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2 years ago

11

What is the range of the relation below?

1 answer:

Greeley [361]2 years ago

5 0

[6, -infinity)

have a good night

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CAN SOMEONE MULTIPLY THIS PLEASE, ITS URGENT

____ [38]

Let's factorise it :

$\: {\qquad \dashrightarrow \sf ( {x}^{3} - 5)(x + 3) }$

$\: {\qquad \dashrightarrow \sf {x}^{3} (x + 3) + [-5(x + 3)] }$

Using Distributive property we get :

$\: {\qquad \dashrightarrow \sf {x}^{3} + {3x}^{3} + ( - 5x - 15) }$

$\: {\qquad \dashrightarrow \sf {x}^{3} + {3x}^{3} - 5x - 15 }$

$\: {\qquad \dashrightarrow \sf 4{x}^{3} - 5x - 15 }$

⠀

Therefore,

$\: {\qquad \dashrightarrow \sf ( {x}^{3} - 5)(x + 3) =4{x}^{3} - 5x - 15}$

8 0

2 years ago

Please help with this one

anyanavicka [17]

Ok this inequality tells you the number of devices you can have before the new plan costs more than the old plan. The new plan expression is $4.50x + $94m = y ( total cost). The old plan is $175m = y (total cost). You can see m (number of months) in both equations, you don't need it this time since we're going to to compare both to one month. Since they're both equal to y you can make them equal to each other. $4.50x + $94 = $175. Now you want to figure when the new plan is less than the old plan you switch the equal sign for a less than sign. $4.50x + $94 < $175; this will help you find the inequality you want. From there just use algebraic steps to find that x has to less than 18 or

x < 18.

3 0

3 years ago

How much difference do a couple of weeks make for birth weight? Late-preterm babies are born with 34 to 36 completed weeks of ge

stiks02 [169]

Answer:

a) 0.3277

b) 0.0128

Step-by-step explanation:

We are given the following information in the question:

N(2750, 560).

Mean, μ = 2750

Standard Deviation, σ = 560

We are given that the distribution of distribution of birth weights is a bell shaped distribution that is a normal distribution.

Formula:

$z_{score} = \displaystyle\frac{x-\mu}{\sigma}$

a) P (less than 2500 grams)

P(x < 2500)

$P( x < 2500) = P( z < \displaystyle\frac{2500 - 2750}{560}) = P(z < -0.4464)$

Calculation the value from standard normal z table, we have,

$P(x < 2500) = P(z < -0.4464) = 0.3277 = 32.77\%$

b) P ((less than 1500 grams)

P(x < 1500)

$P( x < 1500) = P( z < \displaystyle\frac{1500 - 2750}{560}) = P(z < -2.2321)$

Calculation the value from standard normal z table, we have,

$P(x < 1500) = P(z < -2.2321) = 0.0128 = 1.28\%$

3 0

2 years ago

Solve the equation.<br> -6p=48

malfutka [58]

Answer:

p=-8

Step-by-step explanation:

A negative multiplied by a negative is a positive.

3 0

3 years ago

Which is the equation of a hyperbola centered at the origin with x-intercept +\- 3 and asymptote y=2x

Radda [10]

Answer:

${\dfrac{x^{2}}{9} - \dfrac{y^{2}}{36} = 1}$

Step-by-step explanation:

The hyperbola has x-intercepts, so it has a horizontal transverse axis.

The standard form of the equation of a hyperbola with a horizontal transverse axis is $\dfrac{(x - h)^{2}}{a^{2}} - \dfrac{(y - k)^{2}}{b^{2}} = 1$

The center is at (h,k).

The distance between the vertices is 2a.

The equations of the asymptotes are $y = k \pm \dfrac{b}{a}(x - h)$

1. Calculate h and k. The hyperbola is symmetric about the origin, so

h = 0 and k = 0

2. For 'a': 2a = x₂ - x₁ = 3 - (-3) = 3 + 3 = 6

a = 6/2 = 3

3. For 'b': The equation for the asymptote with the positive slope is

$y = k + \dfrac{b}{a}(x - h) = \dfrac{b}{a}x$

Thus, asymptote has the slope of

$\begin{array}{rcl}m& =& \dfrac{b}{a}\\\\2& =& \dfrac{b}{3}\\\\b& =& \mathbf{6}\end{array}$

4. The equation of the hyperbola is

$\large \boxed{\mathbf{\dfrac{x^{2}}{9} - \dfrac{y^{2}}{36} = 1}}$

The attachment below represents your hyperbola with x-intercepts at ±3 and asymptotes with slope ±2.

7 0

2 years ago