Ask question

Ask question

All categories

3 years ago

5

What is the range of this relation?

1 answer:

riadik2000 [5.3K]3 years ago

5 0

The range of the relation is 35,40,45,60,75
Reason being in relation range is the set of all second elements of ordered pairs (y-coordinates).

You might be interested in

A scuba diver is 314 feet below the surface of the water. The angle of depression the diver makes with her boat is 39

dezoksy [38]

Answer:

hi how are you doing today Jasmine

8 0

3 years ago

The exponential function f(x) = 3(5)x grows by a factor of 25 between x = 1 and x = 3. What factor does it grow by between x = 5

notsponge [240]

<h2> Question # 1</h2>

Answer:

25 is the factor which grows by between x = 5 and x = 7.

Step-by-step explanation:

Considering the exponential function

$f\left(x\right)\:=\:3\left(5\right)^x$

The growth factor between $x=1$ and $x=3$ is given by:

$\left[3\left(5\right)^3\right]\div \left[3\left(5\right)^1\right]$

$\mathrm{Calculate\:within\:parentheses}\:\left[3\left(5\right)^3\right]\::\quad 375$

$\mathrm{Calculate\:within\:parentheses}\:\left[3\left(5\right)^1\right]\::\quad 15$

So,

= $375\div \:15$

= $25$

Similarly, growth factor between $x=5$ and $x=7$ is given by:

$\left[3\left(5\right)^7\right]\div \left[3\left(5\right)^5\right]$

= $\frac{3\cdot \:5^7}{3\cdot \:5^5}$

$\mathrm{Divide\:the\:numbers:}\:\frac{3}{3}=1$

= $\frac{5^7}{5^5}$

$\mathrm{Apply\:exponent\:rule}:\quad \frac{x^a}{x^b}=x^{a-b}$

$\frac{5^7}{5^5}=5^{7-5}$

= $5^{7-5}$

= $5^2$

= $25$

Therefore, 25 is the factor which grows by between x = 5 and x = 7.

<h2> Question # 2</h2>

Answer:

The population be in 24 years will be 27000.

Also, the population growth modeled by an exponential function as $y=A\cdot \left(b\right)^t$ is an exponential function.

The graph for $y=A\cdot \left(b\right)^t$ is also shown in attached figure.

Step-by-step explanation:

If a city that currently has a population of 1000 triples in size every 8 years.
what will the population be in 24 years?
Is the population growth modeled by a linear function or an exponential function?

As the city that currently has a population of 1000 triples in size every 8 years.

So, for this case

$y=A\cdot \left(b\right)^t$

where

$A$ = Initial population amount

$b$ = growth rate

$t$ = time

Substituting the values in the function

$y=A\cdot \left(b\right)^t$

$y=1000\cdot \:\:3^{\frac{1}{8}t}$

So, the population be in 24 years

$y=1000\cdot \:\:3^{\frac{1}{8}24}$

As

$3^{\frac{1}{8}\cdot \:24}=3^3$

So

$\:y=3^3\cdot 1000$

$y=1000\cdot \:\:27$

$y=27000$

Therefore, the population be in 24 years will be 27000.

Also, the population growth modeled by an exponential function as $y=A\cdot \left(b\right)^t$ is an exponential function.

<h2 /><h2> Question # 3</h2>

Answer:

the graph of the function will translate horizontally 3/5 units right.

Step-by-step explanation:

We have to find the effect on the graph of the function $f(x)=2x$ when it is replaced by f(x- 3/5).

We already have an idea that rule for horizontal translation:

Given a function $f(x)$ , and a constant c > 0, the function $g(x) = f(x - a)$ represents a horizontal shift c units to the right from f(x). The function $h(x) = f(x + a)$ represents a horizontal shift c units to the left.

As 3/5 > 0, so the graph of the function will translate horizontally 3/5 units right.

Therefore, the graph of the function will translate horizontally 3/5 units right.

Keywords: exponential function, translation function, growth factor

Learn more about exponential function and growth factor form brainly.com/question/10147339

#learnwithBrainly

6 0

3 years ago

Help I need it asap

enyata [817]

Answer:

for no 7 its none for 8 its st

Step-by-step explanation:

I hope this helps

8 0

3 years ago

If 1st and 4tg terms of G.p are 500 and 32 respectively it's second term is ?

bixtya [17]

Answer:

$T_{2} = 200$

Step-by-step explanation:

Given

Geometry Progression

$T_1 = 500$

$T_4 = 32$

Required

Calculate the second term

First, we need to write out the formula to calculate the nth term of a GP

$T_n = ar^{n-1}$

For first term: Tn = 500 and n = 1

$500 = ar^{1-1}$

$500 = ar^{0}$

$500 = a$

$a = 500$

For fought term: Tn = 32 and n = 4

$32 = ar^{4-1}$

$32 = ar^3$

Substitute 500 for a

$32 = 500 * r^3$

Make r^3 the subject

$r^3 = \frac{32}{500}$

$r^3 = 0.064$

Take cube roots

$\sqrt[3]{r^3} = \sqrt[3]{0.064}$

$r = \sqrt[3]{0.064}$

$r = 0.4$

Using: $T_n = ar^{n-1}$

$n = 2$ $r = 0.4$ and $a = 500$

$T_{2} = 500 * 0.4^{2-1}$

$T_{2} = 500 * 0.4^1$

$T_{2} = 500 * 0.4$

$T_{2} = 200$

<em>Hence, the second term is 200</em>

5 0

3 years ago

Help me on this please

Rina8888 [55]

Answer:

x = 3.5

Step-by-step explanation:

Triangle to the right:

4^2 + x^2 = 8^2

16 + x^2 = 64

y^2 = 48

Triangle to the left:

x^2 + 6^2 = 48

x^2 + 36 = 48

x^2 = 12

x = sqrt(12)

x = 3.5

4 0

3 years ago