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3 years ago

Does this graph represent a function? khan academy

Mathematics

2 answers:

Orlov [11]3 years ago

7 0

Answer:

Step-by-step explanation:

It isn't a positive linear association nor a negative linear association. If you were to connect those lines, it would not be a linear function of any kind of association.

Diano4ka-milaya [45]3 years ago

5 0

Answer:

yes

Step-by-step explanation:

its a graph

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19. The population of Jose's town in 1995 was 2400 and the population in 2000 was

Fantom [35]

The linear equation in slope intercept form for population of Jose's town is y = 320x + 2400.

<u>Solution:</u>

Given, the population of Jose's town in 1995 was 2400 and the population in 2000 was 4000,

Let x represent the number of years since 1995.

We have to write a linear equation, in slope- intercept form, that represents this data.

Now, let the change in population per year be p.

Then, population in a year = population change per year $\times$ number of years + population in 1995.

⇒population in a year = p $\times$ x + population in 1995

⇒ population in a year = p $\times$ x + 2400

Here we have an case that, population in 2000 is 4000

Then, number of years since 1995 = 5

So, 4000 = p $\times$ 5 + 2400 ⇒ 5p = 4000 – 2400 ⇒ 5p = 1600 ⇒ p = 320

Then, our equation will be population in a year = 320x + 2400

Considering the population in a year as y ⇒ y = 320x + 2400.

Hence, the linear equation in slope intercept form is y = 320x + 2400.

5 0

3 years ago

Which is the inverse of the function a(d)=5d-3? And use the definition of inverse functions to prove a(d) and a-1(d) are inverse

Drupady [299]

Answer:

$a'(d) = \frac{d}{5} + \frac{3}{5}$

$a(a'(d)) = a'(a(d)) = d$

Step-by-step explanation:

Given

$a(d) = 5d - 3$

Solving (a): Write as inverse function

$a(d) = 5d - 3$

Represent a(d) as y

$y = 5d - 3$

Swap positions of d and y

$d = 5y - 3$

Make y the subject

$5y = d + 3$

$y = \frac{d}{5} + \frac{3}{5}$

Replace y with a'(d)

$a'(d) = \frac{d}{5} + \frac{3}{5}$

Prove that a(d) and a'(d) are inverse functions

$a'(d) = \frac{d}{5} + \frac{3}{5}$ and $a(d) = 5d - 3$

To do this, we prove that:

$a(a'(d)) = a'(a(d)) = d$

Solving for $a(a'(d))$

$a(a'(d)) = a(\frac{d}{5} + \frac{3}{5})$

Substitute $\frac{d}{5} + \frac{3}{5}$ for d in $a(d) = 5d - 3$

$a(a'(d)) = 5(\frac{d}{5} + \frac{3}{5}) - 3$

$a(a'(d)) = \frac{5d}{5} + \frac{15}{5} - 3$

$a(a'(d)) = d + 3 - 3$

$a(a'(d)) = d$

Solving for: $a'(a(d))$

$a'(a(d)) = a'(5d - 3)$

Substitute 5d - 3 for d in $a'(d) = \frac{d}{5} + \frac{3}{5}$

$a'(a(d)) = \frac{5d - 3}{5} + \frac{3}{5}$

Add fractions

$a'(a(d)) = \frac{5d - 3+3}{5}$

$a'(a(d)) = \frac{5d}{5}$

$a'(a(d)) = d$

Hence:

$a(a'(d)) = a'(a(d)) = d$

7 0

2 years ago

15 = x + 4<br> whats the answer

stealth61 [152]

Answer:

Step-by-step explanation:

15-4 = x

15-4 = 11

x = 11

to isolate x, you move 4 to the other side and it becomes 15-4 = x

4 0

3 years ago

The area of the shaded region is 20 cm. Find the value of x

dolphi86 [110]

If you could put more info ab this then i would understand what u want me to put

3 0

2 years ago

Solve the equation. \dfrac18 + c = \dfrac45 8 1 +c= 5 4 start fraction, 1, divided by, 8, end fraction, plus, c, equals, sta

aivan3 [116]

Answer:

$c=\dfrac{27}{40}\\$

Step-by-step explanation:

We are required to solve for c in the equation: $\dfrac{1}{8}+c= \dfrac{4}{5}$

Step 1: Collect like terms

$\dfrac{1}{8}+c= \dfrac{4}{5}\\c= \dfrac{4}{5}-\dfrac{1}{8}$

Step 2: Find the Lowest Common Multiple of the denominators

LCM of 8 and 5 is 40

Step 3: Multiply all through by 40

$40c= \dfrac{4}{5}*40-\dfrac{1}{8}*40$

Step 4: Simplify

40c=32-5

40c=27

Step 5: Divide both sides by 40 and simplify if possible.

$c=\dfrac{27}{40}\\$

4 0

3 years ago