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9 days ago

Jackie bought a 3-pack of tubes of glue. She used 0.27 ounce of the glue from the 3-pack

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Two functions are shown in the table below. Function 1 2 3 4 5 6 f(x) = −x2 + 4x + 12 g(x) = −x + 6 Complete the table on your o

Svet_ta [12734]

For $\fbox{\begin \\\math{x}=6\\\end{minispace}}$ the function $f(x)=-x^{2} +4x+12$ and $g(x)=-x+6$ both yield the same result.

Detailed breakdown:

The functions involved are

$f(x)=-x^{2}+4x+12$

$g(x)=-x+6$

Step 1:

Insert $x=1$ in $f(x)=-x^{2} +4x+12$ to find the value of $f(1)$ .

$f(1)=-1^{2} +4(1)+12\\f(1)=-1+4+12\\f(1)=15$

Insert $x=1$ in $g(x)=-x+6$ to find the value of $g(1)$ .

$g(1)=-1+6\\g(1)=5$

Step 2:

Insert $x=2$ in $f(x)=-x^{2} +4x+12$ to obtain the value of $f(2)$ .

$f(2)=-2^{2} +4(2)+12\\f(2)=-4+8+12\\f(2)=16$

Substitute $x=2$ into $g(x)=-x+6$ to find the value of $g(2)$ .

$g(2)=-2+6\\g(2)=4$

Step 3:

Replace $x=3$ in $f(x)=-x^{2} +4x+12$ to find the value of $f(3)$ .

$f(3)=-3^{2} +4(3)+12\\f(3)=-9+12+12\\f(3)=15$

Also, replace $x=3$ in $g(x)=-x+6$ to find the value of $g(3)$ .

$g(3)=-3+6\\g(3)=3$

Step 4:

Insert $x=4$ in $f(x)=-x^{2} +4x+12$ to find the value of $f(4)$ .

$f(4)=-4^{2} +4(4)+12\\f(4)=-16+16+12\\f(4)=12$

Also, replace $x=4$ in $g(x)=-x+6$ to obtain the value of $g(4)$ .

$g(4)=-4+6\\g(4)=2$

Step 5:

Insert $x=5$ in $f(x)=-x^{2} +4x+12$ to obtain the value of $f(5)$ .

$f(5)=-5^{2} +4(5)+12\\f(5)=-25+20+12\\f(5)=7$

Replace $x=5$ in $g(x)=-x+6$ to find the value of $g(5)$ .

$g(5)=-5+6\\g(5)=1$

Step 6:

Insert $x=6$ into $f(x)=-x^{2} +4x+12$ to find the value of $f(6)$ .

$f(6)=-6^{2} +4(6)+12\\f(6)=-36+24+12\\f(6)=0$

Also, substitute $x=6$ in $g(x)=-x+6$ to obtain the value of $g(6)$ .

$g(6)=-6+6\\g(6)=0$

Step 7:

According to the provided condition $f(x)=g(x)$ .

(a). Insert $f(x)=-x^{2} +4x+12$ and $g(x)=-x+6$ into the previously mentioned equation.

$-x^{2} +4x+12=-x+6$

(b). Multiply through by $-1$ on both sides.

$x^{2} -4x-12=x-6$

(c). Move the term $x-6$ to the left side of the equation.

$x^{2} -4x-12-x+6=0\\x^{2} -5x-6=0$

(d). Divide the middle term so that its sum equals 5 and the product equals 6.

$x^{2} -(6-1)x-6=0\\x^{2} -6x+x-6=0\\x(x-6)+1(x-6)=0\\(x+1)(x-6)=0\\x=-1,6$

From the analysis above, it is noted that for $x=6$ both functions $f(x)$ and $g(x)$ yield the same outcome.

Using a direct approach:

$f(x)=g(x)\\\Leftrightarrow-x^{2} +4x+12=-x+6\\\Leftrightarrow-x^{2} +4x+12+x-6=0\\\Leftrightarrow-x^{2} +5x+6=0\\\Leftrightarrow-x^{2} +6x-x+6=0\\\Leftrightarrow x^{2} -6x+x-6=0\\\Leftrightarrow x(x-6)+1(x-6)=0\\\Leftrightarrow(x+1)(x-6)=0\\\Leftrightarrow x=6,-1$

The table representing function $f(x)=-x^{2} +4x+12$ and $g(x)=-x+6$ is included below.

For more information:

1. What is the y-intercept of the quadratic function f(x) = (x – 6)(x – 2)? (0,–6) (0,12) (–8,0) (2,0)

2. Which is the graph of f(x) = (x – 1)(x + 4)?

6 0

1 month ago

The glee club has $90 to spend on pens and pencils. Each pen costs $0.75 and each pencil costs $0.15. Let x represent the number

Zina [12379]

Answer:

[[IMG_1]]

The relevant equation is: b. 0.75x + 0.15y = 90

8 0

2 months ago

Read 2 more answers

The distribution of the amount of money spent by students for textbooks in a semester is approximately normal in shape with a me

babunello [11817]

The last part of the question is asking;

What was the total amount spent by all (99.7%) students on textbooks in a semester?

Response:

almost all (99.7%) of the students spent between $165 and $315 on textbooks in a semester.

Stepwise clarification:

The standard deviation rule informs that for normally distributed data, approximately 99.7% of observations fall within three standard deviations from the mean.

In this case, we have the given mean as 240, and standard deviation as 25

Thus, calculating three standard deviations below the mean: Mean - 3(standard deviation)

equals 240 - (3 × 25)

yielding 240 - 75 = 165

Now, for three standard deviations above the mean: Mean + 3 (standard deviation) = 240 + (3 × 25)

equals 240 + 75 = 315

Therefore, nearly all (99.7%) of the students spent between $165 and $315 on textbooks in a semester.

7 0

1 month ago

There is a mound of g pounds of gravel in a quarry. Throughout the day, 400 pounds of gravel are added to the mound. Two orders

PIT_PIT [12445]

Question is Incomplete; Complete question is given below;

A pile of g pounds of gravel exists in a quarry. During the day, 400 pounds of gravel is added to the pile. Two orders of 900 pounds are removed, resulting in the mound holding 1,500 pounds of gravel by day's end. Formulate the equation that accurately illustrates this scenario.

Answer:

The equation that best represents the situation is $g+400-1800=1300$ .

Step-by-step explanation:

Given:

Initial gravel amount = ' $g$ '.

Currently Given:

Throughout the day, 400 pounds of gravel is added to the pile.

Thus we can conclude;

Gravel amount after addition = $g+400$

Also given:

Two orders of 900 pounds are sold and the gravel is taken from the pile.

Hence, two orders of 900 pounds = $2\times 900 = 1800 \ pounds$

This leads us to state;

Gravel amount post-removal = $g+400-1800$

Now Given:

Remaining gravel at the end of day = 1300 pounds.

Therefore, we derive that;

$g+400-1800=1300$

Consequently, the equation that best describes the scenario is $g+400-1800=1300$ .

3 0

1 month ago

What is the cosine ratio for angle F?

babunello [11817]

Response:

option D

Step-by-step reasoning:

Provided in the prompt is a right-angled triangle FGH.

Base of triangle = 12

Height of triangle = 5

Hypotenuse of triangle = 13

To solve the question, we'll apply trigonometric identities.

cos(F) = adjacent / hypotenuse

cos(F) = height / hypotenuse

cos(F) = 5 / 13

Thus, the cosine ratio for angle F is 5 / 13

6 0

1 month ago