Ask question

Ask question

All categories

2 years ago

6

the math club has 18 members and 50% are girls. the science club has 25 members and 40% are girls. the principal wants to know w

hich club has more girls. how many girls are in the math club? how many girls are in the science club?

1 answer:

Nuetrik [128]2 years ago

5 0

There’s 9 girls in the math club, and 10 in the science club

You might be interested in

-3x - 3y - z= -6

kifflom [539]

Answer:

Ammm, I think the answer you have is wrong, but what I really think is that the equations you put in are not the right ones because the answer for these is "ugly" (decimals etc...). Want to check your equations?

Step-by-step explanation:

Here is a step by step explanation to your equations (I assumed +3z for the second but you can change it). You can edit it for your equations if you want to.

https://simplisico.com/share/q/QiWKOyN5uYgkSG4iBXOmdG0i

5 0

2 years ago

Select the correct answer.

svp [43]

Step-by-step explanation:

option B is correct as,

10 + x = -18

→ x = -18 -10

→ x = -28

<em><u>hope </u></em><em><u>this </u></em><em><u>answer </u></em><em><u>helps </u></em><em><u>you </u></em><em><u>dear.</u></em><em><u>.</u></em><em><u>.</u></em><em><u>take </u></em><em><u>care</u></em><em><u>!</u></em>

3 0

2 years ago

Read 2 more answers

Buses arrive at a stop every 14 min. Each bus waits 3 min before departing. What is the probability that a person arriving at th

Mnenie [13.5K]

28 I think if you add

3 0

2 years ago

find the coordinates of the point P on the parabola y=1-x^2 with domain 0≤x≤1 that minimize the area of the triangle enclosed by

coldgirl [10]

Let point P be with coordinates $(x_0,y_0).$ Find the equation of the tangent line.

1. If $y=1-x^2,$ then $y'=-2x.$

2. The equation of the tangent line at point P is

$y-y_0=-2x_0(x-x_0).$

Find x-intercept and y-intercept of this line:

when x=0, then $y=y_0+2x_0^2;$
when y=0, then $x=\dfrac{y_0}{2x_0}+x_0=\dfrac{y_0+2x_0^2}{2x_0}.$

The area of the triangle enclosed by the tangent line at P, the x-axis, and y-axis is

$A=\dfrac{1}{2}\cdot (2x_0^2+y_0)\cdot \left(\dfrac{y_0+2x_0^2}{2x_0}\right)=\dfrac{(y_0+2x_0^2)^2}{4x_0}.$

Since point P is on the parabola, then $y_0=1-x_0^2$ and

$A=\dfrac{(1-x_0^2+2x_0^2)^2}{4x_0}=\dfrac{(1+x_0^2)^2}{4x_0}.$

Find the derivative A':

$A'=\dfrac{2(1+x_0^2)\cdot 2x_0\cdot 4x_0-4(1+x_0^2)^2}{16x_0^2}=\dfrac{12x_0^4+8x_0^2-4}{16x_0^2}.$

Equate this derivative to 0, then

$12x_0^4+8x_0^2-4=0,\\ \\3x_0^4+2x_0^2-1=0,\\ \\D=2^2-4\cdot 3\cdot (-1)=16,\ \sqrt{D}=4,\\ \\x_0^2_{1,2}=\dfrac{-2\pm4}{6}=-1,\dfrac{1}{3},\\ \\x_0^2=\dfrac{1}{3}\Rightarrow x_0_{1,2}=\pm\dfrac{1}{\sqrt{3}}.$

And

$y_0=1-\left(\pm\dfrac{1}{\sqrt{3}}\right)^2=\dfrac{2}{3}.$

Answer: two points: $P_1\left(-\dfrac{1}{\sqrt{3}},\dfrac{2}{3}\right), P_2\left(\dfrac{1}{\sqrt{3}},\dfrac{2}{3}\right).$

6 0

2 years ago

Q6. What is the degree of vertex E in this network diagram? *

Reil [10]

Answer:

Step-by-step explanation:

7 0

3 years ago