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3 years ago

Please help me I'm stuck on these questions!

Mathematics

1 answer:

defon3 years ago

8 0

5.) (part 1) is 5x^3, part 2 is 0.25x^2
6.) part 1.) D^2. Part 2.) T^2
7.) part 1.) x^9, part 2.) y^10
9.) part 1.) 15x^2y^3, part 2.) 5ab

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Please answer my question thank you.

Andrews [41]

A. terminating
b. repeating
c. repeating
d. terminating
e. repeating
i think

8 0

3 years ago

Read 2 more answers

Locate the points of discontinuity in the piecewise function shown below.

algol13

Answer:

Step-by-step explanation:

The given piecewise function i

From the given function it is clear that function is divided at x=-1 and x=2. It means we check the discontinuity at x=-1 and x=2.

For x=-1,

LHL:

Since LHL ≠ f(-1), therefore the given function is discontinuous at x=-1.

For x=2,

LHL:

Since LHL ≠ f(2), therefore the given function is discontinuous at x=2.

Therefore, the correct option is A.

8 0

3 years ago

A farmer has 140 bushels of wheat to sell at his roadside stand. He sells an average of 14 bushels

Levart [38]

Answer:

140-(14×7)=42

Step-by-step explanation:

if he sells 14 each day for 7 days. thats 14-7=98 and to find how many he has left, you have to subtract 98 from 140

7 0

3 years ago

7(5-4x) in distributive property

kati45 [8]

Answer:

Distributive Property : <em><u>a(b+c)= ab+ac</u></em>

7(5-4x) = (35-28x) ★

<h3><u>(35-28x)</u> is the right answer.</h3>

3 0

3 years ago

Suppose that we don't have a formula for g(x) but we know that g(3) = −5 and g'(x) = x2 + 7 for all x.

Leni [432]

Answer:

$g(2.9) \approx -6.6$

$g(3.1) \approx -3.4$

The values are too small since $g''$ is positive for both values of $x$ in. I'm speaking of the $x$ values, 2.9 and 3.1.

Step-by-step explanation:

The point-slope of a line is:

$y-y_1=m(x-x_1)$

where $m$ is the slope and $(x_1,y_1)$ is a point on that line.

We want to find the equation of the tangent line of the curve $g$ at the point $(3,-5)$ on $g$ .

So we know $(x_1,y_1)=(3,-5)$ .

To find $m$ , we must calculate the derivative of $g$ at $x=3$ :

$m=g'(3)=(3)^2+7=9+7=16$ .

So the equation of the tangent line to curve $g$ at $(3,-5)$ is:

$y-(-5)=16(x-3)$ .

I'm going to solve this for $y$ .

$y-(-5)=16(x-3)$

$y+5=16(x-3)$

Subtract 5 on both sides:

$y=16(x-3)-5$

What this means is for values $x$ near $x=3$ is that:

$g(x) \approx 16(x-3)-5$ .

Let's evaluate this approximation function for $g(2.9)$ .

$g(2.9) \approx 16(2.9-3)-5$

$g(2.9) \approx 16(-.1)-5$

$g(2.9) \approx -1.6-5$

$g(2.9) \approx -6.6$

Let's evaluate this approximation function for $g(3.1)$ .

$g(3.1) \approx 16(3.1-3)-5$

$g(3.1) \approx 16(.1)-5$

$g(3.1) \approx 1.6-5$

$g(3.1) \approx -3.4$

b) To determine if these are over approximations or under approximations I will require the second derivative.

If $g''$ is positive, then it leads to underestimation (since the curve is concave up at that number).

If $g''$ is negative, then it leads to overestimation (since the curve is concave down at that number).

$g'(x)=x^2+7$

$g''(x)=2x+0$

$g''(x)=2x$

$2x$ is positive for $x>0$ .

$2x$ is negative for $x$ .

That is, $g''(2.9)>0 \text{ and } g''(3.1)>0$ .

So $2x$ is positive for both values of $x$ which means that the values we found in part (a) are underestimations.

6 0

4 years ago