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3 years ago

Plz help... Which set of ordered pairs represents y as a function of x?

1 answer:

6 0

It is a cuz to be a function the x (domain) can only have one y (range) in d it is the number 2 in the 2nd and fourth set of ordered pairs in c it is 3 the first one and the 3rd in b it is -1 its the first and the last one
hope i helped you if u dont mind can i have brainlest <span />

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Help me please !!!!!!!

OLEGan [10]

Answer:

Step-by-step explanation:

because yeah

6 0

3 years ago

What is the lcm of 32 and 45

Talja [164]

1440 will be your least common multiple

7 0

3 years ago

Read 2 more answers

The supplement of an angle is three times the measure of the angle.

stira [4]

Answer: The angle equals 45 ∘ and the supplement is 135 ∘

Explanation:

Since the supplement is three times the angle, we can say s = 3 a

Since we know the supplement is

180 − a , we can plug that in.

180 - a = 3a

180 = 4 a (add a to both sides)

45 = a (divide both sides by 4)

Since we know the angle now, all we have to do is multiply it times 3 to find the supplement.

45 × 3 = 135

4 0

3 years ago

Ok find the value of two numbers if their sum is 70 and their difference is 36 what the answer

cluponka [151]

58 and 22 is the answer.

8 0

3 years ago

Let $f(x) = x^2$ and $g(x) = \sqrt{x}$. Find the area bounded by $f(x)$ and $g(x).$

Anna [14]

Answer:

$\large\boxed{1\dfrac{1}{3}\ u^2}$

Step-by-step explanation:

Let's sketch graphs of functions f(x) and g(x) on one coordinate system (attachment).

Let's calculate the common points:

$x^2=\sqrt{x}\qquad\text{square of both sides}\\\\(x^2)^2=\left(\sqrt{x}\right)^2\\\\x^4=x\qquad\text{subtract}\ x\ \text{from both sides}\\\\x^4-x=0\qquad\text{distribute}\\\\x(x^3-1)=0\iff x=0\ \vee\ x^3-1=0\\\\x^3-1=0\qquad\text{add 1 to both sides}\\\\x^3=1\to x=\sqrt[3]1\to x=1$

The area to be calculated is the area in the interval [0, 1] bounded by the graph g(x) and the axis x minus the area bounded by the graph f(x) and the axis x.

We have integrals:

$\int\limits_{0}^1(\sqrt{x})dx-\int\limits_{0}^1(x^2)dx=(*)\\\\\int(\sqrt{x})dx=\int\left(x^\frac{1}{2}\right)dx=\dfrac{2}{3}x^\frac{3}{2}=\dfrac{2x\sqrt{x}}{3}\\\\\int(x^2)dx=\dfrac{1}{3}x^3\\\\(*)=\left(\dfrac{2x\sqrt{x}}{2}\right]^1_0-\left(\dfrac{1}{3}x^3\right]^1_0=\dfrac{2(1)\sqrt{1}}{2}-\dfrac{2(0)\sqrt{0}}{2}-\left(\dfrac{1}{3}(1)^3-\dfrac{1}{3}(0)^3\right)\\\\=\dfrac{2(1)(1)}{2}-\dfrac{2(0)(0)}{2}-\dfrac{1}{3}(1)}+\dfrac{1}{3}(0)=2-0-\dfrac{1}{3}+0=1\dfrac{1}{3}$

6 0

3 years ago