Ask question

Ask question

All categories

3 years ago

15

I need the help as possible please somebody help me I need help I swear please somebody help me help me help me please someone h

elp me please I swear I need the help from somebody please

1 answer:

muminat3 years ago

3 0

Here is your answer

A. (4,-7)

REASON:

In quadrant IV abscissa is +ve and ordinate is -ve. Since absolute value is 11

Here |4|+|-7|= 4+7= 11

HOPE IT IS USEFUL

You might be interested in

Solve each Inequality. Write your answer 2.) x + 9 > 16 o

MrRissso [65]

Answer:

o< 1/16x + 9/16

Step-by-step explanation:

x+9>16o

Flip the equation.

16o<x+9

Divide both sides by 16.

16o/16 < x+9/16

o< 1/16x + 9/16

5 0

3 years ago

The volume V of an ice cream cone is given by V = 2 3 πR3 + 1 3 πR2h where R is the common radius of the spherical cap and the c

Nuetrik [128]

Answer:

The change in volume is estimated to be 17.20 $\rm{in^3}$

Step-by-step explanation:

The linearization or linear approximation of a function $f(x)$ is given by:

$f(x_0+dx) \approx f(x_0) + df(x)|_{x_0}$ where $df$ is the total differential of the function evaluated in the given point.

For the given function, the linearization is:

$V(R_0+dR, h_0+dh) = V(R_0, h_0) + \frac{\partial V(R_0, h_0)}{\partial R}dR + \frac{\partial V(R_0, h_0)}{\partial h}dh$

Taking $R_0=1.5$ inches and $h=3$ inches and evaluating the partial derivatives we obtain:

$V(R_0+dR, h_0+dh) = V(R_0, h_0) + \frac{\partial V(R_0, h_0)}{\partial R}dR + \frac{\partial V(R_0, h_0)}{\partial h}dh\\V(R, h) = V(R_0, h_0) + (\frac{2 h \pi r}{3} + 2 \pi r^2)dR + (\frac{\pi r^2}{3} )dh$

substituting the values and taking $dx=0.1$ and $dh=0.3$ inches we have:

$V(R_0+dR, h_0+dh) =V(R_0, h_0) + (\frac{2 h \pi r}{3} + 2 \pi r^2)dR + (\frac{\pi r^2}{3} )dh\\V(1.5+0.1, 3+0.3) =V(1.5, 3) + (\frac{2 \cdot 3 \pi \cdot 1.5}{3} + 2 \pi 1.5^2)\cdot 0.1 + (\frac{\pi 1.5^2}{3} )\cdot 0.3\\V(1.5+0.1, 3+0.3) = 17.2002\\\boxed{V(1.5+0.1, 3+0.3) \approx 17.20}$

Therefore the change in volume is estimated to be 17.20 $\rm{in^3}$

4 0

3 years ago

I can define continuous data as:

stira [4]

Answer:

Continuous data are data which can take any values. Examples include time, height and weight. Because continuous data can take any value, there are an infinite number of possible outcomes.

5 0

3 years ago

If tan theta =3, find the value of tan theta + tan (theta+pi) + tan (theta +2pi)

sweet-ann [11.9K]

Answer:

$tan\theta + tan(\theta + \pi) + tan(\theta+2\pi) = 9$

Step-by-step explanation:

Given

$tan\theta = 3$

Required

Find

$tan\theta + tan(\theta + \pi) + tan(\theta+2\pi)$

Calculate $tan(\theta + \pi) \ and\ tan(\theta+2\pi)$

Using tan rule

$tan(\theta + \pi) = \frac{tan\theta + tan\pi}{1 - tan\theta tan\pi}$

So:

$tan(\theta + \pi) = \frac{tan\theta + 0}{1 - tan\theta *0}$

$tan(\theta + \pi) = \frac{tan\theta}{1 }$

$tan(\theta + \pi) = \tan\theta$

$tan(\theta+2\pi)$

$tan(\theta + 2\pi) = \frac{tan\theta + tan2\pi}{1 - tan\theta tan2\pi}$

$tan(\theta + 2\pi) = \frac{tan\theta + 0}{1 - tan\theta *0}$

$tan(\theta + 2\pi) = \tan\theta$ '

'So:

$tan\theta + tan(\theta + \pi) + tan(\theta+2\pi)$ $= tan\theta +tan\theta +tan\theta$

$tan\theta + tan(\theta + \pi) + tan(\theta+2\pi) = 3+3+3$

$tan\theta + tan(\theta + \pi) + tan(\theta+2\pi) = 9$

4 0

2 years ago

Which expression is equivalent to:1/m6

Valentin [98]

Answer:

A

Step-by-step explanation:

6 is a power. So both C and D are wrong because 6 is being treated as an ordinary integer.

So the answer must be either A or B.

B is not correct because to move an expression from the denominator to the numerator changes the sign in the base (which is m in this question).

A is correct. the power is changed from 6 in the denominator to -6 in the numerator. m is shifted to the numerator as well.

8 0

2 years ago