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Ivan
3 years ago
13

Question 11 plz, really need help

Mathematics
2 answers:
madam [21]3 years ago
7 0

Answer:

ummm hold on im trying to figure it out sorry might take long  

fgiga [73]3 years ago
7 0
I’m trying to see ,
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Pat shows that 16 1/2= the square root of 16. What is one possible way pat could have shown this correctly?
kifflom [539]
<h3>Answer: Choice C</h3>

The base 16 is the same as 4*4.

From there, the rule (a*b)^c = a^c*b^c is used to get (4*4)^{1/2} = 4^{1/2}*4^{1/2}

Afterward, the exponents are added getting 1/2+1/2 = 2/2 = 1. The rule is a^b*a^c = a^{b+c} which only works if the bases are both the same.

6 0
3 years ago
A carnival ride holds 30 people at a time. In one hour,
Olegator [25]

Answer:

7

Step-by-step explanation:

8 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:

a)

g(2.9) \approx -6.6

g(3.1) \approx -3.4

b)

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:

a)

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
3 years ago
What decimal is equal to 9/10?
satela [25.4K]
The answer is 0.9 here
3 0
3 years ago
Read 2 more answers
A line includes the points (13,15) and (6,15). What is its equation in slope-intercept form?​
Andrew [12]

Answer:

y=15

Step-by-step explanation:

y=mx+b

slope(y2-y1)/(x2-x1)

(15-15)/(6-13)

0/-7=0

y-intercept 15=0(15)+b

b=15

y=15

6 0
3 years ago
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