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3 years ago

What is the value of x? Enter your answer in the box.

Mathematics

1 answer:

Lynna [10]3 years ago

3 0

That my friend is number 6

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65 Divided by 3459 what is it?

telo118 [61]

Answer: 53.2

Step-by-step explanation:

8 0

3 years ago

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The graph of y=f(x) is shown below. Draw the graph of y=-f(x)

valkas [14]

Answer:

y = -f(x): reflection over x axis

Step-by-step explanation:

y = -f(x): reflection over x axis

8 0

2 years ago

X - y = 2 4x – 3y = 11

swat32

Answer:

For x - y = 2

You need to find x or y in one of the equations and then substitute that into the other.

So we have;

x-y=2

4x-3y=11

We will take the first equation and find x;

x-y=2

add y to both sides;

x-y+y=2+y

x=2+y

Now we take that answer and substitute it forx in the other equation;

4(2+y)-3y=11

8+4y-3y=11

8+y=11

y=3

Now we have what y equals, so we use it in the first equation to find x;

x-3=2

x=5

So we have;

x=5; y=3

Hope you understand!

And for 4x – 3y = 11

Multiply the first equation by 2 and the second by 3 so that there are the same number of y's in each:

8x - 6y = 22 ...(3)

30x + 6y = -3 ...(4)

Now add (3) and (4) term by term:

38x + 0 = 19

38x = 19

or x = 1/2

Put this back into equation (1)

4*(1/2) - 3y = 11

2 - 3y = 11

Subtract 2 from both sides:

-3y = 9

Divide both sides by -3

y = -3

6 0

3 years ago

Read 2 more answers

Is the given point interior , exterior, or on the circle k (x+2)2 + (y-3)2 =18 P (8,4)

Mnenie [13.5K]

One way would be to find the distance from the point to the center of the circle and compare it to the radius

for
$(x-h)^2+(y-k)^2=r^2$
the center is (h,k) and the radius is r

and the distance formula is
distance between $(x_1,y_1)$ and $(x_2,y_2)$ is
$D=\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}$

r=radius
D=distance form (8,4) to center

if r>D, then (8,4) is inside the circle
if r=D, then (8,4) is on the circle
if r<D, then (8,4) is outside the circle

so
$(x+2)^2+(y-3)^2=18$
$(x-(-2))^2+(y-3)^2=(\sqrt{18})^2$
$(x-(-2))^2+(y-3)^2=(3\sqrt{2})^2$

the radius is $3\sqrt{2}$
center is (-2,3)

find distance between (8,4) and (-2,3)

$D=\sqrt{(8-(-2))^2+(4-3)^2}$
$D=\sqrt{(8+2)^2+(1)^2}$
$D=\sqrt{10^2+1}$
$D=\sqrt{100+1}$
$D=\sqrt{101}$

$r=3\sqrt{2}$ ≈4.2
$D=\sqrt{101}$ ≈10.04

do r<D

(8,4) is outside the circle

6 0

3 years ago

1. The exact value of cos60

Ksenya-84 [330]

E) none of the above

5 0

3 years ago