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2 years ago

Give Three Real Life examples of Supplementary Angles.

Mathematics

2 answers:

gregori [183]2 years ago

6 0

Stars, cups, and legos are three real life examples of supplementary angles

devlian [24]2 years ago

3 0

Supplementary angles are made when two angles equal a sum of 180 degrees. Some real life examples of supplementary angles are stars, cups and logos.

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sung buys a 12-month membership to a gym there is no sign up fee and sung makes equal payments each month after three months he

Nataly [62]

The solution is

a) The amount Sung pays per month is $ 65

b) The total amount for the 12-month memberships is $ 780

What is an Equation?

Equations are mathematical statements with two algebraic expressions flanking the equals (=) sign on either side.

It demonstrates the equality of the relationship between the expressions printed on the left and right sides.

Coefficients, variables, operators, constants, terms, expressions, and the equal to sign are some of the components of an equation. The "=" sign and terms on both sides must always be present when writing an equation.

Given data ,

Let the total amount of the 12-month membership fee be represented as A

Now , the value of A is

The amount balance after 3 months = $ 585

The amount balance after 5 months = $ 455

So ,

The difference between the months = 5 months - 3 months

The difference between the months = 2 months

The difference between the amounts = $ 585 - $ 455

The difference between the amounts = $ 130

So , for the period of 2 months , the amount = $ 130

The amount for 1 month of the membership = $ 130 / 2

The amount for 1 month of the membership = $ 65

Now , the equation will be

Total amount of the 12-month membership fee A = 12 x amount for 1 month of the membership

Substituting the values in the equation , we get

Total amount of the 12-month membership fee A = 12 x $ 65

Total amount of the 12-month membership fee A = $ 780

Therefore , the value of A is $ 780

Hence ,

The total membership amount is $ 780 and amount per month is $ 65

To learn more about equations click :

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3 0

1 year ago

BRAINLIEST FOR AN ACTUAL ANSWER

Andreyy89

Answer:

The correct pair of functions is the third one: h(x)=(x−24)^2 and g(x)=x2

Step-by-step explanation:

Example: If we have q(x) = x^2 and its graph, moving the vertex of this graph 24 units to the right results in r(x) = (x - 24)^2.

The correct pair of functions is the third one: h(x)=(x−24)^2 and g(x)=x2

Note: the fourth pair is incorrect, because the " + " sign moves the graph of x^2 24 units to the left.

3 0

2 years ago

HELP ME!!!

larisa [96]

Answer:

-2, 8/3

Step-by-step explanation:

You can consider the area to be that of a trapezoid with parallel bases f(a) and f(4), and width (4-a). The area of that trapezoid is ...

A = (1/2)(f(a) +f(4))(4 -a)

= (1/2)((3a -1) +(3·4 -1))(4 -a)

= (1/2)(3a +10)(4 -a)

We want this area to be 12, so we can substitute that value for A and solve for "a".

12 = (1/2)(3a +10)(4 -a)

24 = (3a +10)(4 -a) = -3a² +2a +40

3a² -2a -16 = 0 . . . . . . subtract the right side

(3a -8)(a +2) = 0 . . . . . factor

Values of "a" that make these factors zero are ...

a = 8/3, a = -2

The values of "a" that make the area under the curve equal to 12 are -2 and 8/3.

_____

<em>Alternate solution</em>

The attachment shows a solution using the numerical integration function of a graphing calculator. The area under the curve of function f(x) on the interval [a, 4] is the integral of f(x) on that interval. Perhaps confusingly, we have called that area f(a). As we have seen above, the area is a quadratic function of "a". I find it convenient to use a calculator's functions to solve problems like this where possible.