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3 years ago

15

For every `8` boys wearing neon blue, there are `2` girls wearing neon blue. If there are `32` boys wearing neon blue, how many

girls are wearing neon blue?

2 answers:

insens350 [35]3 years ago

6 0

Answer:

8:2

32:8

Step-by-step explanation:

It's pretty simple:

If there's 32 boys, which is 4 times the amount you start with, you have to apply that to the number of girls as well since for every 8 boys, there is 2 girls.

8x4

2x4

32:8 Hope this helps

taurus [48]3 years ago

5 0

Answer:

8 girls are wearing neon blue.

Step-by-step explanation:

There are several ways to do this but one of them is to find how many groups of 8 boys are in 32. To do this we divide 32 by 8 which equals 4. We then multiply 2 by 4 which is 8.

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Can someone help me pls

IgorLugansk [536]

250 millilitres

There are 2000 millilitres in 2 litres.
so 2000/8= 250ml

the cup holds 250ml

Hope I helped

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3 years ago

Pizza Perfection sold 93 pizzas on Monday, 67 pizzas on Tuesday, 78 pizzas on Wednesday, 44 pizzas

Nezavi [6.7K]

(93+67+78+44+123)/5=405/5=81 pizzas per day on average

C is your answer.

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3 years ago

Read 2 more answers

Which are vertical angles?<br> O AFE and BFD<br> O BFC and DFE<br> O AFE and CFD<br> O BFC and EFA

mel-nik [20]

the correct answer is a, because a vertical angle is made of 2 intercepting lines.

3 0

3 years ago

The table shows the number of boys and girls who signed up for certain summer classes.

Makovka662 [10]

Answer:

Music and Dance

Step-by-step explanation:

When looking at the numbers for each class u want to add up the numbers. When u get the answers look for the two numbers witch are the closes to each other.

7 0

3 years ago

The set of points (-3, 4), (-1, 1), (-3,-2), and (-5,1) identifies the vertices of a quadrilateral. Which is the most specific d

KIM [24]

Answer:

Option (4). Rhombus

Step-by-step explanation:

From the figure attached,

Distance AB = $\sqrt{(x_{2}-x_{1})^2+(y_{2}-y_{1})^2}$

= $\sqrt{(1-4)^2+(-5+3)^2}$

= $\sqrt{(-3)^2+(-2)^2}$

= $\sqrt{13}$

Distance BC = $\sqrt{(4-1)^2+(-3+1)^2}$

= $\sqrt{9+4}$

= $\sqrt{13}$

Distance CD = $\sqrt{(-2-1)^2+(-3+1)^2}$

= $\sqrt{9+4}$

= $\sqrt{13}$

Distance AD = $\sqrt{(1+2)^2+(-5+3)^2}$

= $\sqrt{9+4}$

= $\sqrt{13}$

Slope of AB ( $m_{1}$ ) = $\frac{y_{2}-y_{1}}{x_{2}-x_{1}}$

= $\frac{4-1}{-3+5}$

= $\frac{3}{2}$

Slope of BC ( $m_{2}$ ) = $\frac{4-1}{-3+1}$

= $-\frac{3}{2}$

If AB and BC are perpendicular then,

$m_{1}\times m_{2}=-1$

But it's not true.

[ $m_{1}\times m_{2}=(\frac{3}{2})(-\frac{3}{2})$ = - $\frac{9}{4}$ ]

It shows that the consecutive sides of the quadrilateral are not perpendicular.

Therefore, ABCD is neither square nor a rectangle.

Slope of diagonal BD = $\frac{4+2}{-3+3}$

= Not defined (parallel to y-axis)

Slope of diagonal AC = $\frac{1-1}{-1+5}$

= 0 [parallel to x-axis]

Therefore, both the diagonals AC and BD will be perpendicular.

And the quadrilateral formed by the given points will be a rhombus.

5 0

3 years ago