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3 years ago

Tell me the coordinates

Mathematics

1 answer:

Reptile [31]3 years ago

3 0

Answer:

1) N/A

2) ( -0.714, 3.857)

Step-by-step explanation:

for question 1 since the slope is the same, they will never intersect

for question 2 they intersect at (-0.714, 3.857)

1) as for the coordinates, for y=3x+2 it is (0,2) (-2/3, 0) it's an upward slope

y=3x-5 coordinates are (0, -5) (1.667, 0) it's an upward slope

2) 4x+y=1 coordinates are (0,1) (1/2, 0) it's a downwards slope

y=3x+6 coordinates are (-2, 0) (0,6) it's a upward slope

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For a test please help thanks

vovangra [49]

Answer:

Step-by-step explanation:

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3 years ago

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LeBron goes to the store with six dollars and buys four packs of gum. James goes to the store with five dollars and buys two pac

lilavasa [31]

G = the cost of one pack of gum
LeBron : 6 - 4g
James : 5 - 2g

what they both have after buying gum..
6 - 4g + 5 - 2g = 11 - 6g

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4 years ago

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Caren held 260 of the shares for herself and sold the rest in equal amounts to six investors. How many shares does each investor

notsponge [240]

Answer:

40 shares each and $3,000 each

Step-by-step explanation:

The computation is shown below:

For the number of shares

Since there are total of 500 shares out of which 260 shares are held by Caren so the remaining shares left is

= 500 shares - 260 shares

= 240 shares

And, there are 6 investors

So the number of shares owned by each investor is

$= \frac{240}{6}$

= 40 shares

Now the value of the shares held by each inventory is

$= 40 shares \times \$75$

= $3,000

Total debt is $65,000

So, the liability per share is

$= \frac{\$65,000}{500}$

= $130

And each investor holds 40 shares

So, each investor is liable for

$= \$130 \times 40\ shares$

= $5,200

As we can see that the value of the shares is less than the amount liable by each investor

So each investor is liable for $3,000

And, the liability of Caren is

$= 260\ shares \times \$75$

= $19,500

5 0

3 years ago

Can I get help solving this graph? I figured out h'(2) does not exist, but am having difficulty finding h'(1) and h'(3).

Mrac [35]

F(x) is a piecewise function defined as:
f(x) = (3/2)x when 0 <= x <= 2
or
f(x) = -(3/2)x+6 when 2 < x <= 4

g(x) is defined as:
g(x) = -(1/4)x+1 when 0 <= x <= 4

Based on the graph or through the equations we can say:
f(1) = (3/2)*1 = 1.5
g(1) = -(1/4)*1+1 = 0.75
f(3) = (3/2)*3 = 4.5
g(3) = -(1/4)*3+1 = 0.25

And the derivative values are:
f ' (1) = 3/2 = 1.5
f ' (3) = -3/2 = -1.5
g ' (1) = -1/4 = -0.25
g ' (3) = -1/4 = -0.25
which are the slopes of each line

So...
h(x) = f(x)*g(x)
h ' (x) = f ' (x)*g(x) + f(x)*g ' (x) ... product rule
h ' (1) = f ' (1)*g(1) + f(1)*g ' (1)
h ' (1) = 1.5*0.75 + 1.5*(-0.25)
h ' (1) = 0.75

and,
h(x) = f(x)*g(x)
h ' (x) = f ' (x)*g(x) + f(x)*g ' (x)
h ' (3) = f ' (3)*g(3) + f(3)*g ' (3)
h ' (3) = -1.5*0.25 + 4.5*(-0.25)
h ' (3) = -1.5

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4 years ago

Whats the answer? Note:I do need this kinda fast because its a timed test.

ella [17]

The answer is 48 np at all

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3 years ago

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