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3 years ago

Quadratic ( standord formula )

Mathematics

1 answer:

Nataly_w [17]3 years ago

5 0

Answer:

The standard form is ax² + bx + c = 0 with a, b, and c being constants, or numerical coefficients, and x is an unknown variable.

Hope this helps :)

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Transformations<br>I need help on this worksheet please.

Neporo4naja [7]

Answer:

Step-by-step explanation:

Question 2

As far as I can see, you got it right. The general transformation for 90 ccw is

(x,y) ===> (-y, x)

What that means is for the x you put in -y changing the sign to the opposite and for the y you put in x and this time you leave the sign alone . The transformation is shown in the left hand diagram.

The two tables are shown below.

Original

(1,4)
(1,2)
(3,2)
(3,5)

The transformed table is

(-4,1)
(-2,1)
(-2,3)
(-5,3)
(-4,1) This is just to let the program know to close the figure For some reason this did not have lines and if I delete it and put the lines in, I won't be able to upload the new diagram.

===========

Four

This one transforms from (x,y) to (-x,-y) which means where you see an x, you put a - x and where you see a y, you put a minus y. It is the middle frame.

Original

(-4,3)
(0,3)
(-2,0)
(-4,3) Here again, this is just to close the figure.

The transformed figure in red I think is

(4,-3)
(0,-3
(2,0)
(4,-3) And this closes the figure as well.

==========

Six

The diagram is on the right

Reflection about the y axis. Here the transformation is (x,y) ====> (-x,y) notice the ys don't change.

(-3,0)
(-5,-2)
(-1,-2)

There is no closure.

Reflection

(3,0)
(5,-2)
(1,-2)

4 0

2 years ago

Find the value of x - Secant and Tangent Angles in Circles

VladimirAG [237]

Answer:

C. 70°

Step-by-step explanation:

The inscribed angle marked 15° intercepts an arc that is double that measure, so the intercepted arc on the right is 2×15° = 30°.

The external angle marked 20° is half the difference of the intercepted arcs, so is ...

20° = (1/2)(x - 30°)

40° = x - 30° . . . . . . multiply by 2

70° = x . . . . . . . . . . . add 30°

The value of x is 70°.

6 0

2 years ago

Which of the following sets contain all roots of the polynomial

Zarrin [17]

All these given roots are contained in option C.

The given equation is :

$9x^{3}+9x^{2}-4x-4=0$

$9x^{2}(x+1)-4(x+1)=0$

$(9x^{2}-4)(x+1)=0$

$9x^{2}-4=0,x+1=0$

$x^{2}=\frac{4}{9},x=-1$

$x=\frac{2}{3},x=-1$

Roots = $-1,\frac{-2}{3} ,\frac{2}{3}$

Hence these lie in between $(\frac{-3}{2} ,\frac{3}{2})$

8 0

2 years ago

When an electric current passes through two resistors with resistance r1 and r2, connected in parallel, the combined resistance,

kondaur [170]

Answer:

The combined resistance of a circuit consisting of two resistors in parallel is given by:

$\frac{1}{R}=\frac{1}{r_1}+\frac{1}{r_2}$

where

R is the combined resistance

$r_1, r_2$ are the two resistors

We can re-write the expression as follows:

$\frac{1}{R}=\frac{r_1+r_2}{r_1r_2}$

$R=\frac{r_1 r_2}{r_1+r_2}$

In order to see if the function is increasing in r1, we calculate the derivative with respect to r1: if the derivative if > 0, then the function is increasing.

The derivative of R with respect to r1 is:

$\frac{dR}{dr_1}=\frac{r_2(r_1+r_2)-1(r_1r_2)}{(r_1+r_2)^2}=\frac{r_2^2}{(r_1+r_2)^2}$

We notice that the derivative is a fraction of two squared terms: therefore, both factors are positive, so the derivative is always positive, and this means that R is an increasing function of r1.

To solve this part, we use again the expression for R written in part a:

$R=\frac{r_1 r_2}{r_1+r_2}$

We start by noticing that there is a limit on the allowed values for r1: in fact, r1 must be strictly positive,

$r_1>0$

So the interval of allowed values for r1 is

$0$

From part a), we also said that the function is increasing versus r1 over the whole domain. This means that if we consider a certain interval

a ≤ r1 ≤ b

The maximum of the function (R) will occur at the maximum value of r1 in this interval: so, at

$r_1=b$

6 0

3 years ago

What is the center and radius of a circle whose equation is x2 + y2 -2x -6y -5 = 0?

ddd [48]

x²+y²-2x-6y-5=0

x²-2x+y²-6y=5

x²-2x+1+y²-6y+9=5+1+9

(x-1)²+(y-3)²=15

(x-1)²+(y-3)²=(V15)²

-> centre of the circle: C(-1,-3)

-> radius of the circle: V15

4 0

1 year ago