If the area of a semicircle is 32 sq. km, then its radius is:

HELP ASAP PLSSSSSSSSSS

inysia [295]

Answer:

Neighborhood Q appears to have a bigger family size

Step-by-step explanation:

Mean = the sum of all data values divided by the total number of data values

Number of families in Neighborhood Q = 9

Mean family size of Neighborhood Q:

= (2 + 5 + 4 + 3 + 2 + 5 + 3 + 6 + 5) ÷ 9

= 35 ÷ 9

= 3.888888...

Number of families in Neighborhood S = 9

Mean family size of Neighborhood S:

= (2 + 3 + 2 + 3 + 7 + 2 + 3 + 3 + 2) ÷ 9

= 27 ÷ 9

= 3

The mean family size of Neighborhood Q is 3.88.. and the mean family size of Neighborhood S is 3. Therefore, Neighborhood Q appears to have a bigger family size as it's average family size is bigger than that of Neighborhood S.

6 0

2 years ago

Can someone help me with these both

Anna [14]

Hello,
so all you have to do is match the abbreviations to the triangles. The abbreviations stand for what is the SAME in both triangles, denoted by similar markings on equal sides and angles.

Abbreviations:
SSS = Side-Side-Side
SAS = Side-Angle-Side
ASA = Angle-Side-Angle
AAS = Angle-Angle-Side
HL = Hypotenuse-Leg

* Note - the angle side angle must go around the triangle in that order. ASA has the side BETWEEN the congruent angles.. SSA does NOT work.

(9.) ASA
(10.) AAS
(11.) SSS
(12.) No way to tell if congruent. (only 3 angles no side)
(13.) ASA
(14.) SAS
(15.) HL

7 0

3 years ago

In the diagram below,

sergey [27]

Can you attach a picture because this is very hard to read

5 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

I am very confused and so are my classmates.

Nadya [2.5K]

Answer:

I used A and got 0.4 and she needs 52200

Step-by-step explanation:

8 0

2 years ago