Ask question

Ask question

All categories

3 years ago

11

At a school, 40% of the 6th-grade students said that hip-hop is their favorite kind of music. If 100 6th grade students prefer h

ip hop music how many 6th grade students are at the school explain or show your reasoning

1 answer:

Alexeev081 [22]3 years ago

8 0

Answer:

250

Step-by-step explanation:

40% of 250 is 100

You might be interested in

In a unit circle, ø = 2pi radians What is the terminal point?

BlackZzzverrR [31]

Answer:

The terminal point is (1, 0) ⇒ D

Step-by-step explanation:

In the unit circle, Ф is the angle between the terminal side and the positive part of the x-xis.

The terminal point on the positive part of the x-axis is (1, 0),which means Ф = 0° or 360° and cosФ = 1, sinФ = 0

The terminal point on the positive part of the y-axis is (0, 1),which means Ф = 90° and cosФ = 0, sinФ = 1

The terminal point on the negative part of the x-axis is (-1, 0),which means Ф = 180° and cosФ = -1, sinФ = 0

The terminal point on the negative part of the y-axis is (0, -1),which means Ф = 270° and cosФ = 0, sinФ = -1

In a unit circle

∵ Ф = 2π radians

∵ 2π radians = 360°

→ By using the 1st rule above

∴ The terminal point is (1, 0)

5 0

3 years ago

What is the first step in writing f(x) = 6x2 + 5 – 42x in vertex form?

podryga [215]

The answer: y= 6(x-7/2)^2-137/2

5 0

3 years ago

The vector (a) is a multiple of the vector (2i +3j) and (b) is a multiple of (2i+5j) The sum (a+b) is a multiple of the vector (

kow [346]

Answer:

$\|a\| = 5\sqrt{13}$ .

$\|b\| = 3\sqrt{29}$ .

Step-by-step explanation:

Let $m$ , $n$ , and $k$ be scalars such that:

$\displaystyle a = m\, (2\, \vec{i} + 3\, \vec{j}) = m\, \begin{bmatrix}2 \\ 3\end{bmatrix}$ .

$\displaystyle b = n\, (2\, \vec{i} + 5\, \vec{j}) = n\, \begin{bmatrix}2 \\ 5\end{bmatrix}$ .

$\displaystyle (a + b) = k\, (8\, \vec{i} + 15\, \vec{j}) = k\, \begin{bmatrix}8 \\ 15\end{bmatrix}$ .

The question states that $\| a + b \| = 34$ . In other words:

$k\, \sqrt{8^{2} + 15^{2}} = 34$ .

$k^{2} \, (8^{2} + 15^{2}) = 34^{2}$ .

$289\, k^{2} = 34^{2}$ .

Make use of the fact that $289 = 17^{2}$ whereas $34 = 2 \times 17$ .

$\begin{aligned}17^{2}\, k^{2} &= 34^{2}\\ &= (2 \times 17)^{2} \\ &= 2^{2} \times 17^{2} \end{aligned}$ .

$k^{2} = 2^{2}$ .

The question also states that the scalar multiple here is positive. Hence, $k = 2$ .

Therefore:

$\begin{aligned} (a + b) &= k\, (8\, \vec{i} + 15\, \vec{j}) \\ &= 2\, (8\, \vec{i} + 15\, \vec{j}) \\ &= 16\, \vec{i} + 30\, \vec{j}\\ &= \begin{bmatrix}16 \\ 30 \end{bmatrix}\end{aligned}$ .

$(a + b)$ could also be expressed in terms of $m$ and $n$ :

$\begin{aligned} a + b &= m\, (2\, \vec{i} + 3\, \vec{j}) + n\, (2\, \vec{i} + 5\, \vec{j}) \\ &= (2\, m + 2\, n) \, \vec{i} + (3\, m + 5\, n) \, \vec{j} \end{aligned}$ .

$\begin{aligned} a + b &= m\, \begin{bmatrix}2\\ 3 \end{bmatrix} + n\, \begin{bmatrix} 2\\ 5 \end{bmatrix} \\ &= \begin{bmatrix}2\, m + 2\, n \\ 3\, m + 5\, n\end{bmatrix}\end{aligned}$ .

Equate the two expressions and solve for $m$ and $n$ :

$\begin{cases}2\, m + 2\, n = 16 \\ 3\, m + 5\, n = 30\end{cases}$ .

$\begin{cases}m = 5 \\ n = 3\end{cases}$ .

Hence:

$\begin{aligned} \| a \| &= \| m\, (2\, \vec{i} + 3\, \vec{j})\| \\ &= m\, \| (2\, \vec{i} + 3\, \vec{j}) \| \\ &= 5\, \sqrt{2^{2} + 3^{2}} = 5 \sqrt{13}\end{aligned}$ .

$\begin{aligned} \| b \| &= \| n\, (2\, \vec{i} + 5\, \vec{j})\| \\ &= n\, \| (2\, \vec{i} + 5\, \vec{j}) \| \\ &= 3\, \sqrt{2^{2} + 5^{2}} = 3 \sqrt{29}\end{aligned}$ .

6 0

3 years ago

Write a phrase as an algebraic expression six less than a number times 11

nadezda [96]

Let the number = x

11x is 11 times the number. You are to take 6 less than this. Be careful how you read six less. It means the 6 is subtracted from the 11x

11x - 6 This is what you finally get.

7 0

3 years ago

5-2.1=? 5-2.17=? 5-2 7/8 solve make sure to re-check your self before answering or no forty points

frozen [14]

Answer: 5-2.1= 2.9

5-2.17= 2.83

5-2 7/8= 2.125

6 0

4 years ago