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3 years ago

Plz explain Math question

Mathematics

1 answer:

CaHeK987 [17]3 years ago

5 0

Answer:

subtract the amount of money ellen has by the amount of money she would have if she did it in her firends bank

Step-by-step explanation:

basically you do two intrest questions subtract and boom also...

GIVE ME BRAINLIEST AND HAVE A NICE DAY

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I need help with this math question

Mkey [24]

M2 = 127
m7= 127
m8= 53

5 0

2 years ago

Pls help me with my math

givi [52]

Answer:

The definition for the given piecewise-defined function is: $\boxed{\displaystyle\sf\ Option\:D:\:\: f(x) = \begin{cases}\displaystyle\sf\ x + 2 & \sf\:{if\:\:x \leq -1} \\\displaystyle\sf\ 2x + 4 & \sf\:{if\:\:x > -1}\end{cases}}$ .

Step-by-step explanation:

<h3>General Concepts:</h3>

Piecewise-defined functions.
Interval notations.

<h3>What is a piecewise-defined function?</h3>

A piecewise-defined function represents specific rules over different intervals of the domain.

<h3>Symbols used in expressing interval notations:</h3>

Open interval: This means that the endpoint is <em>not</em> included in the interval.

We can use the following symbols to indicate the <u>exclusion</u> of endpoints in the interval:

Left or right parenthesis, "( )" (or both).

Greater than (>) or less than (<) symbols.
Open dot " $\circ$ " is another way of expressing the exclusion of an endpoint in the graph of a piecewise-defined function.

Closed interval: This implies the inclusion of endpoints in the interval.

We can use the following symbols to indicate the <u>inclusion</u> of endpoints in the interval:

Open- or closed brackets (or both), "[ ]."

Greater than or equal to (≥) or less than or equal to (≤) symbols.
Closed circle or dot, "•" is another way of expressing the <em>inclusion</em> of the endpoint in the graph of a piecewise-defined function.

<h2>Determine the appropriate function rule that defines different parts of the domain. </h2>

The best way to determine which piecewise-defined function represents the graph is by observing the <u>endpoints</u> and <u>orientation</u> of both partial lines.

Open circle on (-1, 2): The graph shows that one of the partial lines has an <em>excluded</em> endpoint of (-1, 2) extending towards the <u>right</u>. This implies that its domain values are defined when x > -1.

Closed circle on (-1, 1): The graph shows that one of the partial lines has an <em>included</em> endpoint of (-1, 1) extended towards the <u>left</u>. Hence, its domain values are defined when x ≤ -1.

Based on our observations from the previous step, we can infer that x > -1 or x ≤ -1 apply to piecewise-defined functions A or D. However, only one of those two options represent the graph.

<h2>Solution:</h2><h3>a) Test option A:</h3>

$\boxed{\displaystyle\sf Option\:A)\:\:\:f(x) = \begin{cases}\displaystyle\sf\ 2x + 2 & \sf\:{if\:\:x \leq -1} \\\displaystyle\sf\ x + 4 & \sf\:{if\:\:x > -1}\end{cases}}$

<h3>Piece 1: If x ≤ -1, then it is defined by f(x) = 2x + 2. </h3>

We must choose a domain value that falls within the interval of x ≤ -1 whose output is included is included in the graph of the partial line with a <u>closed dot</u>.

Substitute x = -2 into f(x) = 2x + 2:

f(x) = 2x + 2

f(-2) = 2(-2) + 2
f(-2) = -4 + 2
f(-2) = -2 ⇒ <em>False statement</em>.

⇒ The output value of f(-2) = -2 is <u>not</u> included in the graph of the partial line whose endpoint is at (-1, 1).

<h3>Piece 2: If x > -1, then it is defined by f(x) = x + 4. </h3>

We must choose a domain value that falls within the interval of x > -1 whose output is included in the graph of the partial line with an <u>open dot</u>.

Substitute x = 0 into f(x) = x + 4:

f(x) = x + 4
f(0) = (0) + 4
f(0) = 4 ⇒ <em>True statement</em>.

⇒ The output value of f(0) = 4 <u>is</u> included in the graph of the partial line whose endpoint is at (-1, 2).

Conclusion for Option A:

Option A is not the correct piecewise-defined function because one of the pieces, f(x) = 2x + 2, does not specify the interval (-∞, -1].

<h3>b) Test option D:</h3>

$\boxed{\displaystyle\sf Option\:D)\:\:\:f(x) = \begin{cases}\displaystyle\sf\ x + 2 & \sf\:{if\:\:x \leq -1} \\\displaystyle\sf\ 2x + 4 & \sf\:{if\:\:x > -1}\end{cases}}$

<h3>Piece 1: If x ≤ -1, then it is defined by f(x) = x + 2. </h3>

We must choose a domain value that falls within the interval of x ≤ -1 whose output is included is included in the graph of the partial line with a <u>closed dot</u>.

Substitute x = -2 into f(x) = x + 2:

f(x) = x + 2
f(-2) = (-2) + 2
f(-2) = 0 ⇒ <em>True statement</em>.

⇒ The output value of f(-2) = 0 <u>is</u> included the graph of the partial line whose endpoint is at (-1, 1).

<h3>Piece 2: If x > -1, then it is defined by f(x) = 2x + 4.</h3>

We must choose a domain value that falls within the interval of x > -1 whose output is included is included in the graph of the partial line with an <u>open dot</u>.

Substitute x = 0 into f(x) = 2x + 4:

f(x) = 2x + 4
f(0) = 2(0) + 4
f(0) = 0 + 4 = 0 ⇒ <em>True statement</em>.

⇒ The output value of f(0) = 4 <u>is</u> included in the graph of the partial line whose endpoint is at (-1, 2).

<h2>Final Answer: </h2>

We can infer that the piecewise-defined function that represents the graph is:

$\boxed{\displaystyle\sf\ Option\:D:\:\: f(x) = \begin{cases}\displaystyle\sf\ x + 2 & \sf\:{if\:\:x \leq -1} \\\displaystyle\sf\ 2x + 4 & \sf\:{if\:\:x > -1}\end{cases}}$ .

________________________________________

Learn more about piecewise-defined functions here:

brainly.com/question/26145479

8 0

2 years ago

What is the trigonometric ratio for sin C?

Vesna [10]

<h3>Answer: 9/41</h3>

==========================================================

Explanation:

We have a triangle with these three sides.

a = 80
b = unknown
c = 82

Use the pythagorean theorem to find b

a^2+b^2 = c^2

b = sqrt(c^2 - a^2)

b = sqrt(82^2 - 80^2)

b = sqrt(324)

b = 18

This is the missing vertical leg of the triangle. And this is also the side opposite angle C.

We have enough information to compute the sine of the angle.

sin(angle) = opposite/hypotenuse

sin(C) = AB/AC

sin(C) = 18/82

sin(C) = (9*2)/(41*2)

sin(C) = 9/41

6 0

3 years ago

Solve the soultion .explian please<br> x(2x- 5)=0

AysviL [449]

$x(2x-5)=0\iff x=0\ \vee\ 2x-5=0\\\\x=0\ \vee\ 2x=5\\\\x=0\ \vee\ x=\frac{5}{2}\\\\x=0\ \vv\ x=2.5$

7 0

3 years ago

Read 2 more answers

Let g(x) = 2x and h(x)= x^2 -4. find (g o h)(0)

finlep [7]

The answer is -8

====================================================

Explanation:

There are two ways to get this answer

Method 1 will have us plug x = 0 into h(x) to get
h(x) = x^2 - 4
h(0) = 0^2 - 4
h(0) = 0 - 4
h(0) = -4
Then this output is plugged into g(x) to get
g(x) = 2x
g(-4) = 2*(-4)
g(-4) = -8 which is the answer
This works because (g o h)(0) is the same as g(h(0)). Note how h(0) is replaced with -4
So effectively g(h(0)) = -8 which is the same as (g o h)(0) = -8

-----------------------

The second method involves a bit algebra first
Start with the outer function g(x). Then replace every x with h(x). On the right side, we will replace h(x) with x^2-4 because h(x) = x^2-4

g(x) = 2x
g(x) = 2( x )
g(h(x)) = 2( h(x) ) ... replace every x with h(x)
g(h(x)) = 2( x^2-4 ) ... replace h(x) on the right side with x^2-4
g(h(x)) = 2x^2-8
(g o h)(x) = 2x^2-8

Now plug in x = 0
(g o h)(x) = 2x^2-8
(g o h)(0) = 2(0)^2-8
(g o h)(0) = 2(0)-8
(g o h)(0) = 0-8
(g o h)(0) = -8

Regardless of which method you use, the answer is -8

5 0

3 years ago