Ask question

Ask question

All categories

4 years ago

14

In order for students to have a club at school they need a minimum of 23 students if there are 17 students signed up for the clu

b how many more students does the club need?

1 answer:

Ivan4 years ago

3 0

The answer is 6
Explanation: STEP BY STEP

You might be interested in

A 12-ft ladder leaning against a house makes a 64 degree angle with the ground. Will the ladder reach a window sill that is 10.5

Luba_88 [7]

Answer:

Yes, it will

Step-by-step explanation:

Given

Length of Ladder = 12 ft

Angle = 64

Required

Determine if it'll reach a height of 10.5ft from the base

See attachment for proper illustration.

<em>From the attachment, we want to determine that x be greater than 10.5</em>

Using sin formula:

$Sin\ 64 = \frac{x}{12}$

Multiply through by 12

$x = 12 * sin\ 64$

$x = 12 * 0.8988$

$x = 10.79$ (Approximated)

<em>Since 10.79 is greater than 10.5, then the ladder will reach the window</em>

sin64 = x/12

x = 12* sin64

5 0

3 years ago

32. Which of the following is horizontal ellipse

maria [59]

Answer:

b. 16x² + 25y² = 1

Step-by-step explanation:

Only one of them has a < b in ax² + by² = c formula, making its graph horizontal

All the rest are vertical, see attached

6 0

3 years ago

Ms. Vega read 20% of the Star Wars book. She read 8 pages of the book.

never [62]

Answer: 40 pages in total

Step-by-step explanation:

Ms. Vega has read 20% of the book and this translates to 8 pages of the book.

20% = 8 pages

Assume the total number of pages is x:

20% * x = 8

0.2x = 8

x = 8 / 0.2

x = 40 pages in total

4 0

3 years ago

A loaf of bread originally priced at $3.49 has been discounted 10% what is the price now?

amm1812

Hello BusyBunny, this is quite simple. All you would need to do is multiply .10 or 10% * $3.49 which will give you a 10% discount.

5 0

4 years ago

Read 2 more answers

Find the integrate of 2x/sqrt(1-x^2) dx from 0 to 1/2

Ludmilka [50]

Compute the definite integral:

$\large\begin{array}{l}\mathsf{\displaystyle\int_{0}^{\frac{1}{2}}\frac{2x}{\sqrt{1-x^2}}\,dx}\\\\ =\mathsf{\displaystyle\int_{0}^{\frac{1}{2}}\frac{-1}{\sqrt{1-x^2}}\cdot (-2x)\,dx\qquad\quad(i)} \end{array}$

$\large\begin{array}{l} \textsf{Substitute}\\\\ \mathsf{1-x^2=u~~\Rightarrow~~-2x\,dx=du}\\\\\\ \textsf{Finding the new limits of integration:}\\\\ \begin{array}{lcl} \textsf{When }\mathsf{x=0}&~\Rightarrow~&\mathsf{u=1-0^2}\\\\ &&\mathsf{u=1}\\\\\\ \textsf{When }\mathsf{x=\dfrac{1}{2}}&~\Rightarrow~&\mathsf{u=1-\left(\dfrac{1}{2}\right)^2}\\\\ &&\mathsf{u=1-\dfrac{1}{4}}\\\\ &&\mathsf{u=\dfrac{3}{4}} \end{array} \end{array}$

$\large\begin{array}{l} \textsf{Then (i) becomes}\\\\ =\mathsf{\displaystyle\int_{1}^{\frac{3}{4}}\frac{-1}{\sqrt{u}}\,du}\\\\ =\mathsf{\displaystyle\int_{1}^{\frac{3}{4}}\frac{-1}{u^{\frac{1}{2}}}\,du}\\\\ =\mathsf{\displaystyle\int_{1}^{\frac{3}{4}} (-1)\cdot u^{-\frac{1}{2}}\,du}\\\\ =\mathsf{(-1)\cdot \dfrac{u^{-\frac{1}{2}+1}}{-\frac{1}{2}+1}\bigg|_{1}^{\frac{3}{4}}} \end{array}$

$\large\begin{array}{l} =\mathsf{(-1)\cdot \dfrac{~~u^{\frac{1}{2}}~~}{\frac{1}{2}}\bigg|_{1}^{\frac{3}{4}}}\\\\ =\mathsf{(-1)\cdot 2u^{\frac{1}{2}}\Big|_{1}^{\frac{3}{4}}}\\\\ =\mathsf{-2\cdot \sqrt{u}\Big|_{1}^{\frac{3}{4}}}\\\\ =\mathsf{-2\cdot \left(\sqrt{\dfrac{3}{4}}-\sqrt{1}\right)} \end{array}$

$\large\begin{array}{l} =\mathsf{-2\cdot \left(\dfrac{\sqrt{3}}{2}-1\right)}\\\\ =\mathsf{-\diagup\!\!\!\! 2\cdot \dfrac{\sqrt{3}}{\diagup\!\!\!\! 2}+(-2)\cdot (-1)}\\\\ =\mathsf{-\sqrt{3}+2}\\\\ =\mathsf{2-\sqrt{3}} \end{array}$

$\large\begin{array}{l} \boxed{\begin{array}{c}\mathsf{\displaystyle\int_{0}^{\frac{1}{2}}\frac{2x}{\sqrt{1-x^2}}\,dx=2-\sqrt{3}} \end{array}}\qquad\checkmark \end{array}$

<span>If you're having problems understanding this answer, try seeing it through your browser: brainly.com/question/2153950

$\large\textsf{I hope it helps.}$

</span><span><span>Tags: <em>definite integral integrate limits function irrational square root sqrt fraction composite substitution integral calculus</em></span>
</span>

4 0

4 years ago