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vladimir2022 [97]
3 years ago
15

Estimate the sum. 10 1/9 + 5 14/15

Mathematics
1 answer:
quester [9]3 years ago
5 0

Answer:

16.04

Step-by-step explanation:

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The low temperature last week in Iceland was-2.25. The low temperature this week was 1.25. What was the change in temperature fr
viktelen [127]

Answer:

The change in temperature from last week to this week=3.5

Step-by-step explanation:

To get the change in temperature from last week to this week, we need to take last week's temperature and subtract it from this week's temperature.

For example;

Change in temperature=This week's temperature-last week's temperature

where;

This week's temperature=-2.25

last week's temperature=1.25

replacing;

Change in temperature=(1.25)-(-2.25)=1.25+2.25=3.5

The change in temperature from last week to this week=3.5

6 0
3 years ago
AB = 10 and AC = 2 StartRoot 10 EndRoot. What is the perimeter of △ABC?
WITCHER [35]

Answer:

20 + 2 units

Step-by-step explanation:

because the perimeter of ∆ ABC is the answer 20 + 2

4 0
2 years ago
What is the circumference
IRINA_888 [86]
25.1 yards
multiply the radius by 2 and multiply that number with pi and you get your answer of 25.1 yards
8 0
2 years ago
Using the graph below, what is the value of f(6) ?
Dima020 [189]

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f(6) =   - 2

♥️♥️♥️♥️♥️♥️♥️♥️♥️♥️♥️♥️♥️♥️

5 0
3 years ago
Read 2 more answers
Find the measurements (the length L and the width W) of an inscribed rectangle under the line with the 1st quadrant of the x &am
Leni [432]

The question is incomplete. Here is the complete question.

Find the measurements (the lenght L and the width W) of an inscribed rectangle under the line y = -\frac{3}{4}x + 3 with the 1st quadrant of the x & y coordinate system such that the area is maximum. Also, find that maximum area. To get full credit, you must draw the picture of the problem and label the length and the width in terms of x and y.

Answer: L = 1; W = 9/4; A = 2.25;

Step-by-step explanation: The rectangle is under a straight line. Area of a rectangle is given by A = L*W. To determine the maximum area:

A = x.y

A = x(-\frac{3}{4}.x + 3)

A = -\frac{3}{4}.x^{2}  + 3x

To maximize, we have to differentiate the equation:

\frac{dA}{dx} = \frac{d}{dx}(-\frac{3}{4}.x^{2}  + 3x)

\frac{dA}{dx} = -3x + 3

The critical point is:

\frac{dA}{dx} = 0

-3x + 3 = 0

x = 1

Substituing:

y = -\frac{3}{4}x + 3

y = -\frac{3}{4}.1 + 3

y = 9/4

So, the measurements are x = L = 1 and y = W = 9/4

The maximum area is:

A = 1 . 9/4

A = 9/4

A = 2.25

6 0
3 years ago
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