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2 years ago

In your own words make a conjecture about the relation

Mathematics

1 answer:

damaskus [11]2 years ago

3 0

Step-by-step explanation:

The congruent angles of the isocels triangle's opposite sides's are congruent.

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What is the unit rate of the proprrional relationship represented by the equation y=3x

NeX [460]

The unit rate of the proportional relationship y = 3x is 3

<h3>How to determine the unit rate?</h3>

The proportional relationship is given as:

y = 3x

Divide both sides by x

y/x = 3

For a proportional relationship y/x = k;

The unit rate is 3

This means that the unit rate of the proportional relationship y = 3x is 3

Read more about proportional relationship at

brainly.com/question/12242745

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6 0

2 years ago

Seven times the difference of some number and 1 gives the quotient of 70 and 10, find the number

Agata [3.3K]

7(x-1)=70/10
7x-7=7
7x=14
x = 2

The number is 2.

6 0

3 years ago

Y=x^2-5x+11<br> Find vertex form

Marat540 [252]

Answer:

Step-by-step explanation:

The answer to life the universe and everything is 42

8 0

3 years ago

Solve the following equation by factoring:9x^2-3x-2=0

olya-2409 [2.1K]

Answer:

The two roots of the quadratic equation are

$x_1= - \frac{1}{3} \text{ and } x_2= \frac{2}{3}$

Step-by-step explanation:

Original quadratic equation is $9x^{2}-3x-2=0$

Divide both sides by 9:

$x^{2} - \frac{x}{3} - \frac{2}{9}=0$

Add $\frac{2}{9}$ to both sides to get rid of the constant on the LHS

$x^{2} - \frac{x}{3} - \frac{2}{9}+\frac{2}{9}=\frac{2}{9}$ ==> $x^{2} - \frac{x}{3}=\frac{2}{9}$

Add $\frac{1}{36}$ to both sides

$x^{2} - \frac{x}{3}+\frac{1}{36}=\frac{2}{9} +\frac{1}{36}$

This simplifies to

$x^{2} - \frac{x}{3}+\frac{1}{36}=\frac{1}{4}$

Noting that (a + b)² = a² + 2ab + b²

If we set a = x and b = $\frac{1}{6}\right)$ we can see that

$\left(x - \frac{1}{6}\right)^2$ = $x^2 - 2.x. (-\frac{1}{6}) + \frac{1}{36} = x^{2} - \frac{x}{3}+\frac{1}{36}$

$\left(x - \frac{1}{6}\right)^2=\frac{1}{4}$

Taking square roots on both sides

$\left(x - \frac{1}{6}\right)^2= \pm\frac{1}{4}$

So the two roots or solutions of the equation are

$x - \frac{1}{6}=-\sqrt{\frac{1}{4}}$ and $x - \frac{1}{6}=\sqrt{\frac{1}{4}}$

$\sqrt{\frac{1}{4}} = \frac{1}{2}$

So the two roots are

$x_1=\frac{1}{6} - \frac{1}{2} = -\frac{1}{3}$

and

$x_2=\frac{1}{6} + \frac{1}{2} = \frac{2}{3}$

7 0

1 year ago

rectangular piece of paper 4 cm wide length twice its width measuring 3 cm by 7 cm how many sq cm will be visible on top?

aalyn [17]

22 square centimeters

8 0

3 years ago