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2 years ago

ACT help?

Mathematics

1 answer:

timama [110]2 years ago

6 0

Hello!

Firstly, we can complete the linear function. To complete it, we need to find the y-intercept of the equation. To find it, we use the equation: y = -2/3x + b, substitute the point (2, 1) into the equation, and solve for b.

1 = -2/3(2) + b

1 = -4/3 + b (add 4/3 to both sides)

b = 7/3

The completed linear equation is: y = -2/3x + 7/3.

The easiest way to find the ordered pair that is on the graph is to graph the equation and label the points (the graph is posted below).

As seen on the graph, the ordered pair, (8, -3), is graphed. To prove it is correct, we can substitute the ordered pair into the linear function.

-3 = -2/3(8) + 7/3

-3 = -16/3 + 7/3

-3 = -3 | True!

Therefore, the answer is choice D, (8, -3).

I don't know any good sites with ACT practice problems, sorry and good luck!

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2 years ago

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Answer:

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Step-by-step explanation:

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If using the method of completing the square to solve the quadratic equation x2 + 3x - 17 = 0, which number would have to be add

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Answer:

Step-by-step explanation:

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2 years ago

1.Ordenar de menor a mayor los siguientes números

Nataliya [291]

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Espero que esto ayude!

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2 years ago

Read 2 more answers

One of the roots of the quadratic equation dx^2+cx+p=0 is twice the other, find the relationship between d, c and p

scZoUnD [109]

Answer:

$c^2 = 9dp$

Step-by-step explanation:

Given

$dx^2 + cx + p = 0$

Let the roots be $\alpha$ and $\beta$

So:

$\alpha = 2\beta$

Required

Determine the relationship between d, c and p

$dx^2 + cx + p = 0$

Divide through by d

$\frac{dx^2}{d} + \frac{cx}{d} + \frac{p}{d} = 0$

$x^2 + \frac{c}{d}x + \frac{p}{d} = 0$

A quadratic equation has the form:

$x^2 - (\alpha + \beta)x + \alpha \beta = 0$

So:

$x^2 - (2\beta+ \beta)x + \beta*\beta = 0$

$x^2 - (3\beta)x + \beta^2 = 0$

So, we have:

$\frac{c}{d} = -3\beta$ -- (1)

and

$\frac{p}{d} = \beta^2$ -- (2)

Make $\beta$ the subject in (1)

$\frac{c}{d} = -3\beta$

$\beta = -\frac{c}{3d}$

Substitute $\beta = -\frac{c}{3d}$ in (2)

$\frac{p}{d} = (-\frac{c}{3d})^2$

$\frac{p}{d} = \frac{c^2}{9d^2}$

Multiply both sides by d

$d * \frac{p}{d} = \frac{c^2}{9d^2}*d$

$p = \frac{c^2}{9d}$

Cross Multiply

$9dp = c^2$

$c^2 = 9dp$

Hence, the relationship between d, c and p is: $c^2 = 9dp$

8 0

2 years ago