see the numbers well, but the graph is decreasing from (-inf, minimum) and increasing from (minimum, inf). The minimum is the vertex. Hope this helps!

Step-by-step explanation:

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3 years ago

Find the missing length in the right triangle 6.5 and 9.9<br>

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3 years ago

How many radians is -135°

Yuri [45]

Answer:

-3/4 pi radians

Step-by-step explanation:

<em>180° is 1 pi radian</em>

Hence, -<em>135° = -135/180 pi radians</em>

= -0.75 pi radians

= -3/4 pi radians

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3 years ago

The hypotenuse of a right triangle has endpoints A(4, 1) and B(–1, –2). On a coordinate plane, line A B has points (4, 1) and (n

GarryVolchara [31]

Answer:

$(-1,1),(4,-2)$

Step-by-step explanation:

Given: The hypotenuse of a right triangle has endpoints A(4, 1) and B(–1, –2).

To find: coordinates of vertex of the right angle

Solution:

Let C be point $(x,y)$

Distance between points $(x_1,y_1),(x_2,y_2)$ is given by $\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

$AC=\sqrt{(x-4)^2+(y-1)^2}\\BC=\sqrt{(x+1)^2+(y+2)^2}\\AB=\sqrt{(4+1)^2+(1+2)^2}=\sqrt{25+9}=\sqrt{34}$

ΔABC is a right angled triangle, suing Pythagoras theorem (square of hypotenuse is equal to sum of squares of base and perpendicular)

$34=\left [ (x-4)^2+(y-1)^2 \right ]+\left [ (x+1)^2+(y+2)^2 \right ]$

Put $(x,y)=(-1,1)$

$34=\left [ (-1-4)^2+(1-1)^2 \right ]+\left [ (-1+1)^2+(1+2)^2 \right ]\\34=25+9\\34=34$

which is true. So, $(-1,1)$ can be a vertex

Put $(x,y)=(4,-2)$

$34=\left [ (4-4)^2+(-2-1)^2 \right ]+\left [ (4+1)^2+(-2+2)^2 \right ]\\34=9+25\\34=34$

which is true. So, $(4,-2)$ can be a vertex

Put $(x,y)=(1,1)$

$34=\left [ (1-4)^2+(1-1)^2 \right ]+\left [ (1+1)^2+(1+2)^2 \right ]\\34=9+4+9\\34=22$

which is not true. So, $(1,1)$ cannot be a vertex

Put $(x,y)=(2,-2)$

$34=\left [ (2-4)^2+(-2-1)^2 \right ]+\left [ (2+1)^2+(-2+2)^2 \right ]\\34=4+9+9\\34=22$

which is not true. So, $(2,-2)$ cannot be a vertex

Put $(x,y)=(4,-1)$

$34=\left [ (4-4)^2+(-1-1)^2 \right ]+\left [ (4+1)^2+(-1+2)^2 \right ]\\34=4+25+1\\34=30$

which is not true. So, $(4,-1)$ cannot be a vertex

Put $(x,y)=(-1,4)$

$34=\left [ (-1-4)^2+(4-1)^2 \right ]+\left [ (-1+1)^2+(4+2)^2 \right ]\\34=25+9+36\\34=70$

which is not true. So, $(-1,4)$ cannot be a vertex

So, possible points for the vertex are $(-1,1),(4,-2)$

7 0

3 years ago

Read 2 more answers

complete the steps to find the quotient of 492 ÷ 6. enter your answer in the space.step 1( ___÷6) +(180÷ 6)+( ____÷ 6) step 2 __

uranmaximum [27]

I'll go thru the problem and then you fill in the blanks yourself.

8 2
________
6 / 4 9 2
- 4 8
-----------------
1 2
1 2
--------------
0

8 0

3 years ago

Read 2 more answers