Ask question

Ask question

All categories

3 years ago

9

A teacher brings 8 birdies and 32 rackets to a physical education class of 30 students playing badminton. what is the ratio of b

irdies to students for this class?

1 answer:

ipn [44]3 years ago

5 0

To get your ratio of birdies do 8/32/30 and that will give you your ratio

You might be interested in

11. Higher Order Thinking Using what you know

Lynna [10]

Answer:

m∠1 + m∠3 = m∠2 + m∠4

Step-by-step explanation:

In the figure attached,

In the given trapezoid if two sides AB and CD are parallel and AD is a transverse,

m∠1 + m∠3 = 180°

Similarly, if AB and CD are parallel and BC is a transverse,

m∠2 + m∠4 = 180°

Therefore, m∠1 + m∠3 = m∠2 + m∠4 is the relation between these angles.

6 0

3 years ago

Two angles are supplementary. One angle is 16° less than three times the other. Find the measures of the angles

Ivan

Answer:

Step-by-step explanation:

x and y are supplementary angles, so their sum is 180°.

x+y = 180°

x is 16° less than three times y

x = 3y - 16°

(3y-16°) + y = 180°

4y = 196°

y = 49°

x = 3y-16° = 131°

7 0

3 years ago

Carly is traveling to visit family. She drives x miles on the first day and y miles on the second day for a total of 800 miles.

Dmitrij [34]

Answer:

The domain and range of this relationship is (-∞,∞) and (-∞,∞) respectively.

Step-by-step explanation:

We are given that Carly is traveling to visit family.

She drive distance on first day = x

She drive distance on second day = y

We are also given that Carly is travelling a total of 800 miles.

So,x+y=800

We are supposed to find domain and range of this relationship.

y=800-x

Domain : (-∞,∞)

Range:(-∞,∞)

Hence the domain and range of this relationship is (-∞,∞) and (-∞,∞) respectively.

5 0

3 years ago

Read 2 more answers

Prove that: (b²-c²/a)CosA+(c²-a²/b)CosB+(a²-b²/c)CosC = 0

IRISSAK [1]

<u>Prove that:</u>

$\:\:\sf\:\:\left(\dfrac{b^2-c^2}{a}\right)\cos A+\left(\dfrac{c^2-a^2}{b}\right)\cos B +\left(\dfrac{a^2-b^2}{c}\right)\cos C=0$

<u>Proof: </u>

We know that, by Law of Cosines,

$\sf \cos A=\dfrac{b^2+c^2-a^2}{2bc}$
$\sf \cos B=\dfrac{c^2+a^2-b^2}{2ca}$
$\sf \cos C=\dfrac{a^2+b^2-c^2}{2ab}$

<u>Taking</u><u> </u><u>LHS</u>

$\left(\dfrac{b^2-c^2}{a}\right)\cos A+\left(\dfrac{c^2-a^2}{b}\right)\cos B +\left(\dfrac{a^2-b^2}{c}\right)\cos C$

<em>Substituting</em> the value of <em>cos A, cos B and cos C,</em>

$\longmapsto\left(\dfrac{b^2-c^2}{a}\right)\left(\dfrac{b^2+c^2-a^2}{2bc}\right)+\left(\dfrac{c^2-a^2}{b}\right)\left(\dfrac{c^2+a^2-b^2}{2ca}\right)+\left(\dfrac{a^2-b^2}{c}\right)\left(\dfrac{a^2+b^2-c^2}{2ab}\right)$

$\longmapsto\left(\dfrac{(b^2-c^2)(b^2+c^2-a^2)}{2abc}\right)+\left(\dfrac{(c^2-a^2)(c^2+a^2-b^2)}{2abc}\right)+\left(\dfrac{(a^2-b^2)(a^2+b^2-c^2)}{2abc}\right)$

$\longmapsto\left(\dfrac{(b^2-c^2)(b^2+c^2)-(b^2-c^2)(a^2)}{2abc}\right)+\left(\dfrac{(c^2-a^2)(c^2+a^2)-(c^2-a^2)(b^2)}{2abc}\right)+\left(\dfrac{(a^2-b^2)(a^2+b^2)-(a^2-b^2)(c^2)}{2abc}\right)$

$\longmapsto\left(\dfrac{(b^4-c^4)-(a^2b^2-a^2c^2)}{2abc}\right)+\left(\dfrac{(c^4-a^4)-(b^2c^2-a^2b^2)}{2abc}\right)+\left(\dfrac{(a^4-b^4)-(a^2c^2-b^2c^2)}{2abc}\right)$

$\longmapsto\dfrac{b^4-c^4-a^2b^2+a^2c^2}{2abc}+\dfrac{c^4-a^4-b^2c^2+a^2b^2}{2abc}+\dfrac{a^4-b^4-a^2c^2+b^2c^2}{2abc}$

<em>On combining the fractions,</em>

$\longmapsto\dfrac{(b^4-c^4-a^2b^2+a^2c^2)+(c^4-a^4-b^2c^2+a^2b^2)+(a^4-b^4-a^2c^2+b^2c^2)}{2abc}$

$\longmapsto\dfrac{b^4-c^4-a^2b^2+a^2c^2+c^4-a^4-b^2c^2+a^2b^2+a^4-b^4-a^2c^2+b^2c^2}{2abc}$

<em>Regrouping the terms,</em>

$\longmapsto\dfrac{(a^4-a^4)+(b^4-b^4)+(c^4-c^4)+(a^2b^2-a^2b^2)+(b^2c^2-b^2c^2)+(a^2c^2-a^2c^2)}{2abc}$

$\longmapsto\dfrac{(0)+(0)+(0)+(0)+(0)+(0)}{2abc}$

$\longmapsto\dfrac{0}{2abc}$

$\longmapsto\bf 0=RHS$

LHS = RHS proved.

7 0

2 years ago

What value of k will result in the quadratic having one rational solution if 4x2+kx+4=0?

PtichkaEL [24]

The value of 'k' in the quadratic Equation $4x^{2}+kx+4=0$ having one rational solution is k is 8.

<h3>What is a Quadratic Equation?</h3>

Quadratic equations are the polynomial equations of degree 2 in one variable of the type $f(x) = ax^{2} + bx + c = 0$ where a, b, c, ∈ R and a ≠ 0.
The standard form of a quadratic equation is $ax^{2} + bx + c = 0$ .

Here, the given equation is $4x^{2}+kx+4=0$

By comparing the given equation with the standard form of the quadratic equation we get,

a = 4

b = k

c = 4

The given quadratic is having one rational solution, which means $b^{2} -4ac=0$

Therefore,

$k^{2} -4 \times 4\times4=0\\k^{2} -64=0\\k^{2} =64\\k=\sqrt{64} \\k=8$

Therefore, the value of k is 8.

Learn more about the quadratic equation is brainly.com/question/8649555

#SPJ4

7 0

1 year ago