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3 years ago

R(x)=x^3+5x^2-2x-24 Plot the zero(s) of the function.

Mathematics

1 answer:

nevsk [136]3 years ago

6 0

:) Brainliest pls?

Answer:

The zeros are {2, -3, -4} which need to be plotted on the x-axis.

Step-by-step explanation:

I'll find the zeros, aka x-intercepts, and you could probably graph them.

To find the zeros, let's factor this polynomial:

r(x) = (x - 2)(x^2+7x+12)

r(x) = (x - 2)(x + 3)(x + 4)

The zeros are {2, -3, -4} which need to be plotted on the x-axis.

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Write an equation that expresses the following relationship.

ahrayia [7]

Given:

u varies directly with the square of p and inversely with d.

To find:

The equation for the given situation.

Solution:

If y is directly proportional to x, then

$y\propto x$

If y is inversely proportional to x, then

$y\propto \dfrac{1}{x}$

It is given that u varies directly with the square of p and inversely with d. So,

$u\propto \dfrac{p^2}{d}$

It can be written as:

$u=k\dfrac{p^2}{d}$

Where, k is the constant of proportionality.

Therefore, the required equation is $u=\dfrac{kp^2}{d}$ .

6 0

3 years ago

Minimum point of the quadratic x^2+6x-2

WITCHER [35]

Answer:

(-3,-11)

Step-by-step explanation:

Compare the given quadratic equation with the general quadratic equation.

a=1, b=6 and c=-2

$x=-\dfrac{(6)}{2(1)}\\=-3$

Subsitute $-3$ for $x$ in given quadratic equation.

$y=(-3)^2+6(-3)-2\\=9-18-2\\=-11$

The minimum point is (-3,-11).

3 0

3 years ago

Given: △FKL, FK=a, m∠F=45°, m∠L=30° Find: FL

drek231 [11]

The best way to do this is to draw a picture of ΔFKL and include line segment KM that is perpendicular to FL. This creates ΔFKM which is a 45°-45°-90° triangle and ΔLKM which is a 30°-60°-90° triangle.

Find the lengths of FM and ML. Then, FM + ML = FL

FM

ΔFKM (45°-45°-90°): FK is the hypotenuse so FM = $\frac{a}{\sqrt{2}} = \frac{a\sqrt{2} }{2}$

ML

ΔLKM (30°-60°-90°): from ΔFKM, we know that KM = $\frac{a\sqrt{2} }{2}$ , so KL = $\frac{a}{\sqrt{6}} = \frac{a\sqrt{6} }{6}$

FM + ML = FL

$(\frac{3}{3})\frac{a\sqrt{2} }{2} + \frac{a\sqrt{6} }{6}$

= $\frac{3a\sqrt{2} + a\sqrt{6} }{6}$

8 0

3 years ago

Read 2 more answers

The following table shows the number of hours some teachers in two schools expect students to spend on homework each week: Schoo

Lerok [7]

Answer:

For school A: Minimum=6, Q₁=6.5, Median= 14, Q₃=16, Maximum=17, IQR=9.5

For school B: Minimum=5, Q₁=8, Median= 12, Q₃=15.5, Maximum=19, IQR=7.5

No, the box plots are not symmetric.

Step-by-step explanation:

Part A

The given data sets are

School A : 9,14,15,17,17,7,15,6,6

School B : 12,8,13,11,19,15,16,5,8

Arrange the data in ascending order.

School A : 6,6,7,9,14,15,15,17,17

School B : 5,8,8,11,12,13,15,16,19

Divide each data set in four equal parts.

School A : (6,6),(7,9),14,(15,15),(17,17)

School B : (5,8),(8,11),12,(13,15),(16,19)

For school A:

Minimum=6, Q₁=6.5, Median= 14, Q₃=16, Maximum=17

Interquartile range of the data is

$IQR=Q_3-Q_1=16-6.5=9.5$

For school B:

Minimum=5, Q₁=8, Median= 12, Q₃=15.5, Maximum=19

Interquartile range of the data is

$IQR=Q_3-Q_1=15.5-8=7.5$

Part B:

The box plots are not symmetric because the data values are different. Five number summary and IQR of both the data set are different.

4 0

4 years ago

8 minutes whats the distance she walks

UkoKoshka [18]

Answer:

It depend on the speed that she was walking, if she was faster she would have a greater distance. But is commonly considered to be half a mile.

Hope it helps!

3 0

3 years ago