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2 years ago

Which one of these lines is the equation

Mathematics

2 answers:

yaroslaw [1]2 years ago

7 0

Answer:

Line B - red line

Step-by-step explanation:

Marianna [84]2 years ago

3 0

The red
y intercept formula which is y= mx + b
The b represents the y intercept of the graph.
In this case the y intercept/ b would be 2
the red line was the only one to cross the y intercept of 2

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James found a pair of shoes he wanted at two stores. At store A they are $35 and he had a 20% off coupon. At store B they were $

Oxana [17]

Answer: They cost the same amount.

Step-by-step explanation:

35(0.20)=7

35-7=28

40(0.30)=12

40-12=28

28=28

7 0

3 years ago

Help meeeeeeeeeeeeee

Anna71 [15]

Answer:

Area 60 po is 32

Step-by-step explanation:

First u do 6x10 which equal 60area is a he whole and perimeter is what each side and add all sidesis and the perime Is 32So the po is 32 and the . Is 60 hope it. Helps

7 0

2 years ago

Read 2 more answers

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Archy [21]

(a) If the particle's position (measured with some unit) at time t is given by s(t), where

$s(t) = \dfrac{5t}{t^2+11}\,\mathrm{units}$

then the velocity at time t, v(t), is given by the derivative of s(t),

$v(t) = \dfrac{\mathrm ds}{\mathrm dt} = \dfrac{5(t^2+11)-5t(2t)}{(t^2+11)^2} = \boxed{\dfrac{-5t^2+55}{(t^2+11)^2}\,\dfrac{\rm units}{\rm s}}$

(b) The velocity after 3 seconds is

$v(3) = \dfrac{-5\cdot3^2+55}{(3^2+11)^2} = \dfrac{1}{40}\dfrac{\rm units}{\rm s} = \boxed{0.025\dfrac{\rm units}{\rm s}}$

$\dfrac{-5t^2+55}{(t^2+11)^2} = 0 \implies -5t^2+55 = 0 \implies t^2 = 11 \implies t=\pm\sqrt{11}\,\mathrm s \imples t \approx \boxed{3.317\,\mathrm s}$

(d) The particle is moving in the positive direction when its position is increasing, or equivalently when its velocity is positive:

$\dfrac{-5t^2+55}{(t^2+11)^2} > 0 \implies -5t^2+55>0 \implies -5t^2>-55 \implies t^2 < 11 \implies |t|$

In interval notation, this happens for t in the interval (0, √11) or approximately (0, 3.317) s.

(e) The total distance traveled is given by the definite integral,

$\displaystyle \int_0^8 |v(t)|\,\mathrm dt$

By definition of absolute value, we have

$|v(t)| = \begin{cases}v(t) & \text{if }v(t)\ge0 \\ -v(t) & \text{if }v(t)$

In part (d), we've shown that v(t) > 0 when -√11 < t < √11, so we split up the integral at t = √11 as

$\displaystyle \int_0^8 |v(t)|\,\mathrm dt = \int_0^{\sqrt{11}}v(t)\,\mathrm dt - \int_{\sqrt{11}}^8 v(t)\,\mathrm dt$

and by the fundamental theorem of calculus, since we know v(t) is the derivative of s(t), this reduces to

$s(\sqrt{11})-s(0) - s(8) + s(\sqrt{11)) = 2s(\sqrt{11})-s(0)-s(8) = \dfrac5{\sqrt{11}}-0 - \dfrac8{15} \approx 0.974\,\mathrm{units}$

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2 years ago

What will be the cost price of a television when

astra-53 [7]

3+3+4+3+-3+4-66+44+3=34x

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2 years ago

PLEASE HELP ME WITH THIS! ASAP!!!!!!!

marusya05 [52]

Hold on i believe its C but let me go look on google and i'll get back to you to sure

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2 years ago