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2 years ago

Find the y intercept only. 4x – y = 4

Mathematics

2 answers:

Scorpion4ik [409]2 years ago

8 0

Answer:

The y-intercept is the point (0, -4).

Step-by-step explanation:

Recall that x = 0 at the y-intercept. Set x = 0 in 4x - y = 4, obtaining:

4(0) - y = 4, or y = -4.

The y-intercept is the point (0, -4).

faltersainse [42]2 years ago

7 0

Answer:

ok so first we have to convert this into y intercept

4x-y=4

-4x

-y=-4x+4

*-1

y=4x-4

so the y intercept is -4

Hope This Helps!!!

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Solve the compound inequality and graph<br> Please help! :)

frozen [14]

Answer:

Step-by-step explanation:

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8 0

3 years ago

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: $c^2 = 9dp$

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3 years ago

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Alex17521 [72]

Answer:

a is -3

Step-by-step explanation:

negative 3 minus 2 is negative 5 plus 3 is negative 2

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3 years ago

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yaroslaw [1]

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3 years ago

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