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4 years ago

Please help meeeeee :///

Mathematics

2 answers:

eimsori [14]4 years ago

5 0

Wouldn't all the angles equal 180? We've been trying to learn these but I think so. Subtract both numbers by 180 and it should be the answer. Sorry if it's wrong but I'm not sure. I kinda hope this helps you any

xz_007 [3.2K]4 years ago

5 0

6) x=14+132
x=146
The measure angle x is 146°

7)74=38+x
-38 -38
36=x
Therefore the measure angle x us equal to 36°

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Remember to simplify your answer (make improper fractions mixed numbers) 3/4÷1/5

hoa [83]

Answer:

4/15

Step-by-step explanation:

5 0

3 years ago

I need help this please!

sdas [7]

Answer:

I donot know

Step-by-step explanation:

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6 0

3 years ago

If A ABC ~ ARST and m 2 A = 42°, m 2 C = 60°,<br> find the measure of < S.

Free_Kalibri [48]

Answer:

Step-by-step explanation:

∠A + ∠B + ∠C = 180 { Angle sum property of triangle}

42 + ∠B + 60 = 180

∠B + 102 = 180

∠B = 180 - 102 = 78

In similar triangles, corresponding angles are congruent.

∠S ≅ ∠B

∠S = 78°

7 0

3 years ago

The domain of g(x) = 4x –12 is {2, 3, 6, 8}. What is the range

vlada-n [284]

Answer:

Range - {-4,0,12,20}

Step-by-step explanation:

Given that,

The function is :

g(x) = 4x –12

The domain of the function is {2, 3, 6, 8}.

g(2) = 4(2) –12 = -4

g(3) = 4(3) –12 = 0

g(6) = 4(6) –12 = 12

g(8) = 4(8) –12 = 20

Hence, the range of the function is {-4,0,12,20}.

7 0

3 years ago

Please help! Will mark the brainliest!

JulsSmile [24]

The domain is the set of all possible x-values which will make the function valid.
$f(x) = \frac{6}{x+3} \ \ \ \ , \ g(x) = \frac{1}{4-x}$
For the given function The denominator of a fraction cannot be zero

(a)

(1) The domain of f ⇒⇒⇒ R - {-3}

Because ⇒⇒⇒ x+3 = 0 ⇒⇒⇒ x =-3

(2) The domain of g ⇒⇒⇒ R - {4}
Because: 4 - x = 0 ⇒⇒⇒ x = 4

(3) $f + g = \frac{6}{x+3} + \frac{1}{4-x} = \frac{6(4-x)+(x+3)}{(x+3)(4-x)}$
The domain of (f+g) ⇒⇒⇒ R - {-3,4}
because: x+3 = 0 ⇒⇒⇒ x = -3   and    4 - x = 0 ⇒⇒⇒ x = 4

(4) $f - g = \frac{6}{x+3} - \frac{1}{4-x} = \frac{6(4-x)-(x+3)}{(x+3)(4-x)}$
The domain of (f-g) ⇒⇒⇒ R - {-3,4}
because: x+3 = 0 ⇒⇒⇒ x = -3   and    4 - x = 0 ⇒⇒⇒ x = 4

(5) $f * g = \frac{6}{x+3} * \frac{1}{4-x} = \frac{6}{(x+3)(4-x)}$
The domain of (f*g) ⇒⇒⇒ R - {-3,4}
because: x+3 = 0 ⇒⇒⇒ x = -3   and    4 - x = 0 ⇒⇒⇒ x = 4

(6) $f * f = \frac{6}{x+3} * \frac{6}{x+3} = \frac{36}{(x+3)^2}$
The domain of ff ⇒⇒⇒ R - {-3}

Because ⇒⇒⇒ x+3 = 0 ⇒⇒⇒ x =-3

(7) $\frac{f}{g} = \frac{\frac{6}{x+3} }{ \frac{1}{4-x} } = \frac{6(4-x)}{x+3}$
The domain of (f/g) ⇒⇒⇒ R - {-3,4}

because: x+3 = 0 ⇒⇒⇒ x = -3   and    4 - x = 0 ⇒⇒⇒ x = 4
(8) $\frac{g}{f} = \frac{ \frac{1}{4-x} }{ \frac{6}{x+3} } = \frac{x+3}{6(4-x)}$
The domain of (g/f) ⇒⇒⇒ R - {-3,4}

because: x+3 = 0 ⇒⇒⇒ x = -3   and    4 - x = 0 ⇒⇒⇒ x = 4
===================================================
(b)

(9) $(f+g)(x) = \frac{6}{x+3} + \frac{1}{4-x} = \frac{6(4-x)+(x+3)}{(x+3)(4-x)}$

∴ $(f + g)(x) = \frac{24-6x+x+3}{(x+3)(4-x)} = \frac{27-5x}{(x+3)(4-x)}$

(10) $(f - g)(x) = \frac{6}{x+3} - \frac{1}{4-x} = \frac{6(4-x)-(x+3)}{(x+3)(4-x)}$

∴ $(f - g)(x) = \frac{24-6x-x-3}{(x+3)(4-x)} = \frac{21 - 7x}{(x+3)(4-x)}$

(11) $(f * g)(x) = \frac{6}{x+3} * \frac{1}{4-x} = \frac{6}{(x+3)(4-x)}$

(12) $(f * f)(x) = \frac{6}{x+3} * \frac{6}{x+3} = \frac{36}{(x+3)^2}$

(13) $(\frac{f}{g})(x) = \frac{\frac{6}{x+3} }{ \frac{1}{4-x} } = \frac{6(4-x)}{x+3}$

(14) $(\frac{g}{f})(x) = \frac{ \frac{1}{4-x} }{ \frac{6}{x+3} } = \frac{x+3}{6(4-x)}$

===================================================

7 0

3 years ago