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Keith_Richards [23]
3 years ago
10

Omg please answer i am so lost

Mathematics
1 answer:
Soloha48 [4]3 years ago
5 0
A

Because X isn’t repeated in W but there are repeats in the others.
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4.5 + 7.1<br><br> Please help me
Alinara [238K]

Answer:

11.6 is the answer friend.

Step-by-step explanation:

4 0
3 years ago
What is 9/10-3/5 in fractions
Delvig [45]
-3/5 = -6/10
9/10 - 6/10 = 3/10
3 0
3 years ago
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(1 point) In this problem we show that the function f(x,y)=7x−yx+y f(x,y)=7x−yx+y does not have a limit as (x,y)→(0,0)(x,y)→(0,0
polet [3.4K]

Answer:

Step-by-step explanation:

Given that,

f(x, y)=7x−yx+y

We want to show that the limit doesn't exist as (x, y)→(0,0).

Limits typically fail to exist for one of four reasons:

1. The one-sided limits are not equal

2. The function doesn't approach a finite value

3. The function doesn't approach a particular value

4. The x - value is approaching the endpoint of a closed interval

a. Considering the case that y=3x

lim(x,y)→(0,0) 7x−yx+y

Since y=3x

lim(x,3x)→(0,0) 7x−3x(x)+3x

lim(x,3x)→(0,0) 7x−3x(x)+3x

lim(x,3x)→(0,0) 10x−3x²

Therefore,

lim(x,3x)→(0,0) 10x−3x² = 0-0=0

b. Let also consider at y=4x

lim(x,y)→(0,0) 7x−yx+y

Since y=4x

lim(x,4x)→(0,0) 7x−4x(x)+4x

lim(x,4x)→(0,0) 7x−4x(x)+4x

lim(x,4x)→(0,0) 11x−4x²

Therefore,

lim(x,4x)→(0,0) 11x−4x² = 0-0=0

c. Let also consider it generally at y=mx

lim(x,y)→(0,0) 7x−yx+y

Since y=mx

lim(x,mx)→(0,0) 7x−mx(x)+mx

lim(x,mx)→(0,0) 7x−mx(x)+mx

lim(x, mx)→(0,0) (7+m)x−mx²

Therefore,

lim(x, mx)→(0,0) (7+m)x−mx² = 0-0=0

But the limit of the given function exist.

So let me assume the function is wrong and the question meant.

f(x, y)= (7x−y) / (x+y)

So, let analyze again

a. Considering the case that y=3x

lim(x,y)→(0,0) (7x−y)/(x+y)

Since y=3x

lim(x,3x)→(0,0) (7x−3x)/(x+3x)

lim(x,3x)→(0,0) 4x/4x

lim(x,3x)→(0,0) 1

Therefore,

lim(x,3x)→(0,0) 1= 1

So the limit is 1

b. Let also consider at y=4x

lim(x,y)→(0,0) (7x−y)/(x+y)

Since y=4x

lim(x,4x)→(0,0) (7x−4x)/(x+4x)

lim(x,4x)→(0,0) 3x/5x

lim(x,4x)→(0,0) 3/5

Therefore,

lim(x,4x)→(0,0) 3/5 = 3/5

So the limit is 3/5

This show that the limit does not exit.

Since one of the condition given above is met, then the limit does not exist. i.e. The function doesn't approach a particular value

c. Let also consider it generally at y=mx

lim(x,y)→(0,0) (7x−y)/(x+y)

Since y=mx

lim(x,mx)→(0,0) (7x−mx)/(x+mx)

lim(x,mx)→(0,0) (7-m)x/(1+m)x

lim(x, mx)→(0,0) (7-m)/(1+m)

Therefore,

lim(x, mx)→(0,0) (7-m)/(1+m) = (7m)/(1+m)

Then, the limit is (7-m)/(1+m)

So the limit doesn't not have a specific value, it depends on the value of m, so the limit doesn't exist.

7 0
3 years ago
How many solutions are there to the equation 5x+48+7x.=12(x+4)
ELEN [110]
5x+48+7x=12(x+4)\\\\12x+48=12x+48\\\\L=R\\\\There\ are\ infinitly\ many\ solutions.
5 0
3 years ago
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Monica has a 24-inch square frame. There are 4 equal sides. She paints 3/4 of one side of the frame blue. What fraction of the f
Serhud [2]

Answer:

Monica painted \frac{3}{16} of the frame blue.

Step-by-step explanation:

Monica has a 24-inch square frame.There are 4 equal sides. She paints 3/4 of one side of the frame blue.

The perimeter of square is given as = 4s where s is the side length.

Here, perimeter is 24 inches.

So, side will be:

4s=24

Dividing both sides by 4;

s = 6

So, we get the side of the frame as 6 inches.

Now Monica painted 3/4 of 6 inches.

This value becomes : \frac{3}{4}\times6 =4.5 and in fraction it is \frac{9}{2}

This is part of 24 inches.

So, the fraction will be : \frac{\frac{9}{2} }{24}

= \frac{9}{2} \times\frac{1}{24} = 3/16

Hence, Monica painted \frac{3}{16} of the frame blue.

6 0
3 years ago
Read 2 more answers
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