All categories

2 years ago

Help, i just need the answers and no need for explanations.

Mathematics

1 answer:

maria [59]2 years ago

3 0

Answer:

The equation of the quadratic function shown is;

x^2+ 2x -3

Step-by-step explanation:

Here in this question, we need to know the quadratic equation whose graph was shown.

The key to answering this lies in knowing the roots of the equation.

The roots of the equation are the solution to the quadratic equation and can be seen from the graph at the point where the quadratic equation crosses the x-axis.

The graph crosses the x-axis at two points.

These are at the points x = -3 and x = 1

So what we have are;

x + 3 and x -1

Multiplying both will give us the quadratic equation we are looking for.

(x + 3)(x-1) = x(x -1) + 3(x-1)

= x^2 -x + 3x -3 = x^2 + 2x -3

You might be interested in

Twelve pounds of beans are distributed equally into 8 bags to give out at the food bank how many pounds of beans are in each bag

Thepotemich [5.8K]

Answer:

1.5 lbs of beans.................

7 0

3 years ago

Read 2 more answers

you had eight hours until you have to be on set if you will spend 3/4 of an hour driving to the TV station how much time does th

Inessa05 [86]

Answer:

7 hours and 15 minutes

Step-by-step explanation:

8 - 3/4 = 7 1/4 hours

7 0

3 years ago

There was a Rockets game on Friday and Saturday.

katrin [286]

18,055
-
3,138
———-
14,927

14,927
+
18,055
———-
32,982

6 0

2 years ago

Read 2 more answers

Quick Please :) Exponents and powers

MrRissso [65]

Step-by-step explanation:

$( {4}^{ - 3} ) ^{ - 3} = {4}^{9}$

$( { - 4}^{9} ) ^{2} = ( - 4) {}^{18}$

$( {4}^{6} ) ^{ - 3} = \frac{1}{{4}^{18}}$

$( { - 4}^{9} ) {}^{2} = ( { - 4})^{18}$

$({4}^{0} ) = {4}^{0}$

Hope this is correct and helpful

HAVE A GOOD DAY!

6 0

2 years ago

Read 2 more answers

Find [5(cos 330 degrees + I sin 330 degrees)]^3

earnstyle [38]

Given a complex number in the form:
$z= \rho [\cos \theta + i \sin \theta]$
The nth-power of this number, $z^n$ , can be calculated as follows:

- the modulus of $z^n$ is equal to the nth-power of the modulus of z, while the angle of $z^n$ is equal to n multiplied the angle of z, so:
$z^n = \rho^n [\cos n\theta + i \sin n\theta ]$
In our case, n=3, so $z^3$ is equal to
$z^3 = \rho^3 [\cos 3 \theta + i \sin 3 \theta ] = (5^3) [\cos (3 \cdot 330^{\circ}) + i \sin (3 \cdot 330^{\circ}) ]$ (1)
And since
$3 \cdot 330^{\circ} = 990^{\circ} = 2\pi +270^{\circ}$
and both sine and cosine are periodic in $2 \pi$ , (1) becomes
$z^3 = 125 [\cos 270^{\circ} + i \sin 270^{\circ} ]$

6 0

3 years ago