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satela [25.4K]
3 years ago
11

What is the slope-intercept form of the linear equation 2x - 8y = 32?

Mathematics
1 answer:
kiruha [24]3 years ago
3 0

Answer:

y = 1/4x + 4

Step-by-step explanation:

2x - 8y = 32

-2x            -2x

-8y = -2x + 32

divide everything by -8

so that y becomes positive

answer : Y = 1/4x + 4

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Zara has a collection of 4 marbles: an Aggie, a Bumblebee, a Steelie, and a Tiger. She wants to display them in a row on a shelf
KengaRu [80]

Answer:

Step-by-step explanation:

Oops I meant "there are 3! = 6 ways to permute ABTS that have TS."

3 0
3 years ago
How do I find the number of solutions for this equation? 6y = 12x + 36
Soloha48 [4]

Answer:

6y = 12x + 36  is  x= 1 /2 y−3

15y = 45x + 60​   is  x= 1 /3 y+ −4 /3

Step-by-step explanation:

4 0
3 years ago
How do I solve 5x+1=7x+19
notsponge [240]

Answer:

x=-9

Step-by-step explanation:

5x+1=7x+19

1 = 7x-5x +19

1-19 = 2x

-18/2 = x

x = -9

Hope this helps!

Please mark brainliest if you think I helped! Would really appreciate!

5 0
3 years ago
The graph below represents the solution set of which inequality?
natulia [17]

Answer:

option: B (x^2+2x-8) is correct.

Step-by-step explanation:

We are given the solution set as seen from the graph as:

(-4,2)

1)

On solving the first inequality we have:

x^2-2x-8

On using the method of splitting the middle term we have:

x^2-4x+2x-8

⇒  x(x-4)+2(x-4)=0

⇒ (x+2)(x-4)

And we know that the product of two quantities are negative if either one of them is negative so we have two cases:

case 1:

x+2>0 and x-4

i.e. x>-2 and x<4

so we have the region as:

(-2,4)

Case 2:

x+2 and x-4>0

i.e. x<-2 and x>4

Hence, we did not get a common region.

Hence from both the cases we did not get the required region.

Hence, option 1 is incorrect.

2)

We are given the second inequality as:

x^2+2x-8

On using the method of splitting the middle term we have:

x^2+4x-2x-8

⇒ x(x+4)-2(x+4)

⇒ (x-2)(x+4)

And we know that the product of two quantities are negative if either one of them is negative so we have two cases:

case 1:

x-2>0 and x+4

i.e. x>2 and x<-4

Hence, we do not get a common region.

Case 2:

x-2 and x+4>0

i.e. x<2 and x>-4

Hence the common region is (-4,2) which is same as the given option.

Hence, option B is correct.

3)

x^2-2x-8>0

On using the method of splitting the middle term we have:

x^2-4x+2x-8>0

⇒ x(x-4)+2(x-4)>0

⇒ (x-4)(x+2)>0

And we know that the product of two quantities are positive if either both of them are negative or both of them are positive so we have two cases:

Case 1:

x+2>0 and x-4>0

i.e. x>-2 and x>4

Hence, the common region is (4,∞)

Case 2:

x+2 and x-4

i.e. x<-2 and x<4

Hence, the common region is: (-∞,-2)

Hence, from both the cases we did not get the desired answer.

Hence, option C is incorrect.

4)

x^2+2x-8>0

On using the method of splitting the middle term we have:

x^2+4x-2x-8>0

⇒ x(x+4)-2(x+4)>0

⇒ (x-2)(X+4)>0

And we know that the product of two quantities are positive if either both of them are negative or both of them are positive so we have two cases:

Case 1:

x-2 and x+4

i.e. x<2 and x<-4

Hence, the common region is: (-∞,-4)

Case 2:

x-2>0 and x+4>0

i.e. x>2 and x>-4.

Hence, the common region is: (2,∞)

Hence from both the case we do not have the desired region.

Hence, option D is incorrect.




5 0
3 years ago
What greatest three digit number whose hundreds is two less than its ones digit
kodGreya [7K]

Answer:

799

Step-by-step explanation:

Let x = the hundreds digit

Let y = the ones digit

x = y-2

We want to maximize x and y.  The both have to be single digit numbers

Rewriting this

x+2 =y

The largest x can be is 7 if y is a single digit

x is 7 and y is 9

We can pick the tens digit.  Make it as big as possible to make out number as large as possible

799

4 0
4 years ago
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