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JulijaS [17]
3 years ago
15

Can a quadrilateral have 4 obtuse angles?

Mathematics
1 answer:
Viktor [21]3 years ago
7 0

NO.

The sum of their angles is 360 degrees. They can be right angles.

If one of them is obtuse then one must be of acute.

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The Harrison family paid $14,400 in rent last year. The landlord has decided to raise rent 2.5% each year. How much will the Har
skelet666 [1.2K]

Answer: $16,200

Step-by-step explanation:

14,400 rent.

Raise of 2.5% (0.025)/year.

5 years.

Take the rent (14,400) and add 2.5% (0.025) of it (14400) for five (5) years.

14400+0.025(14400)(5)

14400+1800\\=16200

6 0
2 years ago
Find the difference quotient and simplify your answer. f(x) = 4x − x2, f(4 + h) − f(4) h , h ≠ 0
Anon25 [30]

The difference quotient and simplification will be    = [4 -h-2x]

The given equation is as follows:   f(x)= 4x - x²

For finding the quotient and further simplification we must follow the following steps:

[f(x + h) - f(x)] / h = [4(x + h) - (x + h)² - 4x + x²]/ h

<h3>What is simplification of algebraic operations?</h3>

Getting the functions in their lowest terms is known as simplification.

Brackets will get open and solved further;

[f(x + h) - f(x)] / h = [4(x + h) - (x + h)² - 4x + x²]/ h

[f(x + h) - f(x)] / h = [4h - h² - 2x]/ h  

Finally dividing the whole equation with h;

                                = [4 - h - 2x] 

Learn more about algebraic operations,

brainly.com/question/12485460

# SPJ1

7 0
1 year ago
The area of the larger circle is??
nevsk [136]

radius of small circle = 16 m

radius of large circle is doubled so it will be 32 m


Area of the larger circle = π x 32^2 = 1,204π m^2


answer

1,204π m^2

7 0
3 years ago
Read 2 more answers
On a linear X temperature scale, water freezes at −115.0°X and boils at 325.0°X. On a linear Y temperature scale, water freezes
belka [17]

Answer:

The current temperature on the X scale is 1150 °X.

Step-by-step explanation:

Let is determine first the ratio of change in X linear temperature scale to change in Y linear temperature scale:

r = \frac{\Delta T_{X}}{\Delta T_{Y}}

r = \frac{325\,^{\circ}X-(-115\,^{\circ}X)}{-25\,^{\circ}Y - (-65.00\,^{\circ}Y)}

r = 11\,\frac{^{\circ}X}{^{\circ}Y}

The difference between current temperature in Y linear scale with respect to freezing point is:

\Delta T_{Y} = 50\,^{\circ}Y - (-65\,^{\circ}Y)

\Delta T_{Y} = 115\,^{\circ}Y

The change in X linear scale is:

\Delta T_{X} = r\cdot \Delta T_{Y}

\Delta T_{X} = \left(11\,\frac{^{\circ}X}{^{\circ}Y} \right)\cdot (115\,^{\circ}Y)

\Delta T_{X} = 1265\,^{\circ}X

Lastly, the current temperature on the X scale is:

T_{X} = -115\,^{\circ}X + 1265\,^{\circ}X

T_{X} = 1150\,^{\circ}X

The current temperature on the X scale is 1150 °X.

5 0
3 years ago
Given that g(x)=x-3/x+4 find each of the following.
maks197457 [2]

Given that the function g(x)=x-3/x+4, the evaluation gives:

  1. g(9) = 6/13.
  2. g(3) = 0.
  3. g(-4) = undefined.
  4. g(-18.75) = 1.07.
  5. g(x+h) = x+h-3/x+h+4

<h3>How to evaluate the function?</h3>

In this exercise, you're required to determine the value of the function g at different intervals. Thus, we would substitute the given value into the function and then evaluate as follows:

When g = 9, we have:

g(x)=x-3/x+4

g(9) = 9-3/9+4

g(9) = 6/13.

When g = 3, we have:

g(x)=x-3/x+4

g(3) = 3-3/3+4

g(3) = 0/13.

g(3) = 0.

When g = -4, we have:

g(x)=x-3/x+4

g(-4) = -4-3/-4+4

g(-4) = -1/0.

g(-4) = undefined.

When g = -18.75, we have:

g(x)=x-3/x+4

g(-18.75) = -18.75-3/-18.75+4

g(-18.75) = -15.75/-14.75.

g(-18.75) = 1.07.

When g = x+h, we have:

g(x)=x-3/x+4

g(x+h) = x+h-3/x+h+4

Read more on function here: brainly.com/question/17610972

#SPJ1

8 0
2 years ago
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