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3 years ago

The graph of the function f(x) = (x + 2)(x + 6) is shown

Mathematics

1 answer:

Vitek1552 [10]3 years ago

7 0

Answer:

The function is negative for all real values of x where

-6 < x < -2.

Step-by-step explanation:

f(x) = (x +2)(x + 6)

x + 2 = 0 or x + 6 = 0

x = -2 or x = -6

After graphing the function is negative between -6 < x < -2

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Which quantity is proportional to 16⁄4?

Aleks [24]

4.

This is because 4 goes into 16 4 times, hence 4 x 4 being 16.

Therefore, the answer would be four.

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4 years ago

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Two shops sell the same brand of baked beans but in different sizes of tin.

serg [7]

Answer:

Price in shop B is cheaper by 12 p per kg

Step-by-step explanation:

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3 years ago

Find the ordered pairs for the x- and y-intercepts of the equation 3x − 2y = 18 and select the appropriate option below.

olasank [31]

<h2>Greetings!</h2>

Answer:

Step-by-step explanation:

Y intercept:

Simply substitute all the x values with 0:

When x = 0:

3(0) - 2y = 18

Move the - 2y over to the other side making it a +2y:

0 = 18 + 2y

Move the +18 over to the other side making it a -18:

-18 = 2y

Divide both sides by 2:

$\frac{-18}{2} = \frac{2y}{2}$

-9 = y

So y intercept is:

X - intercept:

Simply substitute all the Y's with 0:

3x - 3(0) = 18

3x = 18

Divide both sides by 3:

$\frac{3x}{3} = \frac{18}{3}$

x = 6

So the X intercept is:

This means that your guess of B is correct.

<h2>Hope this helps! </h2>

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3 years ago

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Let ρ = x3 + xe−x for x ∈ (0, 1), compute the center of mass.

hram777 [196]

The center of mass is mathematically given as

$\bar{x}=\left(\frac{44 e-100}{25 e-40}\right)\end{aligned}$

<h3>What is the center of mass.?</h3>

Determine the center of mass in one dimension:

Represent the masses at the respective distances.

$\begin{|c|c|} Masses \ & \ \ \ \ \ \ \ \ \ \ \ \ \ \ Located at \\\rho=x^{3}+x \cdot e^{-x} & \ \ \ \ x \in(0,1)$ \\\end$

We calculate the total mass of the system.

$\begin{aligned}m &=\int_{0}^{1} \rho \cdot d x \\& m =\int_{0}^{1}\left(x^{3}+x \cdot e^{-x}\right) \cdot d x \\&m =\left|\frac{x^{4}}{4}-(x+1) e^{-x}\right|_{0}^{1} \\&m =\left(\frac{5}{4}-\frac{2}{e}\right)\end{aligned}$

Step 03: Calculate the moment of the system.

$\begin{aligned}M &=\int_{0}^{1}(\rho \cdot x) \cdot d x \\& M=\int_{0}^{1}\left(x^{4}+x^{2} \cdot e^{-x}\right) \cdot d x \\&M =\left|\frac{x^{5}}{5}-\left(x^{2}-2 x+2\right) \cdot e^{-x}\right|_{0}^{1} \\&M=\left(\frac{11}{5}-\frac{5}{e}\right)\end{aligned}$

we calculate the center of mass.

$\begin{aligned}\bar{x} &=\left(\frac{M}{m}\right) \\& \bar{x}=\left\{\left(\frac{\left.11-\frac{5}{5}\right)}{\left(\frac{5}{4}-\frac{2}{e}\right)}\right\}\right.\\& \bar{x}=\left(\frac{11 e-25}{5 e}\right) \cdot\left(\frac{4 e}{5 e-8}\right) \\&\bar{x}=\left(\frac{44 e-100}{25 e-40}\right)\end{aligned}$