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e-lub [12.9K]
3 years ago
6

What 2 fingers should you use to properly check your pulse?

Mathematics
2 answers:
lesya692 [45]3 years ago
8 0

Your middle finger and index/pointer.

Dmitry_Shevchenko [17]3 years ago
3 0

The Answer is:

C: Pointer and Middle Finger.

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1. LeAnn has 20 coins in her piggy bank. All the coins are either pennies or dimes. The total value of the coins is 83 cents. H
mafiozo [28]
There is 7 dimes and 13 pennies
8 0
3 years ago
Rahima goes jogging on a course that is 2.75kilometers long. If she completes the full course every morning, how many kilometers
Arada [10]

Answer:

<h2>19.25km</h2>

Step-by-step explanation:

Step one:

given data

we are told that the distance that Rahima covers on a daily is 2.75km long

we also know that a week has 7 days that is Mon-Sunday

Required

the total distance for a week

Step two:

Hence the total distance covered for  7 days is

=2.75*7

=19.25km

<em>Simply multiply the distance covered in a day by 7 to get the distance covered in a week </em>

6 0
2 years ago
Read 2 more answers
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
Help ASAP please please 14 and 15
Yuliya22 [10]
I think 14 is B (idrk) and 15 is definitely C
8 0
3 years ago
There are 6 balls. Half of the balls are blue. One ball is red. The rest are green. Write the fraction for the green balls.
Ipatiy [6.2K]
2/6 of the balls are green or 1/3
7 0
3 years ago
Read 2 more answers
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