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2 years ago

The difference between 8,-7 and -3,-7

Mathematics

2 answers:

Inessa [10]2 years ago

8 0

Answer:

1) is 15.

2) is 4.

Step-by-step explanation:

hope this helps :)

photoshop1234 [79]2 years ago

8 0

Answer:

8,-7 is over in ll

-3,-7 is over in 4

thatsone big difference

lmk if this helped

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Which method would be more efiicient to solve the system y=1/2x and 2x+3y=28

Licemer1 [7]

Probably the subsitution method

y=1/2x
subsitute that fr y

2x+3(1/2x)=28
2x+3/2x=28
times 2 both sides
4x+3x=56
7x=56
divide by 7 both sides
x=8
sub back

y=1/2x
y=1/2(8)
y=4

(x,y)
(8,4)

6 0

2 years ago

Which of the following graphs shows a parabola with a vertex of (1,-9) and solutions of (-2,0) and (4,0)?

Mariulka [41]

Answer:

Hello,

Step-by-step explanation:

Roots are -2 and 4

y=k*(x+2)(x-4)

Vertex = (1,-9) is a point of the parabola

-9=k*(1+2)(1-4) ==> k=1

Equation of the parabola is y=(x+2)(x-4)

But you don' t have given the graphs !!!!

3 0

2 years ago

(10.02)

melisa1 [442]

Answer:

$sin O=\dfrac{3\sqrt{13}}{13}\\cos O=\dfrac{2\sqrt{13}}{13}\\tan O=\dfrac{3}{2}$

Step-by-step explanation:

If the point (2,3) is on the terminal side of an angle in standard position.

Adjacent of O, x=2,

Opposite of O, y=3

Next, we determine the hypotenuse, r using Pythagoras Theorem.

$Hypotenuse =\sqrt{Opposite^2+Adjacent^2} \\r=\sqrt{3^2+2^2} \\r=\sqrt{13}$

Therefore:

$sin O=\dfrac{Opposite}{Hypotenuse} \\sin O=\dfrac{3}{\sqrt{13}} \\$Rationalizing\\sin O=\dfrac{3\sqrt{13}}{13}$

$cos O=\dfrac{Adjacent}{Hypotenuse} \\cos O=\dfrac{2}{\sqrt{13}} \\$Rationalizing\\cos O=\dfrac{2\sqrt{13}}{13}$

$Tan O=\dfrac{Opposite}{Adjacent} \\tan O=\dfrac{3}{2}$

4 0

2 years ago

What Is 0.005 in percent form

Nuetrik [128]

To change 0.005 to a percent, just multiply by 100. $= 0.005 * 100 = 0.5$

Therefore, your answer would be 0.5%

3 0

2 years ago

Read 2 more answers

Which relation is a function

erma4kov [3.2K]

Answer:

Top left one on the attached, must be option A

Step-by-step explanation:

A function has to have unique input values and for each input value has to have an output value.

Analyzing the graphs we can see that only one of them is a function- top left one

All the others have repeat input values, therefore are not functions

All those points marked with red circle

<em>Refer to attached </em>

8 0

2 years ago