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3 years ago

Solve the problem c=3.14d for d

Mathematics

1 answer:

Maksim231197 [3]3 years ago

3 0

Answer:

$c = 3.14d \\ d = \frac{c}{3.14}$

It's change of subject. :)

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The measure of angle 1 is:<br><br> The measure of angle 1 is:

Karolina [17]

Question 1:
We can use the rule of exterior angle of triangle, where the exterior angle is equal to the sum of the other 2 interior angles of the triangle except for the adjacent one.

So, we can calculate angle 1 by adding up the 2 given angles.
Angle1 = 62° + 57°
=119°
=answer C.

Question 2:
We can also apply the rule above, but first we have to calculate the remaining angle of the triangle first.
For this, we have to use the angle sum of triangle. All 3 interior angles of triangle should add up to 180°.

Let that remaining angle be x.
X = 180° - 62° - 57°
X=61°
Now we can apply the rule of exterior angle of triangle.
Angle 1 = 62° + 61°
=123°
=answer D.

4 0

3 years ago

Y _minus 7 equal 10.find y

charle [14.2K]

Answer:

y=17

Step-by-step explanation:

y-7=10

transpose 7

y=10+7

y=17

5 0

3 years ago

What is the area of the real object that the scale drawing models?

Kruka [31]

Answer:

HOPE THIS HELPS!!

Step-by-step explanation:

YOU could find it by doing these steps:

Atual Area = �� square units.

Scale Drawing Area = �� square units.

Value of the Ratio of the Scale Drawing Area to the Actual Area:

7 0

3 years ago

NO LINKS!!! Please help me with these problems

nadezda [96]

<h3>Answers:</h3>

7) Center= (-1,2) Radius= $\boldsymbol{\sqrt{8}}$ Equation: $(x+1)^2+(y-2)^2 = 8$

8) Center= (3,13) Radius= 13 Equation: $(x-3)^2+(y-13)^2 = 169$

=========================================================

Explanation:

Problem 7

Let's find the distance from (-1,2) to (-3,4)

$(x_1,y_1) = (-1,2) \text{ and } (x_2, y_2) = (-3,4)\\\\d = \sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2}\\\\d = \sqrt{(-1-(-3))^2 + (2-4)^2}\\\\d = \sqrt{(-1+3)^2 + (2-4)^2}\\\\d = \sqrt{(2)^2 + (-2)^2}\\\\d = \sqrt{4 + 4}\\\\d = \sqrt{8}\\\\$

This is the radius because it stretches from the center to a point on the circle, so $r = \sqrt{8}$

Squaring both sides will get us $r^2 = 8$

One useful template for a circle is the equation $(x-h)^2+(y-k)^2 = r^2\\\\$

(h,k) is the center

r is the radius

Let's plug in the given center (h,k) = (-1,2) and the r^2 value we found earlier.

$(x-h)^2+(y-k)^2 = r^2\\\\(x-(-1))^2+(y-2)^2 = 8\\\\(x+1)^2+(y-2)^2 = 8\\\\$

You can confirm this by using a tool like Desmos. See below.

------------------------------------------------------------------------

Problem 8

The endpoints of the diameter are (-2,1) and (8,25)

The center is the midpoint of these endpoints.

The midpoint of the x coordinates is (-2+8)/2 = 3

The midpoint of the y coordinates is (1+25)/2 = 13

The center is (h,k) = (3,13)

Now find the distance from the center to one of the points on the circle, let's say to (8,25)

$(x_1,y_1) = (3,13) \text{ and } (x_2, y_2) = (8,25)\\\\d = \sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2}\\\\d = \sqrt{(3-8)^2 + (13-25)^2}\\\\d = \sqrt{(-5)^2 + (-12)^2}\\\\d = \sqrt{25 + 144}\\\\d = \sqrt{169}\\\\d = 13\\\\$