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2 years ago

Ang Name two vertical angles, two supplementary angles in the diagram below.

Mathematics

1 answer:

Ksivusya [100]2 years ago

4 0

Answer:

5 and 3 is vertical. 4 and 6 are vertical. 2 and 1 are supplementary angles.

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If ∠ABC and ∠CBD are supplementary angles, then m∠ABC + m∠CBD = 180° represents the definition of Question 2 options: A) complem

Oksanka [162]

Answer:

I would think it is supplementary.

Step-by-step explanation:

If 2 angles equal 180 degrees, then they are supplementary.

6 0

3 years ago

Find the value of x and the value of y.

Zina [86]

The answer is B. x = 65, y = 10.

6 0

3 years ago

Savvas Texas Algebra II

Elan Coil [88]

Answer:

The number of small balloons = 46

The number of large balloons = 159

The number of bottles of helium gas = 5

The total cost of the supplies = $68.97

Step-by-step explanation:

The volume of each small balloon = 300 in.³

The volume of each large balloon = 1000 in.³

The cost of one package of 50 small balloons = $2.59

The cost of one package of 24 large balloons = $1.99

The volume of one small tank of pressurized helium = 20 ft³

The volume cost of one small tank of pressurized helium = $10.49

By conversion

20 ft³ = 34,560 in.³

The volume of 30 small balloons = 30 × 300 in.³ = 9,000 in.³

The volume of 30 large balloons = 30 × 1000 in.³ = 30,000 in.³

The volume taken by 24 large balloons = 24,000 in.³,

24 large balloons cost = $1.99

The volume taken by 50 small balloons = 15,000 in.³

50 small balloons cost = $2.59

Therefore, we buy more large balloons

We have;

A×300 + B×1000 = C×34,560

Given that all the helium will be used, the volume of helium will be a multiple of 300, therefore, when 5 bottles of helium is used, we have;

A×300 + B×1000 = 5×34,560 = 172,800 in.³

Given that there will be at least 30 of each balloon, gives;

We remove the minimum possible quantity (more than 30) of small balloons from 172,800 in.³ leaving multiples of 1,000s as follows;

Using Excel, the quantities of 36 and 46 small balloons give the same cost

For 46 small balloons

The volume of 46 small balloons = 13,800 in.³

Therefore, 172,800 in.³ - 13,800 in.³ = 159,000 in.³ = The volume of the large balloons

The quantity of the large balloons = 159,000 in.³/(1,000 in.³/balloon) = 159 balloons

Therefore, we have 46 small balloons, and 159 large balloons, with 5 small bottles of helium gas

The number of package of large balloons is 159/24 = 6.625 ≈ 7 packages

The cost of the 7 packages of large balloons = 7 × $1.99 = $13.93

The number of package of small balloons is 46/50= 0.92 ≈ 1 package

The cost of the 1 packages of small balloons = 1 × $2.59 = $2.59

The cost of the five bottles of helium = 5 × $10.49 = $52.45

$68.97

The number of small balloons = 46

The number of large balloons = 159

The number of bottles of helium gas = 5

The total cost of the supplies = $68.97

At the same cost, we can also have;

The number of small balloons = 36

The number of large balloons = 162

The number of bottles of helium gas = 5

The total cost of the supplies = $68.97

Therefore, we can have 36 or 46 small balloons and 162 or 159 large balloons along with 5 bottles of helium gas with a total cost of $13.93 + $2.59 + $52.45 = $68.97.

3 0

3 years ago

An airplane is preparing to land at an airport. It is 44,8000 feet above the ground and is descending at the rate of 3,100 feet

Marina86 [1]

Answer:

Time: 8 minutes

Altitude: 20000ft

Method 1 is easiest

Method 3 is easiest

Step-by-step explanation:

Given

Airplane 1:

$Height = 44800 ft$

$Descending\ Rate = 3100ft/min$

Airplane 2:

$Ascending\ Rate = 2500ft/min$

Required

Determine when both planes would be at the same altitude?

Let the minute be represented by m

For Airplane 1, Its altitude at any height h is:

$Airplane\ 1 = Height - Descending\ Rate * m$

It is minus (-) because the airplane is descending

$Airplane\ 1 = 44800 - 3100 * m$

$Airplane\ 1 = 44800 - 3100m$

For Airplane 2, Its altitude at any height h is:

$Airplane\ 2 = Ascending\ Rate * m$

$Airplane\ 2 = 2500 * m$

$Airplane\ 2 = 2500m$

Method 1:

For both heights to be equal, we have that:

$Airplane\ 1 = Airplane\ 2$

This gives:

$44800 - 3100m = 2500m$

Collect Like Terms

$44800 = 2500m + 3100m$

$44800 = 5600m$

$5600m = 44800$

$m = 44800/5600$

$m = 8\ min$

Hence, the time they will be at the same altitude is 8 minutes

Substitute 8 for m in

$Airplane\ 2 = 2500m$

$Airplane\ 2 = 2500 * 8$

$Airplane\ 2 = 20000\ ft$

Hence, they will be at the same altitude at 20000ft

Method 2:

We have that:

$Airplane\ 1 = 44800 - 3100m$

$Airplane\ 2 = 2500m$

Since they are to be at the same altitude, then

The difference in their altitude must be 0

i.e.

$Airplane\ 1 - Airplane\ 2 = 0$

This gives

$44800 - 3100m - 2500m = 0$

$44800 - 5600m = 0$

Collect Like Terms

$5600m = 44800$

$m = 44800/5600$

$m = 8\ min$

Substitute 8 for m in

$Airplane\ 1 = 44800 - 3100m$

$Airplane\ 1 = 44800 - 3100 * 8$

$Airplane\ 1 = 44800 - 24800$

$Airplane\ 1 = 20000\ ft$

Method 3:

We have that:

$Airplane\ 1 = 44800 - 3100m$

$Airplane\ 2 = 2500m$

Since they are to be at the same altitude, then

The ratio of their altitudes must be 1

i.e.

$\frac{Airplane\ 1}{Airplane\ 2} = 1$

$\frac{44800 - 3100m}{2500m} = 1$

Cross Multiply

$44800 - 3100m = 1 * 2500m$

$44800 - 3100m = 2500m$

Collect Like Terms

$44800 = 2500m + 3100m$

$44800 = 5600m$

$5600m = 44800$

$m = 44800/5600$

$m = 8\ min$

Substitute 8 for m in

$Airplane\ 1 = 44800 - 3100m$

$Airplane\ 1 = 44800 - 3100 * 8$

$Airplane\ 1 = 44800 - 24800$

$Airplane\ 1 = 20000\ ft$

Hence;

Their altitudes must be 20000ft

Though the three methods applied uses the same logic at some point, the first method applied is still the easiest and it is a straight forward method that could be applied in solving the question.

Method 3 is the most difficult.

7 0

3 years ago

Zeke and Scarlett deposit $500.00 into a savings account which earns 5% interest

nordsb [41]

Answer:

578.81

Step-by-step explanation:

PV(1+i)ⁿ

500(1+.05)³

578.8125

which rounds to 578.81

4 0

2 years ago