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goldfiish [28.3K]
3 years ago
10

Show proooof, pls!!!!!!

Mathematics
2 answers:
castortr0y [4]3 years ago
7 0

Answer:

\boxed{\text{38.5 units}^{2}}

Step-by-step explanation:

The polygon is a trapezoid on its side.

The formula for the area of a trapezoid is  

A = ½(a + b)h,

where a and b are the lengths of the parallel sides and h is the distance between them.

AD = a = 2 – (-2) = 2 + 2 = 4 units

BC = b = 2 – (-5) = 2 + 5 = 7 units

AB = h = 3 – (-4) = 3 + 4 = 7 units

A = ½(4 + 7) × 7 = ½ × 11 × 7 = 38.5 units²

\boxed{\text{A = 38.5 units}^{2}}

elena-14-01-66 [18.8K]3 years ago
6 0

answer for this question is 3.85 units squared

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Jason puts $30 worth of gas in his truck. The price per gallon was $2.50. Which equation can be solved to find the number of gal
tankabanditka [31]

So, this is a rate problem, so the forumla we would get is


2.5g=30

(Note this is not one of the choices), so if we look, there is an equivelent equation.

If we divide both sides by g we get

2.5=30/g

B is the correct answer

8 0
3 years ago
The first U.S. census was conducted in 1790. At that time, the population was 3.93 million. Since then, the U.S. population has
Nikolay [14]

Answer:

It's actually A, because the US population got doubled ever year

4 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
|c-10|=3 <br>absolute value
Norma-Jean [14]

The final solutions are C1=7, C2=14


6 0
3 years ago
Simplify the expression. 5–3 • 7^0
aev [14]

Answer:

5-3x7^0

(5-3)7^0

2x1

2

Step-by-step explanation:

7^0 is 1

and do you want to chat

3 0
3 years ago
Read 2 more answers
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