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2 years ago

Determine the interval(s) on which the given function is decreasing?

Mathematics

1 answer:

eduard2 years ago

5 0

Answer:

$(-\infty, -1) \cup (0, \infty)$

Step-by-step explanation:

The function is decreasing if, as x increases, the value of the function (y) gets smaller. This means as you read the graph from left-to-right, the function is decreasing if the graph is falling.

From -infinity to -1, the graph is falling, then the graph rises (f is increasing) until x =0, then the graph falls again from x = 0 to infinity.

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Solve.<br> x² – 8x - 12 = 36

professor190 [17]

sorry answered the wrong one..

7 0

3 years ago

What’s the correct answer?

nikdorinn [45]

Answer:

the triangle above is an isoceles triangle

therefore the base angles are equal

So we have:

X + X + 38= 180

2x = 180-38

2x = 142

X = 71°

6 0

2 years ago

Read 2 more answers

Please solve with explanation

leva [86]

Answer:

i pray you get it right lol

Step-by-step explanation:

just ask a teacher

8 0

2 years ago

The base of a 15-foot ladder is 6 feet from a building. If the ladder reaches the flat roof, how tall is the building?

Vlad [161]

Answer:

the answer would be 13.75 feet

Step-by-step explanation:

The computation of the tall is the building as follows:

Given that

Base is 6 foot

And, the hypotheses is 15 feet

Here we use the Pythagoras theorem

= √15^2 - √6^2

= √225 - √36

= √189

= 13.75 feet

Hence the answer would be 13.75 feet

6 0

3 years ago

Find the linear approximation of the function g(x) = 3 1 + x at a = 0. g(x) ≈ Correct: Your answer is correct. Use it to approxi

EleoNora [17]

Answer:

$3.296x+3$

2.835

3.330

Step-by-step explanation:

$g(x) = 3^{1+x}$

Let the linear approximation be L(x) at a = 0. This is given by

$L(x) = g(a) + g'(a)(x-a)$

$g'(x)$ is the derivative of $g(x)$ . To find this, we use the form

If $f(x) = a^x$ , then $f'(x) =a^x\ln a$

By doing this and applying chain rule for the power (which is a function of x), we have

$g'(x) = 3^{1+x}\ln 3$

Then

$g'(0) = 3^{1+0}\ln 3 = 3.296$

Also $g(0) = 3^{1+0} = 3$

Hence $L(x) = 3+(x-0)\times3.296 = 3.296x + 3$

For $3^{0.95}$ , $1+x = 0.95$ and $x=-0.05$

Using this in $L(x)$ ,

$3^{0.95} = 3.296(-0.05) + 3 = 2.835$

For $3^{1.1}$ , $1+x = 1.1$ and $x=0.1$

Using this in $L(x)$ ,

$3^{1.1} = 3.296(0.1) + 3 = 3.330$

7 0

3 years ago