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3 years ago

The function fx) x2 +5x- 6 is shifted 4 units to the left to create g(x). What is g(x)?

Mathematics

2 answers:

artcher [175]3 years ago

7 0

It's g(x) = (x+4)^2 + 5 (x+4) - 6
I apologize if my answer is incorrect.

vesna_86 [32]3 years ago

4 0

The answer is g(x) = x^2 + 13x + 30

You can get this is multiple ways. Firstly, you could factor to find the zeroes and then move the answers to the left and redistribute the parenthesis.

You could also put the equation into vertex form and shift the number inside the parenthesis by 4.

Either will give you the exact same answer.

You might be interested in

List below are the speeds (mi/h) measured from southbound traffic on I-280 near Cupertino, CA. Use the sample data to construct

zheka24 [161]

Answer:

The 99% confidence interval of the population standard deviation is 1.7047 < σ < 7.485

Step-by-step explanation:

Confidence interval of standard deviation is given as follows;

$\sqrt{\dfrac{\left (n-1 \right )s^{2}}{\chi _{1-\alpha /2}^{}}}< \sigma < \sqrt{\dfrac{\left (n-1 \right )s^{2}}{\chi _{\alpha /2}^{}}}$

s = $\sqrt{\dfrac{\Sigma (x - \bar x)^2}{n - 1} }$

Where:

$\bar x$ = Sample mean

s = Sample standard deviation

n = Sample size = 7

χ = Chi squared value at the given confidence level

$\bar x$ = ∑x/n = (62 + 58 + 58 + 56 + 60 +53 + 58)/7 = 57.857

The sample standard deviation s = $\sqrt{\dfrac{\Sigma (x - \bar x)^2}{n - 1} }$ = 2.854

The test statistic, derived through computation, = ±3.707

Which gives;

$C. I. = 57.857 \pm 3.707 \times \dfrac{2.854}{\sqrt{7} }$

$\sqrt{\dfrac{\left (7-1 \right )2.854^{2}}{16.812}^{}}}< \sigma < \sqrt{\dfrac{\left (7-1 \right )2.854^{2}}{0.872}}$

1.7047 < σ < 7.485

The 99% confidence interval of the population standard deviation = 1.7047 < σ < 7.485.

7 0

2 years ago

Gwen, Tristan, and Keith like to work out at the gym. They each have an app on their phone that estimates the calories they burn

aleksandrvk [35]

Answer:

Part A:

Gwen has already burned 100 calories when Tristan and Keith started, and he continues to burn 10 calories per minute. The equation for Gwen is

y = 100 + 10x

Part B:

Tristan is burning 125 calories every 10 minutes. His burning calories rate is 12.5 calories per minute. The equation for Tristan is

y = 12.5x

Part C:

Keith is burning 300 calories in 30 minutes. His rate per minute is 10 calories per minute. The equation for Keith is

y = 10x

Part D:

Just follow the instructions and do the thing for the app. You can then answer the questions it is very easy.

Part E:

Do the same for E after you do Part D.

Step-by-step explanation:

Hope it helps!

6 0

2 years ago

Read 2 more answers

Find all solutions of the equation: 2cos^2x-cosx=1

Art [367]

If you're using the app, try seeing this answer through your browser: brainly.com/question/3166243

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Solve the trigonometric equation:

$\mathsf{2\,cos^2\,x-cos\,x=1}\\\\ \mathsf{2\,cos^2\,x-cos\,x-1=0}$

Make a substitution:

$\mathsf{cos\,x=t\qquad (-1\le t\le 1)}$

and the equation becomes

$\mathsf{2t^2-t-1=0}$

Rewrite conveniently – t as + t – 2t, and then factor the left-hand side by grouping:

$\mathsf{2t^2+t-2t-1=0}\\\\ \mathsf{t\cdot (2t+1)-1\cdot (2t+1)=0}$

Factor out 2t + 1:

$\mathsf{(2t+1)\cdot (t-1)=0}\\\\ \begin{array}{rcl} \mathsf{2t+1=0}&~\textsf{ or }~&\mathsf{t-1=0}\\\\ \mathsf{2t=1}&~\textsf{ or }~&\mathsf{t=1}\\\\ \mathsf{t=\dfrac{\,1\,}{2}}&~\textsf{ or }~&\mathsf{t=1} \end{array}$

Substitute back for t = cos x:

$\begin{array}{rcl}\mathsf{cos\,x=\dfrac{\,1\,}{2}}&~\textsf{ or }~&\mathsf{cos\,x=1}\\\\ \mathsf{cos\,x=cos\,60^\circ}&~\textsf{ or }~&\mathsf{cos\,x=cos\,0} \end{array}$

Therefore,

$\begin{array}{rcl} \mathsf{x=\pm\,60^\circ+k\cdot 360^\circ}&~\textsf{ or }~&\mathsf{cos\,x=0+k\cdot 360^\circ} \end{array}$

where k is an integer.

Solution set:

$\mathsf{S=\left\{x\in\mathbb{R}:~~x=-\,60^\circ+k\cdot 360^\circ~~or~~x=60^\circ+k\cdot 360^\circ~~or~~x=k\cdot 360^\circ,~~k\in\mathbb{Z}\right\}}$

I hope this helps. =)

3 0

2 years ago

30 years ago zoe was 2/3 as old as luke, 18 years ago zoe was 5/6 as old as luke how old are they now

barxatty [35]

Based on the mathematical statements, Zoe is 56 years old and Luke is 66 years old, now

<h3>How to determine how old they are now?</h3>

From the question, we have the following statements that can be used in our computation:

<u>30 years ago</u>

Zoe was 2/3 as old as Luke

<u>18 years ago</u>

Zoe was 5/6 as old as Luke

Let their present ages be represented as

Zoe = x

Luke = y

So, we have the following representations

<u>30 years ago</u>

Zoe was 2/3 as old as Luke

x - 36 = 2/3(y - 36)

<u>18 years ago</u>

Zoe was 5/6 as old as Luke

x - 18 = 5/6(y - 18)

So, we have the following system of equations

x - 36 = 2/3(y - 36)

x - 18 = 5/6(y - 18)

Make x the subject in x - 18 = 5/6(y - 18)

x = 5/6(y - 18) + 16

Substitute x = 5/6(y - 18) + 16 in x - 36 = 2/3(y - 36)

5/6(y - 18) + 16 - 36 = 2/3(y - 36)

Open the brackets

5/6y - 15 + 16 - 36 = 2/3y - 24

Evaluate the like terms

5/6y - 35 = 2/3y - 24

Multiply through by 6

5y - 210 = 4y - 144

Evaluate the like terms

y = 66

Substitute y = 66 in x = 5/6(y - 18) + 16

x = 5/6(66 - 18) + 16

Evaluate

x = 56

Recall that

Zoe = x

Luke = y

So, we have

Zoe = x = 56

Luke = y = 66

Hence, they are 56 and 66 years, now