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vlada-n [284]
3 years ago
11

Help, please ? Thank you so much in advance and I'd appreciate it!

Mathematics
2 answers:
Darina [25.2K]3 years ago
7 0
First you calculate the area of the big rectangle (ignoring the small one the bottom):

A1 = 10 x 6 = 60.

next you calculate the small one:

A2 = 2 x 2 = 4

Now you subtract the small one from the big one and you get:

Area of the figure = A1 - A2 = 60 - 4 = 56
hoa [83]3 years ago
3 0
To start off, pretend it's a quadrilateral. It makes everything easier and uses all the numbers.

The area of a quadrilateral is base times height. 
10 = base 
6 = height 

6 * 10 = 60 

That's the value of the original, larger, quadrilateral without the mini quadrilateral that we have to subtract. 

The dimensions of the height and base of the mini quadrilateral are 2 by 2. 
2 * 2 = 4 

Subtract to get your answer: 
60 - 4 = 56


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Which expression has a value of 90?
ratelena [41]

Answer:

1. 73.2

2. 90

3. 80

4. 80

Step-by-step explanation:

you will have to divide 3,600 by 50

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3x - 12
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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
Bhairav collected Rs.600 in his piggy bank by putting in Rs.2 and Rs.5 coins. The number of Rs.5 coins are twice as many as Rs.2
Anna [14]

Answer:

100 Rs 5 coins and 50 Rs 2 coins

Step-by-step explanation:

Bhairav collected Rs.600 in his piggy bank by putting in Rs.2 and Rs.5 coins. The number of Rs.5 coins are twice as many as Rs.2 coins

Let x be the number of 5 coins

and y be the number of 2 coins

The number of Rs.5 coins are twice as many as Rs.2 coins

x=2y

5x+2y=600

Replace x with 2y

5x+2y=600\\5(2y)+2y=600\\10y+2y=600\\12y=600\\y=50

x=2y\\x=2(50)\\x=100

So 100 Rs 5 coins and 50 Rs 2 coins

7 0
3 years ago
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