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Katen [24]
3 years ago
5

A car traveled 360 km in 6 h. A train traveled 400 km in 8 h. A boat traveled 375 km in 5 h. Which had the fastest average speed

?
Mathematics
1 answer:
Nezavi [6.7K]3 years ago
7 0

Answer:

boat 75

Step-by-step explanation:

Car--360/6=60

Train--400/8=50

Boat--375/5=75

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What is the absolute value of 2
Anna71 [15]

Answer:The Absolute Value of a number tells us how far the number is from zero, on a Number Line. 2 is 2 units away from zero. Hence |2|=2.hope it helps

Step-by-step explanation:

8 0
3 years ago
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What is PA <br> Enter your answer in the box
SSSSS [86.1K]

Answer:

3

Step-by-step explanation:

P is the in-center

⇒PA=PE=PD because they are in-radius of the in-circle

We know that, tangent segments drawn from a point outside the circle are always equal in length

⇒DK=EK=7.2

In right triangle PKE,

using Pythagoras' Theorem : PK^{2}=PE^{2}+KE^{2}

⇒PE^{2}=PK^{2}-KE^{2}

⇒PE=\sqrt{PK^{2}-KE^{2} }

⇒PE^{2}=\sqrt{7.8^{2}-7.2^{2}} }

⇒PE=3

Therefore, PA=3

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2 years ago
Skcncjnds<br>kjsndjcjknskjckjzxc
timama [110]

skskhfkakakan what is this haha

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
The temperature in the ice rink must stay below 50° F. This morning the temperature was 68°F. The ice rink runs a cooling device
german
The answer is h > 4.
Explanation: 68- 18=50
18/4=4.5. It is asking for below 50, and it had to cool for over 4 hours, which is >. I hope this helps. Please give me Brainliest.
6 0
2 years ago
Read 2 more answers
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