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2 years ago

Need help................

Mathematics

1 answer:

Triss [41]2 years ago

6 0

827537517172763 kahvxjx

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There are 45 students in a class. If 60% of them are girls, find the number of girls and boys

grigory [225]

Number of girls-27
Number of boys-18

60% of 45 is 27 which is the number of girls.So now you have to subtract 27 by the total which will give you 18

8 0

3 years ago

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The coordinates of P and Q are (2,6) and (4,5) respectively. Equation of a line RS is Y= 2X -3. Determine if the lines are paral

Travka [436]

Answer:

Perpendicular

Step-by-step explanation:

Given points P=(2,6) and Q=(4,5)

The line RS is y=2x-3.

Now we know that the line joining any two points (x1,y1) and (x2,y2) has slope m= $\frac{y2-y1}{x2-x1}$

Now the slope of the line PQ is

m1= $\frac{5-6}{4-2}$ = - $\frac{1}{2}$

We know that the slope of the line in the form y=mx+c is m

So the line RS has slope m2=2

Now two lines are parallel if m1=m2

And those two lines are perpendicular if $m1*m2=-1$

here m1= - $\frac{1}{2}$ and m2=2

clearly $m1*m2=-1$

Therefore the two given lines are perpendicular

7 0

2 years ago

Simplify the expression: -3b - 3b

Ksju [112]

Answer:

Step-by-step explanation:

-3b - 3b= (-3b)+(-3b)= -6b

6 0

2 years ago

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What is the percent of 16/25

Lelechka [254]

16/25 = 0.64 = 64%

Another way to do it is to multiply top and bottom by 4
16/25 = (16*4)/(25*4) = 64/100 = 64%

Either way, the answer is 64%

7 0

3 years ago

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NO LINKS!! Please help me with this problem

iogann1982 [59]

${\qquad\qquad\huge\underline{{\sf Answer}}}$

The given figure shows a vertical hyperbola with its centre at origin, and as we observe the figure, we can conclude that :

Length of transverse axis is :

$\qquad \sf \dashrightarrow \: 2b = 12$

$\qquad \sf \dashrightarrow \: b = 6$

length of conjugate axis is :

$\qquad \sf \dashrightarrow \: 2a = 8$

$\qquad \sf \dashrightarrow \: a = 4$

Equation of hyperbola ~

$\qquad \sf \dashrightarrow \: \cfrac{ {y}^{2} }{ {b}^{2} } - \cfrac{ {x}^{2} }{ {a}^{2} } = 1$

plug in the values ~

$\qquad \sf \dashrightarrow \: \cfrac{ {y}^{2} }{ {6}^{2} } - \cfrac{ {x}^{2} }{ {4}^{2} } = 1$

$\qquad \sf \dashrightarrow \: \cfrac{ {y}^{2} }{ {36}^{} } - \cfrac{ {x}^{2} }{ {16}^{} } = 1$

5 0

1 year ago

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