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Aleks04 [339]
2 years ago
6

For each expression, use the distributive property to write an equivalent expression.

Mathematics
1 answer:
Vlada [557]2 years ago
8 0

Here u go refer to the attachment below! v

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What is the slope of a line that passes through the points (-2, 0) and (-4, -3)?
Semmy [17]

Answer:

1/2

Step-by-step explanation:

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If the exchange rate when converting US Dollars to ZA rands is $1=R17
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Answer:

200 US DOLLARS

Step-by-step explanation:

Given; $1=R17

To find: How much is Zar3400 in US DOLLARS

$1=R17

R3400=3400/17 US DOLLARS

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Hope my answer is right thank you.

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PLEASE HELP ASAP!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
SashulF [63]

Answer:

I think it's C

Step-by-step explanation:

5 0
3 years ago
Read 2 more answers
If the function graphed is f(x), find f(1).
Crazy boy [7]
Locate 1 on the x axis. This is the horizontal number line. 

Draw a vertical line through 1 on the x axis. Extend this vertical line as far up and down as you can.

Notice how the vertical line crosses the blue curve. Mark this point. Then draw a horizontal line from this point to the y axis. The horizontal line will touch -4 on the y axis. So that means the point (1,-4) is on the curve.

If the input it is x = 1, then the output is y = -4

So that's why the answer is choice A
7 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
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