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3 years ago

Explain the steps you would take to complete this conversion problem.

Mathematics

1 answer:

UNO [17]3 years ago

3 0

In the picture are the steps on how to solve this conversion problem

Hope this helps and maker me brainless plzzzzz

The answer is
46100000
——————g
221

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Find the side length of the equilateral triangle whose area is the square root of 3 cm2

Alla [95]

Here, Given

Area of Equilateral triangle = 3cm2

therefore, we know;

Area of Equilateral triangle =( $\sqrt{3}$ ÷4)× $a^{2}$

so,

3 =( $\sqrt{3}$ ÷4)× $a^{2}$

3÷( $\sqrt{3}$ ÷4) = $a^{2}$

2.63 cm = a .

The area of an equilateral triangle is the area covered by an equilateral triangle on a two-dimensional plane. An equilateral triangle is a triangle with all sides equal and all angles at 60 degrees. The area of the shape is the number of unit squares that fit inside.

Here, "unit" refers to 1, and a unit square is a square with one side. Alternatively, the area of an equilateral triangle is the total amount of space it surrounds in a two-dimensional plane.

Learn more about Area of triangle here: brainly.com/question/17335144

#SPJ4

6 0

2 years ago

Help please !!!!! asap

AnnyKZ [126]

The answer: m∡BCD = 130° .
_____________________________________
Explanation:
______________________________
m∡BCD = 9x - 5 = our answer.
_____________________________
Note: (9x - 5) + (m∡C IN Δ ACB)= 180 ;
____________________________
Reason: all angles on straight line add up to 180.
___________________________
Note: In Δ ACB; m∡A + m∡B + m∡c = 180.
_________________________________________
Reason: All three angles in any triangle add up to 180.
__________________________________________
Given Δ ACB, we are given:
_____________
m∡C= ?
m∡B = (4x + 5)
m∡A = 65
_____________________
So, given Δ ACB; m∡A + m∡B + m∡c = 180;
→Plug in our known values and rewrite:
___________________________________
Given Δ ACB; 65 + 4x + 5 + (m∡c) = 180;
→Simplify, and rewrite:
___________________________________
Given Δ ACB; 4x + 70 + (m∡c) = 180;
→Subtract "70" from each side of the equation; and rewrite:
___________________________________
Given Δ ACB; 4x + (m∡C) = 110;
→Subtract "4x" from EACH SIDE of the equation; to isolate: "(m∡c)" on one side of the equation; and "solve in terms of "(m∡C)" ;
______________________________________________
Given Δ ACB' m∡C = 110 - 4x ;
__________________________________________
So, we know that: (110 - 4x) + (9x - 5) = 180; (since all angles on a straight line add up to 180.
____________________
We can solve for "x".
____________________
(110 - 4x) + (9x - 5) = 180;
________________________
Rewrite as:
___________
(110 - 4x) + 1(9x - 5) = 180 ; (Note: there is an implied coefficient of "1"; since anything multiplied by "1" equals that same value).
_______________________
Note the "distributive property of multiplication":
_________________
a*(b+c) = ab + ac ; AND:
a*(b - c) = ab - ac .
_______________________
So, +1(9x - 5) = (+1*9x) - (+1*5) = 9x - 5 ;
__________________________
So we can rewrite:
___________________
(110 - 4x) + (9x - 5) = 180 ; as:
________________________
110 - 4x + 9x - 5 = 180 ; We can simplify this by combining "like terms" on the "left-hand side" of the equation:
_________________________________________________
110 - 5 = 105 ;
-4x + 9x = 5x;
______________
So, rewrite as: 5x + 105 = 180; Subtract "105" from EACH side; to get:
_____________________________________
5x = 75 ; Now, divide each side of the equation by "5";
to get: x = 15.
_____________________________________________
Now, we want to know: m∡BCD; which equals:
_____________________________________________
9x - 5 ; let us substitute "15" for "x"; and solve:
______________________
9x - 5 = 9*(15) - 5 = 135 - 5 = 130.
_____________________________
The answer: m∡BCD = 130°
________________________

6 0

3 years ago

Part 3: Write the equation of and graph an ellipse.

tensa zangetsu [6.8K]

Answer:

Step-by-step explanation:

The center is halfway between vertices, at (4, -6).

It is also halfway between foci.

:::::

The vertices are vertically aligned, so the parabola is vertical.

General equation for a vertical ellipse:

(y-k)²/a² + (x-h)²/b² = 1

with

center (h,k)

vertices (h,k±a)

co-vertices (h±b,k)

foci (h,k±c), c² = a²-b²

Apply your data and solve for h, k, a, and b.

center (h,k) = (4, -6)

h = 4

k = -6

vertices (4,-6±a) = (4,-6±9)

a = 9

foci (4,-6±c) = (4,-6±5√2)

b² = a² - c² = 9² - (5√2)² = 31

b = √31

The equation becomes

(y+6)²/81 + (x-4)²/31 = 1

:::::

length of major axis = 2a = 18

length of minor axis = 2b = 2√31

8 0

3 years ago

A silversmith mixed 100 g of a 40% silver alloy with 40 g of

notsponge [240]

Hey, don’t call for that link thing. It’s fake. Here’s how I solved this: x = [35(80%) + 70(25%)]/(35 + 70) = 43.3%

The answer is 43.3%

6 0

3 years ago

Find an equation of the plane that passes through the point P0 = (6, 3, 2) and is parallel to the xy-plane. g

Nutka1998 [239]

The xy-plane has a normal vector of 〈0, 0, 1〉, and any plane parallel to it will have the same normal vector.

Then the equation of the plane through (6, 3, 2) that is parallel to the xy-plane has equation

〈x - 6, y - 3, z - 2〉 • 〈0, 0, 1〉 = 0

==> z - 2 = 0

==> z = 2

3 0

3 years ago