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2 years ago

Which two rooms have an

Mathematics

1 answer:

alexgriva [62]2 years ago

4 0

Answer: room 101 and room 107

Step-by-step explanation:

In room 101, the ratio of boys to girls is 16:12. This is further simplified to its lowest fraction by dividing by 4. It becomes 4:3

In room 104, the ratio of boys to girls is 20:9. This cannot be further simplified to its lowest fraction.

In room 107, the ratio of boys to girls is 12:9. This is further simplified to its lowest fraction by dividing by 3. It becomes 4:3

Therefore, room 101 and room 107 have the same ratio.

You might be interested in

How much of 30% HCI and 45% HCI would be needed to mix and produce 50 liters of 35% HCI

VladimirAG [237]

Answer:

33 1/3 liters of 30%
16 2/3 liters of 45%

Step-by-step explanation:

Let x represent the liters of 45% solution needed. Then the amount of HCl in the mix is ...

0.45x +0.30(50 -x) = 0.35(50)

0.15x = 0.05(50) . . . . . simplify, subtract 0.30(50)

x = (0.05/0.15)(50) = 50/3 = 16 2/3 . . . liters of 45% HCl

33 1/3 liters of 30% and 16 2/3 liters of 45% HCl are needed.

_____

<em>Comment on the solution</em>

You may notice that the general solution to a mixture problem of this sort is that the fraction of the mix that is the highest contributor is ...

(mix % - low %) / (high % - low %) = (.35 -.30) / (.45 -.30) = .05/.15 = 1/3

5 0

3 years ago

Help here too please

lyudmila [28]

What’s the question

7 0

2 years ago

What degree of rotation is represented on this matrix

Korvikt [17]

Answer:

Option B is correct

the degree of rotation is, $-90^{\circ}$

Step-by-step explanation:

A rotation matrix is a matrix that is used to perform a rotation in Euclidean space.

To find the degree of rotation using a standard rotation matrix i.e,

$R = \begin{bmatrix}\cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{bmatrix}$

Given the matrix: $\begin{bmatrix}0 & 1 \\ -1 & 0\end{bmatrix}$

Now, equate the given matrix with standard matrix we have;

$\begin{bmatrix}\cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{bmatrix}$ = $\begin{bmatrix}0 & 1 \\ -1 & 0\end{bmatrix}$

On comparing we get;

$\cos \theta = 0$ and $-\sin \theta =1$

As,we know:

$\cos \theta = \cos(-\theta)$

$\sin(-\theta) = -\sin \theta$

$\cos \theta = \cos(90^{\circ}) = \cos( -90^{\circ})$

we get;

$\theta = -90^{\circ}$

and

$\sin \theta =- \sin (90^{\circ}) = \sin ( -90^{\circ})$

we get;

$\theta = -90^{\circ}$

Therefore, the degree of rotation is, $-90^{\circ}$

7 0

2 years ago

What is the difference in height from the top of the pillar to the bottom of the pillar ?

stiks02 [169]

Answer:

i think the difference in height from the top to the bottom is 8ft

8 0

2 years ago

Let f(x) = 7x^2-5x+3 and g(x) = 2x^2+4x-6

Troyanec [42]

Answer:

$\large\boxed{A.\ f(x)+g(x)=9x^2-x-3}\\\boxed{B.\ f(x)-g(x)=5x^2-9x+9}\\\boxed{g(x)-f(x)=-5x^2+9x-9}$

Step-by-step explanation:

$f(x)=7x^2-5x+3,\ g(x)=2x^2+4x-6\\\\A:\\f(x)+g(x)=(7x^2-5x+3)+(2x^2+4x-6)\\f(x)+g(x)=7x^2-5x+3+2x^2+4x-6\qquad\text{combine like terms}\\f(x)+g(x)=(7x^2+2x^2)+(-5x+4x)+(3-6)\\f(x)+g(x)=9x^2-x-3$

$B:\\f(x)-g(x)=(7x^2-5x+3)-(2x^2+4x-6)\\f(x)+g(x)=7x^2-5x+3-2x^2-4x+6\qquad\text{combine like terms}\\f(x)-g(x)=(7x^2-2x^2)+(-5x-4x)+(3+6)\\f(x)+g(x)=5x^2-9x+9$

$C:\\g(x)-f(x)=(2x^2+4x-6)-(7x^2-5x+3)\\g(x)-f(x)=2x^2+4x-6-7x^2+5x-3\qquad\text{combine like terms}\\g(x)-f(x)=(2x^2-7x^2)+(4x+5x)+(-6-3)\\g(x)-f(x)=-5x^2+9x-9$

3 0

3 years ago