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4 years ago

{1,3,5,7,9,11} find the subsets and proper subsets

Mathematics

1 answer:

vaieri [72.5K]4 years ago

6 0

Answer:

Step-by-step explanation:

now gimme a brainliest

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For the Relation (3,0), (4,-1), (5,-2), (6,-3) what is the range?

GaryK [48]

did you got the answer??

7 0

3 years ago

When bands play at the arena the money made from ticket sales is split in an agreed ratio. When Phillip Cage played, £21,000 was

natulia [17]

Answer:

£6000

Step-by-step explanation:

Given:

Phillip Cage played, £21,000 was split at a ratio of 2:5 between the arena and Phillip Cage.

Question asked:

How much money did the arena make from Phillip Cage's concert?

Solution:

Ratio Arena and Philip cage in which money distributed = 2 : 5

Let ratio be $x$

Money earned by Arena = $2x$

Money earned by Phillip Cage = $5x$

Phillip Cage played total for = £21,000

Money earned by Arena + Money earned by Phillip Cage = £21,000

$2x+5x=21000\\7x=21000\\$

Dividing both sides by 7

$x=3000$

Money earned by Arena = $2x$ = $2\times3000=6000$

Thus, Arena made £6000 from Phillip Cage's concert.

5 0

3 years ago

Hillary’s family is thinking of relocating to a different city to save money. They set up a budget comparing the cost of living

mr_godi [17]

the answer is santa fe

3 0

3 years ago

Read 2 more answers

According to the distributive property,6(a+b)=

olga55 [171]

Answer:

Step-by-step explanation:

The Distributive property says that a(b+c) = ab+ac

In my example you distribute the value on the outside to the things on the inside

In your problem, you have a six on the outside that needs to be distributed to an a and b

1) 6 multiplied by a equals 6a giving you the first part of your answer!

2) 6 multiplied by b equals 6b giving you the second part of your answer!

Dont forget that there is a + sign in between a and b, dont drop it!

Put the two values together to get 6a+6b

Let me know if you have any further questions :)

8 0

3 years ago

What are the answers to these 2 question 4&5 PLZ HELP

Ahat [919]

The first problem:

y = -2 x² ⇒⇒⇒ x² = -y/2 →(1)33 x² + y² = 27 → (2)
By substitution from (1) at (2) with the value of x²
∴ 33 (-y/2) + y² = 27
∴ y² - 16.5 y - 27 = 0
a = 1 , b = -16.5 , c = -27
∴ $y= \frac{-b \pm \sqrt{b^2-4ac} }{2a} = \frac{16.5 \pm \sqrt{(-16.5)^2-4*1*(-27)} }{2*1}$
∴ y = 18 or y = -3/2
By substitution from at (1) with the value of y
for y = 18 ⇒⇒⇒ x² = -18/2 = -9 (unacceptable)
for y = -3/2 ⇒⇒⇒ x² = -(-3/2)/2 = 3/4
∴ $x= \pm \sqrt{ \frac{3}{4} } = \pm \frac{ \sqrt{3} }{2} 
$
The correct options are 2 , 7
Solution of the system of equations is
$( \frac{ \sqrt{3} }{2} , \frac{-3}{2} )$
and
$( -\frac{\sqrt{3} }{2} , \frac{-3}{2} )$
==================================
The second problem:
The general equation of the hyperbole is
$\frac{x^2}{a^2} - \frac{y^2}{b^2} =1

Transverse axis is horizontal 

The equation if the asymptotes are y = \pm \frac{b}{a}x$
For the given equation:
$\frac{x^2}{225} - \frac{y^2}{36} = 1$
a² = 225 ⇒⇒⇒ a = √225 = 15
b² = 36 ⇒⇒⇒ b = √36 = 6
∴ the slope of the asymptotes = b/a and -b/a

b/a = 6/15 = 2/5

-b/a = -6/15 = -2/5

∴ m = 2/5 and m = -2/5

3 0

3 years ago