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4 years ago

Write the domain and range as inequalities.

Mathematics

1 answer:

Andrews [41]4 years ago

6 0

Answer:

domain: -∞ < x < ∞
range: -2 ≤ y < ∞

Step-by-step explanation:

<u>Domain</u>

The domain is the horizontal extent of the function, the set of input values for which it is defined. A parabola that opens up or down is defined for all x values, so its domain is ...

-∞ < x < ∞ . . . . . . (-∞, ∞) in interval notation

<u>Range</u>

The range is the vertical extent of the function, the set of output values it produces. The vertex of the parabola is one extreme of the range. Depending on the direction it opens, the other extreme will be +∞ or -∞. The range includes the y-value of the vertex. Here, the range is ...

-2 ≤ y < ∞ . . . . . . [-2, ∞) in interval notation

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Alejandro was picked by this teacher to find the integer that has the square root closest to 7, without going over. he wrote 52

andrey2020 [161]

Considering that the square root of 49 is 7, and 52 > 49, Alejandro was not correct, as the correct integer is of 48.

<h3>What is the integer that has the square root closest to n, without going over?</h3>

The integer that has square root n is given by:

S(n) = n².

Hence the closest is given by:

C(n) = n² - 1.

In this problem, we have n = 7, hence:

C(7) = 7² - 1 = 48.

Which means that Alejandro was not correct.

More can be learned about square roots at brainly.com/question/27678604

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2 years ago

The outside temperature was 73°f at 1.p.m and decreases at a rate of 1.5°f each hour. What expression can be used to determine t

Nonamiya [84]

Your expression will be h = -1.5°x + 73°

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3 years ago

Find the area of each parallelogram. What is the relationship between the areas?

castortr0y [4]

Answer:

Area of parallelogram is given by:

$A = bh$

where b is the base and h is the height of parallelogram.

In parallelogram TQRS.

Coordinate of TQRS are;

T(8, 16), Q(4, 4), R(16, 4) and S(20, 16)

Coordinate of T'Q'R'S' are;

T'(2, 4), Q'(1, 1), R'(4, 1) and S'(5, 4)

Find the length of QR and PT:

Using distance(D) formula:

$D = \sqrt{(x_1-x_2)^2+(y_1-y_2)^2}$

$QR = \sqrt{(4-16)^2+(4-4)^2} =\sqrt{(-12)^2+0}= \sqrt{144}= 12$ units

Similarly;

For PT:

From the graph:

P(8, 4) and T(8, 16), then

$PT = \sqrt{(8-8)^2+(4-16)^2} =\sqrt{(-0)^2+(-12)^2}= \sqrt{144}= 12$ units

In parallelogram TQRS

PT represents the height and QR represents the base of the parallelogram respectively.

then;

Area of parallelogram TQRS = $QR \cdot PT$

⇒Area of parallelogram TQRS = $12 \cdot 12 = 144$ unit square.

Now, in parallelogram T'Q'R'S'

Q'R' represents the base and P'T' represents the height of the parallelogram respectively.

here, P'(2, 1)

Find the length of Q'R' and P'T':

$Q'R' = \sqrt{(1-4)^2+(1-1)^2} =\sqrt{(-3)^2+0}= \sqrt{9}= 3$ units

$P'T' = \sqrt{(2-2)^2+(1-4)^2} =\sqrt{(-0)^2+(-3)^2}= \sqrt{9}= 3$ units

Then;

Area of parallelogram T'Q'R'S' = $Q'R' \cdot P'T'$

Area of parallelogram T'Q'R'S' = $9 \cdot 9= 81$ unit square.

Now, we have to find the relationship between the areas.

$\frac{\text{Area of parallelogram TQRS}}{\text{Area of parallelogram T'Q'R'S'}} = \frac{144}{81}$

then;

the relationship between the areas of TQRS and T'Q'R'S' is:

$\text{Area of parallelogram TQRS} = \frac{144}{81} \cdot {\text{Area of parallelogram T'Q'R'S'}$

7 0

3 years ago

I need help with 8 and 9 it’s due at 8:00 am this morning. ??

-Dominant- [34]

Answer:

8 no answer is 0.53

(divide no of male going to college by total no of males).

9 no answer is r=-0.97

Step-by-step explanation:

6 0

3 years ago

SOMEONE HELP ME PLEASEEEEE

Sergeu [11.5K]

The answer is b if you do the equation right

5 0

3 years ago

Read 2 more answers