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3 years ago

Confused on this help plss

Mathematics

1 answer:

ValentinkaMS [17]3 years ago

8 0

Answer:

of iv

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x+2({\color{#c92786}{4-x}})=2x+6-3x

x+2(4−x)=2x+6−3x

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Step-by-step explanation:

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Show that x=1 is not a solution to 4x+2 2 = 13.

likoan [24]

Answer:

Replacing x = 1 in the equation above, we have: (4x1) + 22 = 13

=> 5 +22 = 13 => 27 $\neq$ 13

So 27 is not equal to 13 => x=1 is not a solution to 4x+22 = 13.

Step-by-step explanation:

4 0

3 years ago

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Find the product AB if possible A = (1 -6 1 -8 -1 -9 -4 -1 -9) B = (-4 -5 1 -3 3 -4 -1 4 9)

allsm [11]

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3 years ago

Identify each expression that represents the slope of a tangent to the curve y=1/x+1 at any point (x,y) .

tankabanditka [31]

Answer:

Slope of a tangent to the curve = $f'(x) = \frac{-1 }{(x+1)^{2} }$

Step-by-step explanation:

Given - y = 1/x+1

To find - Identify each expression that represents the slope of a tangent to the curve y=1/x+1 at any point (x,y) .

Proof -

We know that,

Slope of tangent line = f'(x) = $\lim_{h \to 0} \frac{f(x+h) - f(x)}{h}$

We have,

f(x) = y = $\frac{1}{x+1}$

So,

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

Now,

Slope = f'(x)

And

$f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h} \\= \lim_{h \to 0} \frac{\frac{1}{x+h+1} - \frac{1}{x+1} }{h}\\= \lim_{h \to 0} \frac{x+1 - (x+h+1) }{h(x+1)(x+h+1)}\\= \lim_{h \to 0} \frac{x +1 - x-h-1 }{h(x+1)(x+h+1)}\\= \lim_{h \to 0} \frac{-h }{h(x+1)(x+h+1)}\\= \lim_{h \to 0} \frac{-1 }{(x+1)(x+h+1)}\\= \frac{-1 }{(x+1)(x+0+1)}\\= \frac{-1 }{(x+1)(x+1)}\\= \frac{-1 }{(x+1)^{2} }$

∴ we get

Slope of a tangent to the curve = $f'(x) = \frac{-1 }{(x+1)^{2} }$

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3 years ago

Please no link answers you will be reported

rewona [7]

Answer:

Step-by-step explanation:

You can use the key features (such as the vertex) to determine which is the correct equation

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3 years ago

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kogti [31]

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