If i had 4 pounds of sugar for $2 how much for 1 lb

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Maksim231197 [3]

Answer:

D. x=1

Step-by-step explanation:

Image can be rewritten as 3x + 6 = 9.

Solve for x.

3x +6 = 9

Isolate x by subtracting 6 from both sides.

3x = 3

Get rid of the 3 by dividing by 3 on both sides.

x = 1

Hope this helped.

8 0

3 years ago

Y = f(x) = 2x Answer as a decimal

skad [1K]

Answer:

insufficient data

Step-by-step explanation:

please give complete question

4 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

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

(c) The particle is at rest when its velocity is zero:

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

(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

An observer (O) spots a plane flying at a 42° angle to his horizontal line of sight. If the plane is flying at an altitude of 15

NeTakaya

Answer:

D. 22,417 feet

Step-by-step explanation:

Fine the diagram in the attachment for proper elucidation. Using the SOH, CAH, TOA trigonometry identity to solve for the distance (x) from the plane (P) to the observer (O), the longest side x is the hypotenuse and the side facing the angle of elevation is the opposite.

Hypotenuse = x and Opposite = 15,000feet

According to SOH;

$sin 42^0 = \frac{Opposite}{Hypotenuse} \\\\Sin42^0 = \frac{15000}{x}\\ \\x = \frac{15000}{sin42^0}\\\\ \\$