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blagie [28]
3 years ago
11

Find the missing side length. Round to the nearest tenth.15 cm9 cm​

Mathematics
1 answer:
vazorg [7]3 years ago
5 0
Using Pythagorean’s theorem
a^2 +b^2= c^2
We know
c=15
b=9
a=?
Sub that in the Pythagorean’s theorem
a^2+9^2 = 15^2
a^2 + 81 = 225
a^2= 225-81
a^2 = 144
a= square root of 144
a= 12

Therefore the answer is 12
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Jet001 [13]

Answer:

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Hi,so I've tried solving this problem for a while and I just don't get it,if you could give me some ideas I would appreciate it.
ziro4ka [17]

Answer:

187 cm²

Step-by-step explanation:

The bottom rectangle area is easy, it is 15*9 = 135 cm².

To find the area of the triangle, you only need the base width and its height (you don't need the hypotenuse). You can then use the formula: area triangle is base times half height.

The base width is 15-7 = 8 cm

The height is 22-9 = 13 cm

So the area of the triangle is 13*8/2 = 52 cm²

Together with the 135 of the rectangle that sums to 52+135 = 187 cm².

4 0
3 years ago
(22+x-7)=(x-5)how do u do long division <br>​
photoshop1234 [79]

Answer:

there is no solution to this problem?

Step-by-step explanation:

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3 0
3 years ago
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can someone plz help me on this plz I beg u
STALIN [3.7K]

Answer:

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6 0
3 years ago
Read 2 more answers
i f N(-7,-1) is a point on the terminal side of ∅ in standard form, find the exact values of the trigonometric functions of ∅.
Natali [406]

Answer:

sin\ \varnothing = \frac{-1}{10}\sqrt 2

cos\ \varnothing = \frac{-7}{10}\sqrt 2

tan\ \varnothing = \frac{1}{7}

cot\ \varnothing = 7

sec\ \varnothing = \frac{-5}{7}\sqrt 2

csc\ \varnothing = -5\sqrt 2

Step-by-step explanation:

Given

N = (-7,-1) --- terminal side of \varnothing

Required

Determine the values of trigonometric functions of \varnothing.

For \varnothing, the trigonometry ratios are:

sin\ \varnothing = \frac{y}{r}       cos\ \varnothing = \frac{x}{r}       tan\ \varnothing = \frac{y}{x}

cot\ \varnothing = \frac{x}{y}       sec\ \varnothing = \frac{r}{x}       csc\ \varnothing = \frac{r}{y}

Where:

r^2 = x^2 + y^2

r = \sqrt{x^2 + y^2

In N = (-7,-1)

x = -7 and y = -1

So:

r = \sqrt{(-7)^2 + (-1)^2

r = \sqrt{50

r = \sqrt{25 * 2

r = \sqrt{25} * \sqrt 2

r = 5 * \sqrt 2

r = 5 \sqrt 2

<u>Solving the trigonometry functions</u>

sin\ \varnothing = \frac{y}{r}

sin\ \varnothing = \frac{-1}{5\sqrt 2}

Rationalize:

sin\ \varnothing = \frac{-1}{5\sqrt 2} * \frac{\sqrt 2}{\sqrt 2}

sin\ \varnothing = \frac{-\sqrt 2}{5*2}

sin\ \varnothing = \frac{-\sqrt 2}{10}

sin\ \varnothing = \frac{-1}{10}\sqrt 2

cos\ \varnothing = \frac{x}{r}

cos\ \varnothing = \frac{-7}{5\sqrt 2}

Rationalize

cos\ \varnothing = \frac{-7}{5\sqrt 2} * \frac{\sqrt 2}{\sqrt 2}

cos\ \varnothing = \frac{-7*\sqrt 2}{5*2}

cos\ \varnothing = \frac{-7\sqrt 2}{10}

cos\ \varnothing = \frac{-7}{10}\sqrt 2

tan\ \varnothing = \frac{y}{x}

tan\ \varnothing = \frac{-1}{-7}

tan\ \varnothing = \frac{1}{7}

cot\ \varnothing = \frac{x}{y}

cot\ \varnothing = \frac{-7}{-1}

cot\ \varnothing = 7

sec\ \varnothing = \frac{r}{x}

sec\ \varnothing = \frac{5\sqrt 2}{-7}

sec\ \varnothing = \frac{-5}{7}\sqrt 2

csc\ \varnothing = \frac{r}{y}

csc\ \varnothing = \frac{5\sqrt 2}{-1}

csc\ \varnothing = -5\sqrt 2

3 0
2 years ago
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