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Brilliant_brown [7]

3 years ago

7

2x + 6 = 2x - 2 ??????????

1 answer:

svp [43]3 years ago

6 0

2x-6=2x+2

We move all terms to the left:

2x-6-(2x+2)=0

We get rid of parentheses

2x-2x-2-6=0

We add all the numbers together, and all the variables

-8!=0

There is no solution for this equation

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An equation is shown. 2x-4y=8 solve the equation for x in terms of y

horrorfan [7]

2x-4y=8

Add 4y to each side to separate the variables, since you want to solve for x let be on the left

2x=4y+8

Divide the entire equation by 2 to allow x to stand alone

x=2y+4

And I have to type some more so it will let me answer the question so here ya go

7 0

3 years ago

Assume the equation has a solution for z.<br> -cz+ 6z = tz + 83<br>z=?

solong [7]

Answer:

z = 83 / (- c + 6 - t)

Step-by-step explanation:

Given:

-cz+ 6z = tz + 83

z=?

-cz+ 6z = tz + 83

Collect like terms

-cz+ 6z - tz = 83

Factorize

z(- c + 6 - t) = 83

Divide both sides by (- c + 6 - t)

z(- c + 6 - t) / (- c + 6 - t) = 83 / (- c + 6 - t)

z = 83 / (- c + 6 - t)

5 0

3 years ago

Because of their connection with secant lines, tangents, and instantaneous rates, limits of the form ModifyingBelow lim With h

Gre4nikov [31]

Answer:

$\dfrac{1}{2\sqrt{x}}$

Step-by-step explanation:

$f(x) = \sqrt{x} = x^{\frac{1}{2}}$

$f(x+h) = \sqrt{x+h} = (x+h)^{\frac{1}{2}}$

We use binomial expansion for $(x+h)^{\frac{1}{2}}$

This can be rewritten as

$[x(1+\dfrac{h}{x})]^{\frac{1}{2}}$

$x^{\frac{1}{2}}(1+\dfrac{h}{x})^{\frac{1}{2}}$

From the expansion

$(1+x)^n=1+nx+\dfrac{n(n-1)}{2!}+\ldots$

Setting $x=\dfrac{h}{x}$ and $n=\frac{1}{2}$ ,

$(1+\dfrac{h}{x})^{\frac{1}{2}}=1+(\dfrac{h}{x})(\dfrac{1}{2})+\dfrac{\frac{1}{2}(1-\frac{1}{2})}{2!}(\dfrac{h}{x})^2+\tldots$

$=1+\dfrac{h}{2x}-\dfrac{h^2}{8x^2}+\ldots$

Multiplying by $x^{\frac{1}{2}}$ ,

$x^{\frac{1}{2}}(1+\dfrac{h}{x})^{\frac{1}{2}}=x^{\frac{1}{2}}+\dfrac{h}{2x^{\frac{1}{2}}}-\dfrac{h^2}{8x^{\frac{3}{2}}}+\ldots$

$x^{\frac{1}{2}}(1+\dfrac{h}{x})^{\frac{1}{2}}-x^{\frac{1}{2}}=\dfrac{h}{2x^{\frac{1}{2}}}-\dfrac{h^2}{8x^{\frac{3}{2}}}+\ldots$

$\dfrac{x^{\frac{1}{2}}(1+\dfrac{h}{x})^{\frac{1}{2}}-x^{\frac{1}{2}}}{h}=\dfrac{1}{2x^{\frac{1}{2}}}-\dfrac{h}{8x^{\frac{3}{2}}}+\ldots$

The limit of this as $h\to 0$ is

$\lim_{h\to0} \dfrac{f(x+h)-f(x)}{h}=\dfrac{1}{2x^{\frac{1}{2}}}=\dfrac{1}{2\sqrt{x}}$ (since all the other terms involve $h$ and vanish to 0.)

8 0

3 years ago

What is the area of this shape?

Elena L [17]

Answer:

12mm (triangle)

165mm (whole shape)

Step-by-step explanation:

The triangle is 6 x 4 = 24 ÷ 2= 12

<em>(</em><em>divide</em><em> </em><em>by</em><em> </em><em>2</em><em> </em><em>because</em><em> </em><em>its</em><em> </em><em>a</em><em> </em><em>teiangle</em><em>)</em>

<em>2nd</em><em> </em><em>shap</em><em>e</em><em> </em><em>is</em><em> </em><em>1</em><em>0</em><em> </em><em>x</em><em> </em><em>8</em><em> </em><em>=</em><em> </em><em>8</em><em>0</em>

<em>3rd</em><em> </em><em>sha</em><em>pe</em><em> </em><em>is</em><em> </em><em>6</em><em> </em><em>x</em><em> </em><em>1</em><em>2</em><em> </em><em>=</em><em> </em><em>7</em><em>2</em>

<em>7</em><em>2</em><em> </em><em>+</em><em> </em><em>8</em><em>0</em><em> </em><em>+</em><em> </em><em>1</em><em>2</em><em> </em><em>=</em><em> </em><em>165mm</em>

8 0

3 years ago

If the original price is $89.00 and the markdown is 33% what is the sale price

vladimir1956 [14]

Answer:

$29.37

Step-by-step explanation:

6 0

3 years ago

Read 2 more answers