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Elina [12.6K]
3 years ago
11

Is the table proportional? 1,5 2,10 3,15​

Mathematics
1 answer:
elixir [45]3 years ago
7 0

Answer:

no becuz its a whole number not a fraction

Step-by-step explanation:

ax+b

a=5

b=0

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Find tan theta if theta is an angle in standard position and the point with coordinates (4,3) lies on the terminal side of the a
Karolina [17]

Answer: 0.75

Step-by-step explanation:

We can do this by drawing a terminal side in a standard position and then drawing the associated triangle. We then use the Pythagoras theorem to find the missing side the hypotenuse. Then, this will give a Pythagorean Triplet of 3, 4 and 5.

We now have a triangle with the values of:

x = 4

y =3

h^2= 4^2 + 3^2

h^2 = 16 + 9

h^2 = 25

h = 5

h = 5

Tangent = opposite/adjacent

= y/x

= 3/4

= 0.75

8 0
3 years ago
FOR BRAINLIEST ANSWER HURRY HELP THANKS If (a,b) is a point in quadrant IV, what must be true about a? What must be true about b
Natali5045456 [20]

Answer:

Well if (a,b) is in Quadrant IV which is the last quadrant the a or x is a positive number and the b or y is a negative number.

7 0
3 years ago
Read 2 more answers
(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 do you add <br> 2 2/3 + 6/7
Korolek [52]

Answer:

3 11/21

Step-by-step explanation:

2 2/3=(2*3+2)/3=<u>8/3</u>

8/3+6/7=56/21+18/21=74/21 or 3 11/21

5 0
3 years ago
Read 2 more answers
1) If the temperature outside was 73 3/4 degrees at 5:00pm, but it fell 8 degrees by 10:00pm, what is the temperature
vladimir2022 [97]

Answer:

Step-by-step explanation:

Temp 5 pm:     73.75       3/4 = 0.75

<u>Temp 10 pm      -8   </u>         falling temperatures are - numbers.

Temp 10 pm =  65.75

5 0
2 years ago
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