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Colt1911 [192]
3 years ago
13

Pls help me ill give brainly just pls solve both parts of the question No bots or ill report

Mathematics
2 answers:
Allisa [31]3 years ago
8 0

Answer:

E: start at the rate of 30

F: starts at the rate of 20

Step-by-step explanation:

because if you look at the one u go up from that to what ever number hits it first

Vinil7 [7]3 years ago
6 0

Answer:

THEY START AT A DIFFERENT PLase sorry about all caps

Step-by-step explanation:

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Easy points let’s gooo
Alina [70]

Answer:

12 x 10 = 120

120 divided by 2 = 60

Formula of a triangle:

B(base) x H(height) x 1/2(basically dividing by 2)

8 0
3 years ago
Question and choices Are below
Vika [28.1K]

Answer:

9ft

Step-by-step explanation:

Use the pythagorean theorem.

15^2 = 12^2 + x^2

solve for x

x^2 = 81

x = 9

7 0
3 years ago
45<br> 2<br> If<br> = 2, what is the value of<br> A) O<br> B) 1<br> C) 2<br> D) 4.
eimsori [14]

Answer:

Tangent = opposite / adjacent

Tangent of 45º equals 1

Step-by-step explanation:

7 0
3 years ago
Please help me with this !
SSSSS [86.1K]

Answer:

mE=mC

mD=mB

Step-by-step explanation:

thats should be correct

3 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
4 years ago
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