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Cloud [144]
3 years ago
9

The area of a rectangular pool is 6992 m².

Mathematics
1 answer:
Nina [5.8K]3 years ago
3 0

Step-by-step explanation:

area=6992m2

length=92m

Now,

area of rectangle=6992

You might be interested in
Solve the system of equations: y= x^2 + 2x - 3 and y=3x+ 3.
Semenov [28]

Answer:

(-2, -3)

(3, 12)

Step-by-step explanation:

To solve this, we're gonna get rid of the y's with substitution

x² + 2x - 3 = 3x + 3

Let's make this equation equal to zero

Subtract 3 from both sides

x² + 2x - 3 = 3x + 3

            - 3         - 3

x² + 2x - 6 = 3x

Subtract 3x from both sides

x² + 2x - 6 = 3x

    - 3x        - 3x

x² - x - 6 = 0

Factor the equation

(x - 3)(x + 2) = 0

This means x can be -2 or 3

Let's solve it with -2 first, plug the new x in y = 3x + 3

y = 3(-2) + 3

y = -6 + 3 = -3

Do the same for x = 3

y = 3(3) + 3

y = 9 + 3 = 12

3 0
2 years ago
Part 3 - Discussion/Explanation Question
SpyIntel [72]

Step-by-step explanation:

Vertical asymptote can be Identites if there is a factor only in the denominator. This means that the function will be infinitely discounted at that point.

For example,

\frac{1}{x - 5}

Set the expression in the denominator equal to 0, because you can't divide by 0.

x - 5 = 0

x = 5

So the vertical asymptote is x=5.

Disclaimer if you see something like this

\frac{(x - 5)(x + 3)}{(x - 5)}

x=5 won't be a vertical asymptote, it will be a hole because it in the numerator and denominator.

Horizontal:

If we have a function like this

\frac{1}{x}

We can determine what happens to the y values as x gets bigger, as x gets bigger, we will get smaller answers for y values. The y values will get closer to 0 but never reach it.

Remember a constant can be represent by

a \times  {x}^{0}

For example,

1 = 1 \times  {x}^{0}

2 =  2 \times {x}^{0}

And so on,

and

x =  {x}^{1}

So our equation is basically

\frac{1 \times  {x}^{0} }{ {x}^{1} }

Look at the degrees, since the numerator has a smaller degree than the denominator, the denominator will grow larger than the numerator as x gets larger, so since the larger number is the denominator, our y values will approach 0.

So anytime, the degree of the numerator < denominator, the horizontal asymptote is x=0.

Consider the function

\frac{3 {x}^{2} }{ {x}^{2}  + 1}

As x get larger, the only thing that will matter will be the leading coefficient of the leading degree term. So as x approach infinity and negative infinity, the horizontal asymptote will the numerator of the leading coefficient/ the leading coefficient of the denominator

So in this case,

x =  \frac{3}{1}

Finally, if the numerator has a greater degree than denominator, the value of horizontal asymptote will be larger and larger such there would be no horizontal asymptote instead of a oblique asymptote.

8 0
2 years ago
3) If a = -19, d = 4 then find t2<br>​
geniusboy [140]

Answer:

-15.

Step-by-step explanation:

Arithmetic sequence:-

t2 = a + d

= -19 + 4

= -15.

8 0
3 years ago
Which are correct representations of the inequality –3(2x-5) &lt; 5(2 - x)? Select two options.​
Igoryamba

Answer:  All the ways I solved this the answer is the same,

x>5 or (5,∞) That being the Interval Notation {The second one}

Step-by-step explanation:

3 0
3 years ago
Please help explanation if possible
poizon [28]

Step-by-step explanation:

Hi there!

The given equation is:

y = -2x + 5………………(i)

Comparing the equation with y = mx+c, we get;

m1 = -2

Also another equation of the line which passes through point (-4,2), we get;

(y-2) = m2(X+4)............(ii) { using the formula (y-y1) = m2(x-x2)}

According to the question, they are perpendicular to eachother, So according to the condition of perpendicular lines;

m1*m2 = -1

-2*m2 = -1

or, m2 = 1/2.

Therefore, m2= 1/2.

Now, keeping the value of m2 in equation (ii).

(y-2) = 1/2(x+4)

y = (1/2)x + 4

Therefore, the required equation is: y = (1/2)x + 4.

<u>Hope</u><u> it</u><u> helps</u><u>!</u>

3 0
3 years ago
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