Write the sentence as an equation. 328 is the quotient of 36 and b FAST

The temperature fell 5°C each hour for 3h. a. Write an integer multiplication to represent the temperature change, b. What was t

MakcuM [25]

Answer:

Step-by-step explanation:

a) 3 x 5

b) 15 Celcius

8 0

3 years ago

Triangle EFG is dilated by a scale factor of 3 centered at (0, 1) to create triangle E'F'G'. Which statement is true about the d

lesya [120]

Answer:

segment EH and segment E prime H prime both pass through the center of dilation.

Step-by-step explanation:

Center of dilation is point (0.1), same as H. Both, E(0,5) and H(0,1) are placed over y-axis, then E' and H' are also located at y-axis.

After dilation, H' is placed at (0,1) because it coincides with the center of dilation

Distance between E and center of dilation is 4 units, then E' should be at 4*3=12 units from the center of dilation and over y-axis. Therefore, E' is placed at (0, 13)

So, segment E'H' goes from (0,1) to (0,13), and pass through the center of dilation, like segment EH.

4 0

3 years ago

In the equation 3x+ 2 = 11 which number is the constant?

grigory [225]

Answer:

The constant is 2 and 11

Step-by-step explanation:

A constant is a number that always has the same value. In this equation 2 and 11 are the constants because they stay the same. 3x doesn't stay the same because 3 is multiplying x.

Hope This Helps :)

6 0

3 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

3 years ago

-23z7+(-3a) simplify

Brums [2.3K]

The answer would be -161z-3a

3 0

3 years ago