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andre [41]
3 years ago
15

Last question of the day lol:Find the Slope and Y-intercept of each lines​

Mathematics
2 answers:
jeka57 [31]3 years ago
6 0

Answer:

“Be strong and courageous. Do not fear or be in dread of them, for it is the LORD your God who goes with you. He will not leave you or forsake you.” [Deuteronomy 31:6]

Step-by-step explanation:

Nina [5.8K]3 years ago
5 0

Answer:

The first one would be -4 and the second one would be 0

Step-by-step explanation:

look at where they cross the y line

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Find the differential of the function w=xsin(2yz^4).
SCORPION-xisa [38]

The differential of

w = x \sin(2yz^4)

is

\mathrm dw = \dfrac{\partial w}{\partial x}\,\mathrm dx + \dfrac{\partial w}{\partial y}\,\mathrm dy + \dfrac{\partial w}{\partial z}\,\mathrm dz

where the partial derivatives are

\dfrac{\partial w}{\partial x} = \sin(2yz^4) \\\\ \dfrac{\partial w}{\partial y} = x \cos(2yz^4) \times 2z^4 = 2xz^4 \cos(2yz^4) \\\\ \dfrac{\partial w}{\partial z} = x \cos(2yz^4) \times 8yz^3 = 8xyz^3 \cos(2yz^4)

So the differential d<em>w</em> is

\mathrm dw = \boxed{\sin(2yz^4)}\,\mathrm dx + \boxed{2xz^4 \cos(2yz^4)}\,\mathrm dy + \boxed{8xyz^3 \cos(2yz^4)}\,\mathrm dz

5 0
3 years ago
You want to rent a television Company A charges $50 per week plus a one-time fee of $15. Company B charges 130 per week plus a o
emmasim [6.3K]
Company A: y = 50x + 15
Company B: y = 130x + 55

Set the equations equal to each other and solve for x (number of weeks).
50x + 15 = 130x + 55

Subtract 50x from both sides:
15 = 80x + 55

Subtract 55 from both sides:
-40 = 80x

Did you have a typo? The companies will never have the same cost because company B costs so much more and weeks cannot be negative.
8 0
3 years ago
The line plot shows the ages of participants in two group activities at a science center.
Anni [7]
I believe the answer is a
5 0
4 years ago
Read 2 more answers
Roots of the quadratic equation 0=2x^2+12x-14
cluponka [151]

Answer:

x=-7,1

Step-by-step explanation:

7 0
3 years ago
Read 2 more answers
A force of 12 lb is required to hold a spring stretched 8 in. beyond its natural length. How much work W is done in stretching i
sattari [20]

Answer:

Work done in stretching the spring = 7.56 lb-ft.

Step-by-step explanation:

Normal length of the spring = 8 in or \frac{8}{12} ft

= \frac{2}{3} ft

If the spring has been stretched to 11 inch then the stretched length of the spring is = 11 in

= \frac{11}{12} ft

Force applied to stretch the spring = 12 lb

By Hook's law,

F = kx [where k is the spring constant and x = length by which the spring is stretched]

12 = k(\frac{2}{3})

k = 18

Work done (W) to stretch the spring by \frac{11}{12} ft will be

W = \int\limits^\frac{11}{12} _0 {kx} \, dx

    = \int\limits^\frac{11}{12} _0 {(18x)} \, dx

    = 18[\frac{x^{2}}{2}]^{\frac{11}{12}}_0

    = 9(\frac{11}{12})²

    = 7.56 lb-ft

6 0
3 years ago
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