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3 years ago

Is it possible to draw a quadrilateral that is not a rectangle but has at least one right angle?

Mathematics

2 answers:

olasank [31]3 years ago

8 0

Answer: Triangle or Square

Step-by-step explanation:

Each have one right angle

joja [24]3 years ago

4 0

The answer is yes.

Explanation :

The only requirement for a quadrilateral is that it has four sides. It is not necessary for them to be equal sides. This implies that as long as the angles total up to 360, you may have any angles you desire.

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For questions 8 – 14, use the following functions:

pogonyaev

we are given

$f(x)=5$

$g(x)=\sqrt{x+2}$

(8)

$(f+g)(x)=f(x)+g(x)$

we can plug it

$(f+g)(x)=5+\sqrt{x+2}$

(9)

$(f-g)(x)=f(x)-g(x)$

we can plug it

$(f-g)(x)=5-\sqrt{x+2}$

(10)

$(f*g)(x)=f(x)*g(x)$

we can plug it

$(f*g)(x)=5\sqrt{x+2}$

(11)

$(\frac{f}{g} )(x)=\frac{f(x)}{g(x)}$

we can plug it

$(\frac{f}{g} )(x)=\frac{5}{\sqrt{x+2}}$

(12)

$(\frac{g}{f} )(x)=\frac{g(x)}{f(x)}$

we can plug it

$(\frac{g}{f} )(x)=\frac{\sqrt{x+2}}{5}$

(13)

$(fog)(x)=f(g(x))$

$f(x)=5$

we can replace g(x)

we get

$(fog)(x)=5$

(14)

$(gof)(x)=g(f(x))$

$f(x)=5$

we can replace f(x)

$(gof)(x)=\sqrt{f(x)+2}$

we get

$(gof)(x)=\sqrt{5+2}$

$(gof)(x)=\sqrt{7}$

8 0

3 years ago

if a point is chosen inside the large circle what is the probability that it will also be inside the small circle?

Ira Lisetskai [31]

Answer:

0.25

Step-by-step explanation:

hope that helps! ^w^

8 0

3 years ago

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Calculate the area of the red triangle to find the area of the garden. Show your work.

sergiy2304 [10]

Area of a triangle = [B(h)] : 2 = (ab) : 2 = [90ft x 120ft] : 2 = 5400ft squared.

5 0

2 years ago

Ohhhhhhh! I have a Brainliest

blondinia [14]

In week 2 it took 25 minutes also the mode is 75.

If correct brainliest please

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4 0

3 years ago

Find the Arc Length AND Sector Area from the circle. Use 3.14 for pi. Do not round your answer.

Leokris [45]

Answer:

Arc Length : 43.98

Sector Area : 175.92

Step-by-step explanation:

Formula for Arc Length : 2·π·r·Ф / 360

Formula for Sector Area : π· $r^{2}$ ·Ф / 360

- Where r is the radius

- Where Ф is the angle

Arc Length : 2 · π · 8 · 315 / 360 = 43.98

Sector Area : π · $8^{2}$ · 315 / 360 = 175.92

hope this helps !! :D

7 0

3 years ago

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