find the product of: (4x-4y)(2x+y)

the correct answer is option (b) 3x² - 7.

The coefficient is the numerical factor in front of the variable. Let's calculate the coefficient for each option:

a) The coefficient of x in 2x is 2, not 3.

b) The coefficient of x² in 3x² is 3, which matches our requirement.

c) The coefficient of x in 4x is 4, not 3.

d) There is no x term in x³ + 2, so the coefficient for x is 0.

Therefore, the correct answer is option (b) 3x² - 7.

In option (b), the term with the coefficient of 3 is 3x². The coefficient of this term is indeed 3 because it's the number multiplied by x squared. The other terms in the expression do not have a coefficient of 3.

In options (a) and (c), the coefficients for the x terms are 2 and 4, respectively, which do not match the requirement.

Option (d) doesn't have an x term; it only has x raised to the power of 3 (x³), and the coefficient for this term is 0.

So, by process of elimination and calculation, we determine that option (b) is the correct answer with the term 3x² having a coefficient of 3.

complete question

Which algebraic expression has a term with a coefficient of 3?

a) 2x + 5  

b) 3x² - 7  

c) 4x - 3  

d) x³ + 2

The value of x is 7 .

Hence the correct option B .

Given:  " x³ = 343  

Solve for x,

Rewriting x³ = 343

∛(343) = 7

7³  =  7 × 7 × 7 = 343 .

7³  =  7 × 7 × 7

= 49 * 7

= 343

Therefore correct option is B .

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we know that

the perimeter of a polygon is the sum of the length sides

in this problem we have five vertices

so

the polygon has five sides

Let

[tex]A(-1,3)\\B(-1,6)\\C(2,10)\\D(5,6)\\E(5,3)[/tex]

the perimeter is equal to

[tex]P=AB+BC+CD+DE+AE[/tex]

The formula to calculate the distance between two points is equal to

[tex]d=\sqrt{(y2-y1)^{2}+(x2-x1)^{2}}[/tex]

Step 1

Find the distance AB

[tex]A(-1,3)\\B(-1,6)[/tex]

substitute the values in the formula

[tex]d=\sqrt{(6-3)^{2}+(-1+1)^{2}}[/tex]

[tex]d=\sqrt{(3)^{2}+(0)^{2}}[/tex]

[tex]dAB=3\ units[/tex]

Step 2

Find the distance BC

[tex]B(-1,6)\\C(2,10)[/tex]

substitute the values in the formula

[tex]d=\sqrt{(10-6)^{2}+(2+1)^{2}}[/tex]

[tex]d=\sqrt{(4)^{2}+(3)^{2}}[/tex]

[tex]dBC=5\ units[/tex]

Step 3

Find the distance CD

[tex]C(2,10)\\D(5,6)[/tex]

substitute the values in the formula

[tex]d=\sqrt{(6-10)^{2}+(5-2)^{2}}[/tex]

[tex]d=\sqrt{(-4)^{2}+(3)^{2}}[/tex]

[tex]dCD=5\ units[/tex]

Step 4

Find the distance DE

[tex]D(5,6)\\E(5,3)[/tex]

substitute the values in the formula

[tex]d=\sqrt{(3-6)^{2}+(5-5)^{2}}[/tex]

[tex]d=\sqrt{(-3)^{2}+(0)^{2}}[/tex]

[tex]dDE=3\ units[/tex]

Step 5

Find the distance AE

[tex]A(-1,3)\\E(5,3)[/tex]

substitute the values in the formula

[tex]d=\sqrt{(3-3)^{2}+(5+1)^{2}}[/tex]

[tex]d=\sqrt{(0)^{2}+(6)^{2}}[/tex]

[tex]dAE=6\ units[/tex]

Step 6

Find the perimeter

the perimeter is equal to

[tex]P=AB+BC+CD+DE+AE[/tex]

substitute the values

[tex]P=3+5+5+3+6=22\ units[/tex]

therefore

the answer is

the perimeter of the polygon is [tex]22\ units[/tex]

The problem asks to solve a system of two equations by representing each condition as an equation. By simplifying and substituting, we can find the values of the two unknowns.

The question involves solving a system of two equations. Let's denote the first number as x and the second number as y. The given conditions can be represented as follows:

To simplify, we can rewrite the equations as:

We can substitute equation (1) into equation (2) to get:

x = 2(x - 18) + 32

Solving the equation will give us the value of x and substituting x into equation (1), we will get the value of y.

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Answer: -8

Step-by-step explanation: Substitute the value of the variable into the expression and then simplify you should get the answer -8

BC= (-4)(2)=-8

The coefficient of kinetic friction is 0.58

To determine the coefficient of kinetic friction [tex](\(\mu_k\))[/tex], we can use the following approach:

1. Identify the horizontal component of the pulling force [tex](\(F_{\text{pull, horizontal}}\))[/tex] and set it equal to the force of kinetic friction [tex](\(f_k\))[/tex].

2. The equation for the horizontal component of the pulling force is:

[tex]\[ F_{\text{pull, horizontal}} = F_{\text{pull}} \cdot \cos(30^\circ) \][/tex]

3. The equation for the force of kinetic friction is:

[tex]\[ f_k = \mu_k \cdot N \][/tex]

4. Since the sled is on a horizontal surface, the normal force [tex](\(N\))[/tex] is equal to the weight of the sled [tex](\(m \cdot g\))[/tex], where m is the mass of the sled and g is the acceleration due to gravity.

Putting these equations together and setting them equal to each other because the sled is moving at constant velocity, we get:

[tex]\[ F_{\text{pull}} \cdot \cos(30^\circ) = \mu_k \cdot m \cdot g \][/tex]

Now, substitute the given values:

[tex]\[ 80 \cdot \cos(30^\circ) = \mu_k \cdot 20 \cdot 9.8 \][/tex]

Solve for [tex]\(\mu_k\)[/tex]:

 [tex]\[ \mu_k = \frac{80 \cdot \cos(30^\circ)}{20 \cdot 9.8} \][/tex]

Calculating this expression yields:

[tex]\[ \mu_k \approx 0.58 \][/tex]

So, the calculated coefficient of kinetic friction is approximately 0.58, which matches the provided answer.

Complete question:

A 20 kg sled is being pulled across a horizontal force at a constant velocity the pulling force has a magnitude of 80 N and it's directed at an angle of 30° above the horizontal determine the coefficient of kinetic friction.

Answer:

Formula that gives coordinates of the mid point of the segment connecting is [tex](\frac{a+c}{2},\frac{b+d}{2})[/tex]

Step-by-step explanation:

Given point are ( a , b ) and ( c , d )

To find : Coordinates of the mid point of the segment connecting given points.

We know that if point are [tex](x_1,y_1)\:\:and\:\:(x_2,y_2)[/tex]

then coordinate of the midpoints given by [tex](\frac{x_1+x_2}{2},\frac{y_1+y_2}{2})[/tex]

So, Coordinates of the given points = [tex](\frac{a+c}{2},\frac{b+d}{2})[/tex]

Therefore, Formula that gives coordinates of the mid point of the segment connecting is [tex](\frac{a+c}{2},\frac{b+d}{2})[/tex]

Algebra tiles would not be a good tool to use to factor the trinomial because you would need to drag an x-squared tile, 18 x-tiles, and 80 plus tiles. That is a total of 99 tiles. It would take a lot of time to drag that many tiles on the board and there might not be enough space to hold all of them. You would also need to determine how to arrange all those tiles to make a rectangle. Since 80 is a multiple of 10, you might recognize that 10 and 8 are the factors that add to 18, which would give you the values to use in the X method.

The probability that the product of the numbers they roll is even is, 2/3

We have to given that,

Geoff and Trevor each roll a fair six-sided die.

Here, Number of possible outcomes for Geoff,

1, 2, 3, 4, 5, 6

And, Number of outcomes for Trevor,

1, 2, 3, 4, 5, 6

Hence, Possible outcomes for product of the numbers they roll is even are,

2 x 1 = 2

2 x 2 = 4

2 x 3 = 6

2 x 4 = 8

2 x 5 = 10

2 x 6 = 12

3 x 2 = 6

3 x 4 = 12

3 x 6 = 18

4 x 1 = 2

4 x 2 = 8

4 x 3 = 12

4 x 4 = 16

4 x 5 = 20

4 x 6 = 24

5 x 2 = 10

5 x 4 = 20

5 x 6 = 30

6 x 1 = 6

6 x 2 = 12

6 x 3 = 18

6 x 4 = 24

6 x 5 = 30

6 x 6 = 36

Hence, Total possible outcomes = 24

And., Total outcomes = 6x 6 = 36

So, the probability that the product of the numbers they roll is even is,

P = 24 / 36

P = 12/18

P = 2/3

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Answer:

[tex]y-1 = \frac{2}{3}(x-3)[/tex]

Step-by-step explanation:

To find the equation of graphed line we pick two points from the line

(3,1) and (-3,-3)

Equation of a line is y=mx+b

where 'm' is the slope and b is the y intercept

slope [tex]m= \frac{y2-y1}{x2-x1} =\frac{1+3}{3+3} = \frac{2}{3}[/tex]

we use point slope formula

y-y1= m(x-x1)

here point(x1,y1) is (3,1)  and slope m = 2/3

Plug in all the value

[tex]y-1 = \frac{2}{3} (x-3)[/tex]

Answer:  Third option is correct.

Step-by-step explanation:

Since we have given that

1800(1.2).

We need to find the equivalent form:

we can rewrite 1.2 as [tex]\dfrac{12}{10}[/tex]

After find the lowest term which is [tex]\dfrac{6}{5}[/tex]

[tex]1800(1.2)\\\\=18000(\dfrac{12}{10})\\\\=18000(\dfrac{6}{5})[/tex]

Hence, it is equivalent to [tex]18000(\dfrac{6}{5})[/tex]

Thus, Third option is correct.

To determine if expressions are equivalent, we simplify them or substitute variables with actual values to see if they yield the same result. We simplify by eliminating terms where possible, grouping like terms, and applying mathematical operations.

Without the original mathematical expressions, it's difficult to provide direct answers to your question. Normally, we determine whether expressions are equivalent by simplifying them using algebraic rules or substituting variable with actual values to see if they produce the same results.

For example, the expressions 2(3 + 5) and 6 + 10 are equivalent because they both simplify to 16. We identify equivalent expressions by simplifying which includes eliminating terms where possible, grouping like terms, and applying mathematical operations.

Ultimately, the exact process may vary depending on the complexity and specific form of the expressions in question.

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The expressions equivalent to [tex]$\sqrt[3]{128}^x$[/tex] are (c) [tex]$(4 \sqrt[3]{2})^x$[/tex] and (d) [tex]$\left(4\left(2^{\frac{1}{3}}\right)\right)^x$[/tex] due to their compatible structures with the original expression.

To determine which expressions are equivalent to [tex]$\sqrt[3]{128}^x$[/tex], let's simplify the given expression first:

[tex]\[\sqrt[3]{128}^x = 2^x \times \sqrt[3]{2}^x\][/tex]

Now, let's compare this with the provided options:

a. [tex]$128^{\frac{x}{3}}$[/tex] - Not equivalent.

b. [tex]$128^{\frac{3}{x}}$[/tex] - Not equivalent.

c. [tex]$(4 \sqrt[3]{2})^x$[/tex] - Equivalent, as [tex]$4 = 2^{\frac{3}{2}}$[/tex].

d. [tex]$\left(4\left(2^{\frac{1}{3}}\right)\right)^x$[/tex] - Equivalent, as [tex]$2^{\frac{1}{3}} = \sqrt[3]{2}$[/tex].

e. [tex]$(2 \sqrt[3]{4})^x$[/tex] - Not equivalent.

Therefore, the expressions (c) [tex]$(4 \sqrt[3]{2})^x$[/tex] and (d) [tex]$\left(4\left(2^{\frac{1}{3}}\right)\right)^x$[/tex] are equivalent to [tex]$\sqrt[3]{128}^x$[/tex].