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oksian1 [2.3K]
3 years ago
9

HELP ME PLEASE IT IS DUE TODAY I DON'T UNDERSTAND ANY OF THIS

Mathematics
2 answers:
natulia [17]3 years ago
4 0
I really hate math wish teachers would give us a break
Kryger [21]3 years ago
3 0

Answer:

first pic: A and C

second pic: i dont know for A, B = w + 1.4, C = 3y = 5.1, D = z + 1/3 = 1/2

third pic: a. 2x = 14.62 b. there are 2 x's on one side, and 1 number on the other, which can be taken as the answer to 2x. if we can do 2x = 14.62, we can do 14.62 divided by 2 to get the other number x, 14.62 divided by 2 = 7.31. c. i don't know

after this

Step-by-step explanation:

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The distance of the point from the origin = 9.29 units.

Step-by-step explanation:

Given point:

(7,-6)

The angle lies such that the terminal side of the angle contains the given point.

To draw the angle and find the distance from the origin to the given point.

Solution:

The terminal side of the angle is where the angle ends with the initial side being the positive side of the x-axis.

So, we can plot the point (7,-6) by moving 7 units on the x-axis horizontally and -6 units on the y-axis vertically.

We can find the distance of the point from the origin by find the hypotenuse of the triangle formed.

Applying Pythagorean theorem.

Hypotenuse^2=Shorter\ Leg^2+Shortest\ Leg^2

Hypotenuse^2 = (7)^2+(-6)^2

Hypotenuse^2=49+36\\Hypotenuse^2=85

Taking square root both sides :

\sqrt{Hypotenuse^2}=\sqrt{85}

Hypotenuse = 9.29\ units

Thus, the distance of the point from the origin = 9.29 units.

The figure is shown below.

5 0
3 years ago
A circle is centered at J(3, 3) and has a radius of 12.
stealth61 [152]

Answer:

(-6,\, -5) is outside the circle of radius of 12 centered at (3,\, 3).

Step-by-step explanation:

Let J and r denote the center and the radius of this circle, respectively. Let F be a point in the plane.

Let d(J,\, F) denote the Euclidean distance between point J and point F.

In other words, if J is at (x_j,\, y_j) while F is at (x_f,\, y_f), then \displaystyle d(J,\, F) = \sqrt{(x_j - x_f)^{2} + (y_j - y_f)^{2}}.

Point F would be inside this circle if d(J,\, F) < r. (In other words, the distance between F\! and the center of this circle is smaller than the radius of this circle.)

Point F would be on this circle if d(J,\, F) = r. (In other words, the distance between F\! and the center of this circle is exactly equal to the radius of this circle.)

Point F would be outside this circle if d(J,\, F) > r. (In other words, the distance between F\! and the center of this circle exceeds the radius of this circle.)

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\begin{aligned}d(J,\, F) &= \sqrt{(x_j - x_f)^{2} + (y_j - y_f)^{2}}\\ &= \sqrt{(3 - (-6))^{2} + (3 - (-5))^{2}} \\ &= \sqrt{145}  \end{aligned}.

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