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2 years ago

21 and 22 are complementary angles. The measure of Z1 is 60°. The

Mathematics

1 answer:

Lemur [1.5K]2 years ago

3 0

Answer:

The value of x is 7

Step-by-step explanation:

Complementary angles have a sum of 90°, therefore the equation to solve for x is:

∠1+∠2=90°

60°+5(x-1)°=90°

5(x-1)=30°

x-1=6

x=7

So the value of x is 7

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In 1925, Zbigniew Morón published a rectangle that could be dissected into nine different sized squares as shown in the diagram.

Vlada [557]

Answer:

Area of the rectangle = 1056 square units

Step-by-step explanation:

Rectangle published could be dissected into nine different squares with different sizes.

Area of the squares with different measure of sides will be,

For side length = 1 unit

Area of the square = (Side)²

= 1²

= 1 square units

For side length = 4 units

Area of the square = 4² = 16 units²

For side length = 7 units

Area of the square = 7² = 49 units²

For side length = 8 units

Area of the square = 8² = 64 units²

For side length = 9 units

Area of the square = 9² = 81 square units

For side length = 10 units

Area of the square = 10² = 100 square units

For side length = 14 units

Area of the square = 14² = 196 square units

For side length = 15 units

Area of the square = 15² = 225 square units

For side length = 18 units

Area of the square = 18² = 324 square units

Total area of the rectangle by adding these 9 squares

= 1 + 16 + 49 + 64 + 81 + 100 + 196 + 225 + 324

= 1056 square units

8 0

3 years ago

You are designing a rectangular garden. the garden will be enclosed by fencing on three sides and by a house on the fourth side.

RUDIKE [14]

Suppose the dimensions of the rectangle is x by y and let the side enclosed by a house be one of the sides measuring x, then the sides that is to be enclosed are two sides measuring y and one side measuring x.

Thus, the length of fencing needed is given by

P = x + 2y

The area of the rectangle is given by xy,

i.e.

$xy = 288 \\ \\ \\ \\ \Rightarrow y= \frac{288}{x}$

Substituting for y into the equation for the length of fencing needed, we have

$P=x+2\left( \frac{288}{x} \right)=x+ \frac{576}{x}$

For the amount of fencing to be minimum, then

$\frac{dP}{dx} =0 \\ \\ \Rightarrow1- \frac{576}{x^2} =0 \\ \\ \Rightarrow \frac{576}{x^2} =1 \\ \\ \Rightarrow x^2=576 \\ \\ \Rightarrow x=\sqrt{576}=24$

Now, recall that

$y= \frac{288}{x} = \frac{288}{24} =12$

Thus, the length of fencing needed is given by

P = x + 2y = 24 + 2(12) = 24 + 24 = 48.

Therefore, 48 feets of fencing is needed to enclose the garden.

8 0

2 years ago

Find the mean of the data in the bar chart below.

Snowcat [4.5K]

<h2>Answer:</h2>

<h2>Step-by-step explanation:</h2>

<u><em>The mean is also the average.</em></u>

Add up all the number of students all together:

12+13+14+9

Now, Divide the total amount of students (48) by the amount of numbers you added (in this case, 48/4 because we added 4 numbers wich are 12+13+14+9)

The average amount of students is 12

7 0

2 years ago

Does ANYONE Know How To Solve THIs…Brainliest Will be awarded!!!

emmainna [20.7K]

Answer:

w = 8
x = 8√2
y = 8√2
z = 16

Step-by-step explanation:

A 45-45-90 right triangle is isosceles. That means both legs are the same length, and its hypotenuse is √2 times the leg length. You can figure these facts using trigonometry or the Pythagorean theorem, or you can rely on your memory of them.

w = 8 . . . . . . . . . . . same as the leg length shown

x = y = 8√2 . . . . . hypotenuse of smaller triangle; both legs are the same

z = (8√2)√2 = 16 . . . . hypotenuse of larger triangle

3 0

2 years ago

The scale on a map is 1: 75 000. (a) Calculate the ACTUAL distance, in km, between two places which are 16 cm apart on the map.

finlep [7]

Answer:

a. 1 200 000 km

b. 0.000 773 cm

c. 67 500 000 000 km²

d. 0.000 000 016 cm²

Step-by-step explanation:

a. Calculate the ACTUAL distance, in km, between two places which are 16 cm apart on the map.

Since the scale is 1 : 75000 that is 1 cm : 75000 km, we want to find the actual distance of 16 cm on the map. Let x be the actual distance in km.

So, 1 cm : 75000 km = 16 cm : x km

So, 1 cm/75000 km = 16 cm/x km

Thus x km = 16 cm × 75000 km/1 cm = 1 200 000 km

(b) Calculate length of a road on the map which is 0.58 km.

Let y be the length of road on the map. So, using our scale

1 cm : 75000 km = y cm : 0.58 km

So, 1 cm/75000 km = y cm/0.58 km

Thus y cm = 0.58 km × 1 cm/75000 km = 0.000 773 cm

(c) Calculate, in km², the ACTUAL area of a lake which is 12 cm2 on the map.

Since the scale is 1 : 75000 , the scale for the area would be the square of this. So, 1² : 75000² = 1 : 5 625 000 000 which is 1 cm² : 5 625 000 000 km²

Let z be the actual area of the lake.

So, 1 cm² : 5 625 000 000 km² = 12 cm² : z km²

So, 1 cm²/5 625 000 000 km² = 12 cm²/z km²

Thus z km² = 12 cm² × 5 625 000 000 km²/1 cm² = 67 500 000 000 km²

(d) Calculate the area of a play field on the map which has has an actual area of 90 km².

Let a represent the area of the play field on the map. Using our scale for the area,

1 cm² : 5 625 000 000 km² = a cm² : 90 km²

So, a cm² = 90 km² × 1/5 625 000 000 km² = 0.000 000 016 cm²

4 0

2 years ago