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ivolga24 [154]
3 years ago
11

Is 7.0001 a terminating or repeating decimal

Mathematics
2 answers:
ANTONII [103]3 years ago
6 0
The answer would be repeating.
dedylja [7]3 years ago
3 0
It is a terminating decimal
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(a) Each student in a pottery class is responsible for cleaning the area around his or her workspace at the end of the lesson. E
ArbitrLikvidat [17]
Ok this is not as hard as it seems

(4.5 feet)^2 times the number of students (13) 

20.25 x 13

263.25 is the correct answer



I hope it helps :<
5 0
3 years ago
If f(x) =1/9 x-2, what is f-1(x)?
Sav [38]

Answer:

f^-1(x) = 9(x+2)

Step-by-step explanation:

To find the inverse function, exchange x and y and then solve for y

y = 1/9 x -2

Exchange x and y

x = 1/9 y-2

Solve for y

Add 2 to each side

x+2 = 1/9 y-2+2

x+2 = 1/9y

Multiply each side by 9

9(x+2) = 9*1/9y

9(x+2) = y

The inverse function

f^-1(x) = 9(x+2)

7 0
3 years ago
Read 2 more answers
Evaluate the expression if y=4.4 and z= 12<br> z2+ 8y
Evgen [1.6K]

Answer: <em>59.2</em>

Step-by-step explanation:

<em>Take your equation</em>

<em>z2+8y</em>

<em>Now plug in the values given</em>

<em>(12)2+8(4.4)</em>

<em>24+35.2</em>

<em>59.2</em>

5 0
3 years ago
What is the value of x
VLD [36.1K]

Answer:

value of x = 8

Step-by-step explanation:

if the given lines are parallel, then the shown pair of angles are corresponding angles,

and we know that the the pair of corresponding angles are equal so,

=》16x - 8 = 120

=》16x = 128

=》x = 128 ÷ 16

=》x = 8

6 0
2 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
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