Laura & Mary are doing their homework together.

1 answer:

8 0

Laura is correct.

A function is a relation in which every x-value corresponds with no more than one y-value. In this problem, the x-value 1 has two different answers; this means it is not a function.

Additionally, if you graphed this, it would not pass the vertical line test.

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Mrs. Mott called two companies about getting new uniforms for the soccer team. The first company she called charges $70 per unif

Fittoniya [83]

Answer:

For less than 7 uniforms.

Step-by-step explanation:

The first company she called charges $70 per uniform.

So, the cost of x uniforms will be $70x.

The second company she called charges $280 plus $30 per uniform.

So, the cost of x uniform will be $(280 + 30x).

Now, if the total cost of purchasing x number of uniforms from the first company is less than that from the second company then, we can write the inequality equation as

70x < 280 + 30x

⇒ 70x - 30x < 280

⇒ 40x < 280

⇒ x < 7

Therefore, for less than 7 uniforms the cost from the first company will be less than the cost from the second company. (Answer)

5 0

3 years ago

Write a quadratic equation given the roots -1/3 and 5, show your work

Travka [436]

$\boxed{(x - a)(x - b) = 0}$

The equation above is the intercept form. Both a-term and b-term are the roots of equation.

$x = - \frac{1}{3} \\ x = 5$

These are the roots of equation. Therefore we substitute a = - 1/3 and b = 5 in the equation.

$(x + \frac{1}{3} )(x - 5) = 0$

Here we can convert the expression x+1/3 to this.

$x + \frac{1}{3} = 0 \\ 3x + 1 = 0$

Rewrite the equation.

$(3x + 1)(x - 5) = 0$

Simplify by multiplying both expressions.

$3 {x}^{2} - 15x + x - 5 = 0 \\ 3 {x}^{2} - 14x - 5 = 0$

Answer Check

Substitute the given roots in the equation.

$3 {(5)}^{2} - 14(5) - 5 = 0 \\ 75 - 70 - 5 = 0 \\ 75 - 75 = 0 \\ 0 = 0$

$3( - \frac{1}{3} )^{2} - 14( - \frac{1}{3}) - 5 = 0 \\ 3( \frac{1}{9} ) + \frac{14}{3} - 5 = 0 \\ \frac{1}{3} + \frac{14}{3} - \frac{15}{3} = 0 \\ \frac{15}{3} - \frac{15}{3} = 0 \\ 0 = 0$