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3 years ago

The point P(x, y) lies on the terminal side of an angle a = Ū in standard position. What are the signs of the values of x and y?

Mathematics

1 answer:

Akimi4 [234]3 years ago

8 0

I think the answer is both x and y are positive

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Solve for x^2-2x+1=0

seraphim [82]

Answer:

$x=1$

Step-by-step explanation:

$x^2-2x+1=0\\$

We will start by using the quadratic formula.

$\frac{-(-2)+-\sqrt{(-2)^2-4(1)(1)}}{2(1)}$

We will simplify that and get:

$\frac{-(-2)}{2(1)}$

So our answers would be:

$x=1$

7 0

3 years ago

How do i find the square root of pi

SpyIntel [72]

Answer:

3.141592654...

Step-by-step explanation:

it just keeps going and going and never stops.

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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: $c^2 = 9dp$

8 0

3 years ago

What is the greatest common factor of 48 and 84? A.8 B.12 C.4 D.24

LiRa [457]

48 = 2*2*2*2*3
84 = 2*2*3*7

Looking for common factors we get the GCF to be 2*2*3 = 12

Its B

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4 years ago

(6+9)×[(120-90)-9]÷7

belka [17]

45 is the answer to this

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3 years ago

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