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2 years ago

14. Solve for the value of x in this equation. 2x - (4x + 1) = 3x - 6

Mathematics

2 answers:

____ [38]2 years ago

5 0

Step-by-step explanation:

2x-(4x+1)=3x-6

2x-4x-1=3x-6
-2x-1=3x-6
3x+2x=6-1
5x=5
x=5/5
x=1

nekit [7.7K]2 years ago

5 0

Answer:

-1.4 = x

Step-by-step explanation:

Step 1: Combine like terms and subtract the 3x from 2x.

-x - (4x + 1) = - 6

Step 2: Combine more like terms. (and remove parentheses)

-5x + 1 = - 6

Step 3: Combine different like terms.

-5x = -7

Step 4: Divide it out.

x = - 1.4

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Whats the answer to x^3 y^0 z^-2

Elina [12.6K]

Answer:

$\frac{x^3}{z^2}$

Step-by-step explanation:

There are no given values, therefore we can't determine a definite answer. However, we are able to simplify the expression.

As we all know that any variable or number to the power of 0 is equal to 1. This means that $y^0 = 1$ .

We also know that a negative power can bring the variable down and make it a fraction. This means that $z^-^2=\frac{1}{z^2}$

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2 years ago

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Answer:

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Step-by-step explanation:

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2 years ago

Read 2 more answers

F(x) = 6X-3 - 3 What is the graph equation

TiliK225 [7]

Answer:

See attachment

Step-by-step explanation:

Assuming the equation is

$f(x) = {6}^{x - 3} - 3$

Then we have a transformed exponential function whose parent function is

$g(x) = {6}^{x}$

We can see that:

There is a transformation of the form:

$f(x) = g(x - 3) - 3$

The parent function has been shifted to the right 4 units and down 3 units.

The is shown in attachment.

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3 years ago

Find the lengths of all segments in the diagram. AB = BC = BD = AC =

Dmitrij [34]

Answer:

Since we know that AD =12

AC=1/2*AD

AC=6

AB=BC=1/2*AC

AB=BC=3

hope you like the answer

please Mark as brainlist

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2 years ago

Find the values of m and b that make the following function differentiable. the piecewise function f of x equals x cubed when x

aalyn [17]

$f(x)=\begin{cases}x^3&\text{for }x\le1\\mx+b&\text{for }x>1\end{cases}$

$f$ must be continuous in order to be differentiable, so we need to have

$\displaystyle\lim_{x\to1^-}f(x)=\lim_{x\to1^+}f(x)=f(1)$

By its definition, $f(1)=1^3=1$ , and

$\displaystyle\lim_{x\to1^-}f(x)=\lim_{x\to1}x^3=1$

$\displaystyle\lim_{x\to1^+}f(x)=\lim_{x\to1}(mx+b)=m+b$

so that $\boxed{m+b=1}$ .

We want the derivative to exist at $x=1$ , which requires that we pick an appropriate value for $f'(1)$ so that $f'(x)$ is also continuous. At the moment, we know

$f'(x)=\begin{cases}3x^2&\text{for }x1\end{cases}{/tex]so we need to pick [tex]f'(1)$ such that

$\displaystyle\lim_{x\to1^-}f'(x)=\lim_{x\to1^+}f'(x)$

We have

$\displaystyle\lim_{x\to1^-}f'(x)=\lim_{x\to1}3x^2=3$

$\displaystyle\lim_{x\to1^+}f'(x)=\lim_{x\to1}m=m$

so that $\boxed{m=3}$ (which means we need to pick $f'(1)=3$ ) and so $m+b=1\implies\boxed{b=-2}$

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3 years ago