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salantis [7]
3 years ago
9

How to I find this? I tried using Pythagorean theorem, but it didn't work

Mathematics
2 answers:
jarptica [38.1K]3 years ago
4 0
The Pythagorean theorem works.

12^2+9^2=c^2
144+81=c^2
225=c^2
15=c

Therefore the perimeter is 12 + 9 + 15 = 36 in. (B)
pentagon [3]3 years ago
4 0
Then you applied it incorrectly

for a right triangle with legs legnth a and b and hypotonuse c,
a^2+b^2=c^2
hypotonuse=longest side or side oposite right angle



so we see that
hypotnuse=c=unknown
legs are legnth a and b or12 and 9
plug in
a=12
b=9
c=c
a^2+b^2=c^2
12^2+9^2=c^2
144+81=c^2
125=c^2
sqrt both sides
15=c

the legnth of the hypotnuse=15

to find the perimiter, you add all sides up
a+b+c=
12+9+15=
36

answer is 36 inches


B is answer
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Answer: 126^{\circ}

Step-by-step explanation:

If we draw RP and PS, then in the quadrilateral formed, we know that \angle PRQ=\angle PSQ=90^{\circ}, and since angles in a quadrilateral add to 360 degrees, this means \angle SPR=126^{\circ}.

Thus, arc RS also measures 126 degrees

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Look at the figure below: an image of a right triangle is shown with an angle labeled y If sin y° = 7 divided by q and tan y° =
Andre45 [30]

Answer:

Cosy° = \frac{r}{q}

Step-by-step explanation:

From the picture attached,

By applying all trigonometric ratios in the given right triangle,

Siny° = \frac{7}{q} = \frac{\text{Opposite side}}{\text{Hypotenuse}}

tany° = \frac{7}{r} = \frac{\text{Opposite side}}{\text{Adjacent side}}

Therefore, Cosy° = \frac{\text{Adjacent side}}{\text{Hypotenuse}}

Cosy° = \frac{r}{q}

Therefore, Cosy° = \frac{r}{q} will be the answer.

5 0
3 years ago
Find the area of a sector with a central angle of 120° and a diameter of 7.3 cm. Round to the nearest tenth.
Andru [333]

Answer:14cm^{2}

Step-by-step explanation:

diameter=7.3cm

radius=\frac{diameter}{2}

          =\frac{7.3}{2}

          =3.65cm

angle =120°

area=\frac{angle (tita)}{360} × \frac{\pi r^{2} }{1}

=\frac{120}{360} × \frac{22}{7} ×3.65^{2}

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6 0
3 years ago
Solve 5x^2 − 3x + 17 = 9.
Vedmedyk [2.9K]

Answer:

The roots of the equation are x = \frac{3+\sqrt{151}i}{10} and x = \frac{3-\sqrt{151}i}{10}

and there are no real roots of the equation given above

Step-by-step explanation:

To solve:

5x² − 3x + 17 = 9

or

⇒ 5x² − 3x + 17 - 9 = 0

or

⇒ 5x² − 3x + 8 = 0

Now,

the roots of the equation in the form ax² + bx + c = 0 is given as:

x = \frac{-b\pm\sqrt{b^2-4ac}}{2a}

in the above given equation

a = 5

b = -3

c = 8

therefore,

x = \frac{-(-3)\pm\sqrt{(-3)^2-4\times5\times8}}{2\times5}

or

x = \frac{3\pm\sqrt{9-160}}{10}

or

x = \frac{3+\sqrt{-151}}{10} and x = \frac{3-\sqrt{-151}}{10}

or

x = \frac{3+\sqrt{151}i}{10} and x = \frac{3-\sqrt{151}i}{10}

here i = √(-1)

Hence,

The roots of the equation are x = \frac{3+\sqrt{151}i}{10} and x = \frac{3-\sqrt{151}i}{10}

and there are no real roots of the equation given above

7 0
2 years ago
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