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bogdanovich [222]

3 years ago

13

Which sequence is geometric and has 1/4 as its fifth term and 1/2 as the common ratio?

2 answers:

marta [7]3 years ago

4 0

B , because the all go by half until it goes up more

vladimir1956 [14]3 years ago

4 0

Answer:

B. 1/4, 1/2, 1, 2

Step-by-step explanation:

It goes half

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Mike shoots a large marble (Marble A, mass: 0.05 kg) at a smaller marble (Marble B, mass: 0.03 kg) that is sitting still. Marble

Nadusha1986 [10]

<u>Answer</u>

1.333 m/s

<u>Explanation</u>

The momentum before collision and after collision is always conserved.

m₁v₁ = m₂v₂

(0.05×0.6) + (0.03×0) = (0.03 × v) + (0.05 × -0.2)

0.03 + 0 = 0.03v - 0.01

0.03v = 0.03+0.01

0.03v = 0.04

v = 0.04/0.03

= 4/3

=<em> 1.333 m/s</em>

6 0

4 years ago

(√x+1/√x)^2find dy/ dx and substitute 2 in dy/ dx

lions [1.4K]

Hello,
That's better.

I suppose $y=( \sqrt{x} + \dfrac{1}{ \sqrt{x} } )^2 \\

y'=2*( \sqrt{x} + \dfrac{1}{ \sqrt{x} } )*( \dfrac{1}{2 \sqrt{x} } - \dfrac{3}{2x \sqrt{x} } )$

If x=2 then
$y'=2*( \sqrt{2} + \dfrac{1}{ \sqrt{2} } )*( \dfrac{1}{2 \sqrt{2} } - \dfrac{3}{2*2* \sqrt{2} } )\\
=2* \dfrac{2+ \sqrt{2} }{ \sqrt{2} } * \dfrac{1}{4*\sqrt{2} } *(-1)\\
=- \dfrac{2+ \sqrt{2} }{4}$

3 0

3 years ago

Pam has 90 m of fencing to enclose an area in a petting zoo with two dividers to separate three types of young animals. The thre

andrew-mc [135]

Answer:

The area function is

$A=\frac{135}{2}x-\frac{9}{2}x^2$ .

The domain and range of A is $(0,15m)$ and $(0, 253.125 m^2]$ .

Step-by-step explanation:

The given length of fencing is $90 m$ .

Let the length and width of each pen be $x$ and $y$ respectively as shown in the figure.

As there are 3 pens, so, the total area,

$A= 3 xy \;\cdots (i)$

From the figure the total length of fencing is $6x+4y$ .

Here, for a significant area for the animals, $x>0$ as well as $y>0$ as $x$ and $y$ are the sides of ben.

From the given value:

$6x+4y=90\;\cdots (ii)$

$\Rightarrow y=\frac {45}{2}-\frac{3x}{2}$

Now, from equation (i)

$A=3x\left(\frac {45}{2}-\frac{3x}{2}\right)$

$\Rightarrow A=\frac{135}{2}x-\frac{9}{2}x^2\;\cdots (iii)$

This is the required area function in the terms of variable $x$ .

For the domain of area function, from equation (ii)

$x=15-\frac{2y}{3}$

$\Rightarrow x$ [as y>0]

So, the domain of area function is $(0,15m)$ .

For the range of area function:

As $x \rightarrow 0$ or $y\rightarrow 0$ , then $A\rightarrow 0$ [from equation (i)]

$\Rightarrow A>0$

Now, differentiate the area function with respect to $x$ .

$\frac {dA}{dx}=\frac{135}{2}-9x$

Equate $\frac {dA}{dx}$ to zero to get the extremum point.

$\frac {dA}{dx}=0$

$\Rightarrow \frac{135}{2}-9x=0$

$\Rightarrow x=\frac{15}{2}$

Check this point by double differentiation

$\frac {d^2A}{dx^2}=-9$

As, $\frac {d^2A}{dx^2}$ , so, point $x=\frac{15}{2}$ is corresponding to maxima.

Put this value back to equation (iii) to get the maximum value of area function. We have

$A=\frac{135}{2}\times \frac {15}{2}-\frac{9}{2}\times \left(\frac {15}{2}\right)^2$

$\Rightarrow A=253.125 m^2$

Hence, the range of area function is $(0, 253.125 m^2]$ .

4 0

4 years ago

Match the equations of the lines with their graphs.

xxMikexx [17]

Answer:

Step-by-step explanation:

1. y=2x-3

2. y=2x+3

3. y=-2x+3

4. y=-2x-3

3 0

3 years ago

The lines shown below are perpendicular. If the green line has a slope of

klasskru [66]

Answer: 4

Step-by-step explanation:

Perpendicular lines have reciprocal slopes

8 0

3 years ago