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marishachu [46]
3 years ago
10

Find the difference 84.28 - 8.37

Mathematics
2 answers:
valentina_108 [34]3 years ago
5 0

Answer:

75.91

Step-by-step explanation:

Use long subtraction to evaluate.

klasskru [66]3 years ago
3 0

Answer:

75.91

i use brains okay?

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the height of an equivalent triangle is 15cm and it's perimeter is 36cm find the area of the triangle​
OlgaM077 [116]

Answer:

The area is 72 cm².

Step-by-step explanation:

Since it is equivalent triangle

h = b

so,

b + b + b (since all sides are equal)= 36 cm

3b = 36 cm

or, b = 36/3

so, b = 12 cm

so

area of triangle = (1/2)×b×h

= (1/2)×12cm×12cm

= 6cm × 12cm

= 72 cm²

4 0
3 years ago
Which of the following expressions is equivalent to -5.4 - (2.1 + 1.8)?
Paul [167]

Answer:

its the first one

Step-by-step explanation:

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3 years ago
The equation that represents the canned goods order is 24x + 64y = 384, where x = number of minutes spent producing fruit cans,
Lubov Fominskaja [6]
So wee need to find x

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3 years ago
Read 2 more answers
Draw the following scenario using a vertical number line.
nekit [7.7K]
You would move your point 24 up from the starting point then move it three down from the 24 so if you start at zero go to 24 then move it down to 21.

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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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