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3 years ago

[Angles question]

Mathematics

1 answer:

11111nata11111 [884]3 years ago

8 0

Here is my friendly solution i'll skip the proof part for now. I am working on the others.

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Solve the system using elimination.<br><br> 5x + 4y = 12<br> 3x - 3y = 18

weeeeeb [17]

Point form: (4,-2)
Equation form: x=4, y=-2

5 0

3 years ago

Square ABCD and isosceles triangle BUC are drawn to create trapezoid AUCD. Square A B C D and triangle B U C are attached at sid

Art [367]

Answer:

d) 135º

Step-by-step explanation:

Note that the angle DCU is the sum of the angles DCB and BCU. The angle DCB is 90º because A B C D is a square, then all its angles are equal to 90º.

After attaching B U C to A B C D, we obtain a trapezoid A U C D. Since A U C D has at least one pair of parallel sides, then AU should be parallel to CD, thus the angle CBU must be 90º.

B U C is isoceles, so we conclude that other two angles must have the same size, and due to the sum of the angles of a triangle being 180º, then both BUC and BCU are equal to 45º

As a result, the angle DCU is equal to 90º+45º = 135º. Option d is the correct one.

5 0

4 years ago

Read 2 more answers

The corners of a meadow are shown on a coordinate grid. Ethan wants to fence the meadow. What length of fencing is required?

Nuetrik [128]

Answer:

34.6 units

Step-by-step explanation:

The lenght of fencing required is the total distance between point A to B, B to C, C to D, and D to A. That is the distance between all 4 corners of the meadow.

The coordinates of the corners of the meadow is shown on a coordinate plane in the attachment. (See attachment below).

Let's use the distance formula to calculate the distance between the 4 corners of the meadow using their coordinates as follows:

Distance between point A(-6, 2) and point B(2, 6):

$AB = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$

Let,

$A(-6, 2)) = (x_1, y_1)$

$B(2, 6) = (x_2, y_2)$

$AB = \sqrt{(2 - (-6))^2 + (6 - 2)^2}$

$AB = \sqrt{(8)^2 + (4)^2}$

$AB = \sqrt{64 + 16} = \sqrt{80}$

$AB = 8.9$ (nearest tenth)

Distance between B(2, 6) and C(7, 1):

$BC = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$

Let,

$B(2, 6) = (x_1, y_1)$

$C(7, 1) = (x_2, y_2)$

$BC = \sqrt{(7 - 2)^2 + (1 - 6)^2}$

$BC = \sqrt{(5)^2 + (-5)^2}$

$BC = \sqrt{25 + 25} = \sqrt{50}$

$BC = 7.1$ (nearest tenth)

Distance between C(7, 1) and D(3, -5):

$CD = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$

Let,

$C(7, 1) = (x_1, y_1)$

$D(3, -5) = (x_2, y_2)$

$CD = \sqrt{(3 - 7)^2 + (-5 - 1)^2}$

$CD = \sqrt{(-4)^2 + (-6)^2}$

$CD = \sqrt{16 + 36} = \sqrt{52}$

$CD = 7.2$ (nearest tenth)

Distance between D(3, -5) and A(-6, 2):

$DA = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$

Let,

$D(3, -5) = (x_1, y_1)$

$A(-6, 2) = (x_2, y_2)$

$DA = \sqrt{(-6 - 3)^2 + (2 - (-5))^2}$

$DA = \sqrt{(-9)^2 + (7)^2}$

$DA = \sqrt{81 + 49} = \sqrt{130}$

$DA = 11.4$ (nearest tenth)

Length of fencing required = 8.9 + 7.1 + 7.2 + 11.4 = 34.6 units

8 0

3 years ago

How many 3/4 cups are in 6 cups of container of ceral

Kitty [74]

There are 8 3/4 cups in 6 cups of cereal because 3/4 times 8 equals 6

8 0

3 years ago

The RideEm Bicycles factory can produce 170 bicycles in a day at a total cost of $10,900. It can produce 190 bicycles in a day a

KatRina [158]

If we assume that when producing 190 bicycles vs 170 bicycles the difference in total daily cost is purely attributed to the extra 20 bicycles, THEN the cost to produce 20 bicycles (190 - 170) = 11600 - 10900 = 700

So the cost per bicycle is 700/20 = 35/bike

Now we can subtract the cost of bicycle production to determine the daily fixed cost that is not related to the number of bicycles.

i.e. 170 bicycles x 35 = 5950

So on a day that RideEm produces 170 bicycles, subtract 5950 from total day cost of 10900, which equals 4950.

So the fixed daily costs are $4950, and on top of this, the company pays 35/bike to produce.

4 0

3 years ago