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2 years ago

How can you solve real-world and mathematical problems with numerical and algebraic equations and inequalities?

1 answer:

7 0

You can solve real-world and mathematical problems with numerical and algebraic equations and inequalities. Algebra can be applied to the temperature in different places that both change, height of a growing child over time, the speed of a car that changes over time (mph), and the age of people that increases over year. Algebra is used in parts of our everyday life, we just don’t realize it.

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HELP!!!!!!!Choose the graph that matches the following system of equations:

PolarNik [594]

The last one. And I did answer for you too.

4 0

2 years ago

I need help with pythagoras’ theorem

fomenos

Answer:

Pythagoras’ theorem is a way to find a side or hypothesis when you have 2 sides.

The formula is: a^2 + b^2 = c^2

a and b are sides

c is the hypothesis

<u>Ex: A triangle has a leg that is 5 inches and a leg that is 7 inches. Find the hypothesis using Pythagoras' theorem. </u>

A leg is another way of saying a side.

5^2 + 7^2 = c^2

25 + 49 = x^2

sqrt(74) = sqrt(x^2)

sqrt(74) inches = hypothesis

<u>Ex: A triangle has a leg that is 9 feet and a hypothesis that is 25 feet. Find the other leg using Pythagoras' theorem. </u>

9^2 + b^2 = 25^2

81 + b^2 - 81 = 625 - 81

sqrt(b^2) = sqrt(544)

b = sqrt(554)

Do you understand more?

4 0

2 years ago

Avocados cost $3 per pound. write and graph an equation in two variables that represents the cost of buying avocados.

mezya [45]

Answer:

y = 3x

Step-by-step explanation:

sorry cant do the graph

6 0

2 years ago

Please help fastttt i need to make sure i pass this please help me

qaws [65]

Answer:

5x+y=1

Step-by-step explanation:

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2 years ago

Lamar is writing a coordinate proof to show that a segment from the midpoint of the hypotenuse of a right triangle to the opposi

Zanzabum

1. N is a midpoint of the segment KL, then N has coordinates

$\left(\dfrac{x_K+x_L}{2},\dfrac{y_K+y_L}{2} \right) =\left(\dfrac{0+2a}{2},\dfrac{2b+0}{2} \right) =(a,b).$

2. To find the area of △KNM, the length of the base MK is 2b, and the length of the height is a. So an expression for the area of △KNM is

$A_{KMN}=\dfrac{1}{2}\cdot \text{base}\cdot \text{height}=\dfrac{1}{2}\cdot 2b\cdot a=ab.$

3. To find the area of △MNL, the length of the base ML is 2a and the length of the height is b. So an expression for the area of △MNL is

$A_{MNL}=\dfrac{1}{2}\cdot \text{base}\cdot \text{height}=\dfrac{1}{2}\cdot 2a\cdot b=ab.$

4. Comparing the expressions for the areas you have that the area $A_{KMN}$ is equal to the area $A_{MNL}$ . This means that the segment from the midpoint of the hypotenuse of a right triangle to the opposite vertex forms two triangles with equal areas.

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2 years ago

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