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2 years ago

Carter ran 7/8 mile in a race. Taylor ran 1/4 mile. How much further for carter run than Taylor?

Mathematics

2 answers:

Artist 52 [7]2 years ago

3 0

Answer:

1 17/20 is the answer

r-ruslan [8.4K]2 years ago

3 0

Answer:

5/8

Step-by-step explanation:

1/4=2/8

7/8-2/8=5/8

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How do you write this in standard form? (6-7i)/(1-2i)

xenn [34]

$\frac { \left( 6-7i \right) }{ \left( 1-2i \right) } \cdot 1\\ \\ =\frac { \left( 6-7i \right) }{ \left( 1-2i \right) } \cdot \frac { \left( 1+2i \right) }{ \left( 1+2i \right) } \\ \\ =\frac { 6+12i-7i-14{ i }^{ 2 } }{ 1+2i-2i-4{ i }^{ 2 } }$

$\\ \\ =\frac { 6+5i-14\left( -1 \right) }{ 1-4\left( -1 \right) } \\ \\ =\frac { 6+5i+14 }{ 1+4 } \\ \\ =\frac { 20+5i }{ 5 } \\ \\ =\frac { 20 }{ 5 } +\frac { 5i }{ 5 } \\ \\ =4+i$

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3 years ago

4267 divided by 31 in mixed number form

Rainbow [258]

Mixed Number Form:
137 $\frac{20}{31}$

8 0

3 years ago

5x+4y= 10 –5x+y = 15

Ivahew [28]

Answer:

what exactly are we trying to solve for

6 0

3 years ago

The product of -2 and a number x is greater than 30 Write the word sentence as a inequality, then solve inequality, thank you!

anygoal [31]

-2x > 30

x > -15 (??)

7 0

2 years ago

NO LINKS OR FILES!

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

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

7 0

2 years ago