All categories

3 years ago

Which expression is equivalent to –3(u+7) PLEASE ANSWER THE QUESTION

Mathematics

1 answer:

Andreyy893 years ago

5 0

Answer: (u+7) x -3

I like to use substitution, for example u=2

(2+7) x -3 = -27

-3(2+7) = -27

You might be interested in

Y'all i need help immediately

adoni [48]

Answer:

Option B

Step-by-step explanation:

Don't trust answers teachers don't bother to spell correctly. Plus, finding the 100th term is very easy.

However, the explicit formula will take a long time to use, because it's explicit.

Option C is just wrong. You're supposed to multiply in the recursive formula.

Option D might be correct, but never trust those answers. Teachers do that to trip you up

3 0

3 years ago

I NEED THIS DONE ASAPPPP<br> expand 4(m+2)<br><br> 4(m+2) =

77julia77 [94]

4(m + 2) expanded is 4 x m and 4 x 2

simplified: 4m + 8

3 0

3 years ago

What is the equation for the plane illustrated below?

TiliK225 [7]

Answer:

Hence, none of the options presented are valid. The plane is represented by $3 \cdot x + 3\cdot y + 2\cdot z = 6$ .

Step-by-step explanation:

The general equation in rectangular form for a 3-dimension plane is represented by:

$a\cdot x + b\cdot y + c\cdot z = d$

Where:

$x$ , $y$ , $z$ - Orthogonal inputs.

$a$ , $b$ , $c$ , $d$ - Plane constants.

The plane presented in the figure contains the following three points: (2, 0, 0), (0, 2, 0), (0, 0, 3)

For the determination of the resultant equation, three equations of line in three distinct planes orthogonal to each other. That is, expressions for the xy, yz and xz-planes with the resource of the general equation of the line:

xy-plane (2, 0, 0) and (0, 2, 0)

$y = m\cdot x + b$

$m = \frac{y_{2}-y_{1}}{x_{2}-x_{1}}$

Where:

$m$ - Slope, dimensionless.

$x_{1}$ , $x_{2}$ - Initial and final values for the independent variable, dimensionless.

$y_{1}$ , $y_{2}$ - Initial and final values for the dependent variable, dimensionless.

$b$ - x-Intercept, dimensionless.

If $x_{1} = 2$ , $y_{1} = 0$ , $x_{2} = 0$ and $y_{2} = 2$ , then:

Slope

$m = \frac{2-0}{0-2}$

$m = -1$

x-Intercept

$b = y_{1} - m\cdot x_{1}$

$b = 0 -(-1)\cdot (2)$

$b = 2$

The equation of the line in the xy-plane is $y = -x+2$ or $x + y = 2$ , which is equivalent to $3\cdot x + 3\cdot y = 6$ .

yz-plane (0, 2, 0) and (0, 0, 3)

$z = m\cdot y + b$

$m = \frac{z_{2}-z_{1}}{y_{2}-y_{1}}$

Where:

$m$ - Slope, dimensionless.

$y_{1}$ , $y_{2}$ - Initial and final values for the independent variable, dimensionless.

$z_{1}$ , $z_{2}$ - Initial and final values for the dependent variable, dimensionless.

$b$ - y-Intercept, dimensionless.

If $y_{1} = 2$ , $z_{1} = 0$ , $y_{2} = 0$ and $z_{2} = 3$ , then:

Slope

$m = \frac{3-0}{0-2}$

$m = -\frac{3}{2}$

y-Intercept

$b = z_{1} - m\cdot y_{1}$

$b = 0 -\left(-\frac{3}{2} \right)\cdot (2)$

$b = 3$

The equation of the line in the yz-plane is $z = -\frac{3}{2}\cdot y+3$ or $3\cdot y + 2\cdot z = 6$ .

xz-plane (2, 0, 0) and (0, 0, 3)

$z = m\cdot x + b$

$m = \frac{z_{2}-z_{1}}{x_{2}-x_{1}}$

Where:

$m$ - Slope, dimensionless.

$x_{1}$ , $x_{2}$ - Initial and final values for the independent variable, dimensionless.

$z_{1}$ , $z_{2}$ - Initial and final values for the dependent variable, dimensionless.

$b$ - z-Intercept, dimensionless.

If $x_{1} = 2$ , $z_{1} = 0$ , $x_{2} = 0$ and $z_{2} = 3$ , then:

Slope

$m = \frac{3-0}{0-2}$

$m = -\frac{3}{2}$

x-Intercept

$b = z_{1} - m\cdot x_{1}$

$b = 0 -\left(-\frac{3}{2} \right)\cdot (2)$

$b = 3$

The equation of the line in the xz-plane is $z = -\frac{3}{2}\cdot x+3$ or $3\cdot x + 2\cdot z = 6$

After comparing each equation of the line to the definition of the equation of the plane, the following coefficients are obtained:

$a = 3$ , $b = 3$ , $c = 2$ , $d = 6$

Hence, none of the options presented are valid. The plane is represented by $3 \cdot x + 3\cdot y + 2\cdot z = 6$ .

8 0

3 years ago

what is the quotient 5-x/x^2 3x-4 divided by x^2-2x-15/x^2 5x 4 in simplifed form state any restrictions on the varible

zheka24 [161]

The quotient when $\frac{5-x}{x^2+3x-4} /\frac{x^2-2x - 15}{x^2+5x+4}$ in simplified form is $\frac{-(x+1)}{(x-1)(x+3)}$

<h3>What is an equation?</h3>

An equation is an expression that shows the relationship between two or more numbers and variables.

Given that equation:

$\frac{5-x}{x^2+3x-4} /\frac{x^2-2x - 15}{x^2+5x+4}$

$=\frac{5-x}{(x+4)(x-1)} /\frac{(x-5)(x+3)}{(x+4)(x+1)} \\\\=\frac{5-x}{(x+4)(x-1)} * \frac{(x+4)(x+1)}{(x-5)(x+3)}\\\\\frac{-(x-5)}{(x+4)(x-1)} * \frac{(x+4)(x+1)}{(x-5)(x+3)}\\\\=\frac{-(x+1)}{(x-1)(x+3)}$

The quotient when $\frac{5-x}{x^2+3x-4} /\frac{x^2-2x - 15}{x^2+5x+4}$ in simplified form is $\frac{-(x+1)}{(x-1)(x+3)}$

Find out more on equation at: brainly.com/question/2972832

#SPJ1

8 0

2 years ago

Solve for x. PLZ HELP ASAP!!!

ycow [4]

X is a vertical angle to the angle marked as 100 degrees.

Vertical angles are the same so x = 100 degrees

Answer: 100 degrees

3 0

3 years ago