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2 years ago

What is the domain of the function {(-1,-1), (0, 1), (2, -1)}?

Mathematics

2 answers:

DochEvi [55]2 years ago

8 0

Answer:

The domain would be {-1,0,2}.

Step-by-step explanation:

7nadin3 [17]2 years ago

7 0

I think that the answer is {-1,0,2}

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Nisha collects comic books, and she convinced her friend Hakim to start collecting as well. Every week, they go to the store tog

andreyandreev [35.5K]

By using the given table, we will get the linear equation:

h = n - 9

<h3>How to find the equation for Hakim?</h3>

We know that they always get comic books together, so always that Nisha's collection increases by one, Hakim's collection will also increase by one, so we will have a linear equation.

Now, if you look at the given table, you can see that Hakim's collection is always 9 units less than Nisha's collection, then the linear equation that models the relation between the two collections, h and n, is really straightforward, it will be:

Hakim's collection = Nisha's collection - 9

Using the variables, we get:

h = n - 9

That is the linear equation Hakim wanted.

If you want to learn more about linear equations:

brainly.com/question/1884491

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10 months ago

PLZ HELP I PROMISE I WILL GIVE BRAINLIEST!!!! ASAP

irga5000 [103]

Answer:

$4c^{7}d^{13}$

Step-by-step explanation:

$(2cd^{4} )^{2} *(cd)^{5}$

$4c^{2}d^{8} *c^{5} d^{5}$

$4c^{7}d^{13}$

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2 years ago

A juggler tosses a ball into the air . The balls height, h and time t seconds can be represented by the equation h(t)= -16t^2+40

malfutka [58]

PART A

The given equation is

$h(t) = - 16 {t}^{2} + 40t + 4$

In order to find the maximum height, we write the function in the vertex form.

We factor -16 out of the first two terms to get,

$h(t) = - 16 ({t}^{2} - \frac{5}{2} t) + 4$

We add and subtract

$- 16(- \frac{5}{4} )^{2}$

to get,

$h(t) = - 16 ({t}^{2} - \frac{5}{2} t) + - 16( - \frac{5}{4})^{2} - -16( - \frac{5}{4})^{2} + 4$

We again factor -16 out of the first two terms to get,

$h(t) = - 16 ({t}^{2} - \frac{5}{2} t + ( - \frac{5}{4})^{2} ) - -16( - \frac{5}{4})^{2} + 4$

This implies that,

$h(t) = - 16 ({t}^{2} - \frac{5}{2} t + ( - \frac{5}{4}) ^{2} ) + 16( \frac{25}{16}) + 4$

The quadratic trinomial above is a perfect square.

$h(t) = - 16 ( t- \frac{5}{4}) ^{2} +25+ 4$

This finally simplifies to,

$h(t) = - 16 ( t- \frac{5}{4}) ^{2} +29$

The vertex of this function is

$V( \frac{5}{4} ,29)$

The y-value of the vertex is the maximum value.

Therefore the maximum value is,

$29$

PART B

When the ball hits the ground,

$h(t) = 0$

This implies that,

$- 16 ( t- \frac{5}{4}) ^{2} +29 = 0$

We add -29 to both sides to get,

$- 16 ( t- \frac{5}{4}) ^{2} = - 29$

This implies that,

$( t- \frac{5}{4}) ^{2} = \frac{29}{16}$

$t- \frac{5}{4} = \pm \sqrt{ \frac{29}{16} }$

$t = \frac{5}{4} \pm \frac{ \sqrt{29} }{4}$

$t = \frac{ 5 + \sqrt{29} }{4} = 2.60$

or

$t = \frac{ 5 - \sqrt{29} }{4} = - 0.10$

Since time cannot be negative, we discard the negative value and pick,

$t = 2.60s$