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marin [14]
3 years ago
6

Is this positive or negative? ( slopes ) y= -2/3 x - 1

Mathematics
1 answer:
timofeeve [1]3 years ago
4 0

Answer:

positive

Step-by-step explanation:

negative times negative is positive

positive times positive is positive

positive times negative is negative

negative times positive is negative

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Suppose g (x) is increasing and concave up everywhere and g(A)=7, gprime(A)=13, h=.01
rewona [7]
First we need a point (x,y) : (A, 7) 
<span>Now slope (from f'(A)) = 15 </span>
<span>Next, the equation (using point slope formula) </span>

<span>y - 7 = 15 (x -A) </span>
<span>y = 15 (x - A) + 7 </span>

<span>Now in the x spot we put 'A-.01' </span>
<span>y = 15 ( A - .01 - A) +7= 15(-.01) +7 = -.15+ 7 = 6.85 
hope this helps</span>
8 0
3 years ago
How much more interest is earned on an investment of $12,585 at 3.5% interest over 5 years if the interest is compounded annuall
Nina [5.8K]

Answer:

It is B.$159.65


Step-by-step explanation:

I got it the test

7 0
3 years ago
Read 2 more answers
Which property is demonstrated?
Korolek [52]

Answer:

Multiplicative Inverse



4 0
3 years ago
QUESTION 3 [10 MARKS] A bakery finds that the price they can sell cakes is given by the function p = 580 − 10x where x is the nu
Andru [333]

Answer:

(a)Revenue=580x-10x²,

Marginal Revenue=580-20x

(b)Fixed Cost, =900

Marginal Cost,=300+50x

(c)Profit Function=280x-900-35x²

(d)x=4

Step-by-step explanation:

The price, p = 580 − 10x where x is the number of cakes sold per day.

Total cost function,c = (30+5x)²

(a) Revenue Function

R(x)=x*p(x)=x(580 − 10x)

R(x)=580x-10x²

Marginal Revenue Function

This is the derivative of the revenue function.

If R(x)=580x-10x²,

R'(x)=580-20x

(b)Total cost function,c = (30+5x)²

c=(30+5x)(30+5x)

=900+300x+25x²

Therefore, Fixed Cost, =900

Marginal Cost Function

This is the derivative of the cost function.

If c(x)=900+300x+25x²

Marginal Cost, c'(x)=300+50x

(c)Profit Function

Profit, P(x)=R(x)-C(x)

=(580x-10x²)-(900+300x+25x²)

=580x-10x²-900-300x-25x²

P(x)=280x-900-35x²

(d)To maximize profit, we take the derivative of P(x) in (c) above and solve for its critical point.

Since P(x)=280x-900-35x²

P'(x)=280-70x

Equate the derivative to zero

280-70x=0

280=70x

x=4

The number of cakes that maximizes profit is 4.

5 0
3 years ago
Let f (x) = 3x − 1 and ε &gt; 0. Find a δ &gt; 0 such that 0 &lt; ∣x − 5∣ &lt; δ implies ∣f (x) − 14∣ &lt; ε. (Find the largest
s344n2d4d5 [400]

Answer:

This proves that f is continous at x=5.

Step-by-step explanation:

Taking f(x) = 3x-1 and \varepsilon>0, we want to find a \delta such that |f(x)-14|

At first, we will assume that this delta exists and we will try to figure out its value.

Suppose that |x-5|. Then

|f(x)-14| = |3x-1-14| = |3x-15|=|3(x-5)| = 3|x-5|< 3\delta.

Then, if |x-5|, then |f(x)-14|. So, in this case, if 3\delta \leq \varepsilon we get that |f(x)-14|. The maximum value of delta is \frac{\varepsilon}{3}.

By definition, this procedure proves that \lim_{x\to 5}f(x) = 14. Note that f(5)=14, so this proves that f is continous at x=5.

3 0
3 years ago
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