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3 years ago

A3/8

Mathematics

1 answer:

mart [117]3 years ago

5 0

Answer:

A.)130 80

Step-by-step explanation:

1y/3x

-8x/1xy

answer:130 80

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A player scored 92 92 points in a single professional basketball game. he made a total of 58 58 baskets, consisting of field go

gregori [183]

Hello,

x → field goals
y → foul shorts
x + y = 58
+( -2x + y = 92 )
--------------------------
-x + 0y = -34
x = 34
y = 58 - 34 = 24

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3 years ago

I need some help with this

arsen [322]

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3 years ago

Rectangle ABCD has vertex coordinates A(1,-2), B(4, -2), C(4,-4), and D(1,

dimulka [17.4K]

Answer:(7,-3)

Step-by-step explanation:

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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

Hi guys can I have some help here with year 6 rounding? <br><br> (Send me screen shot)

alexira [117]

Answer:

nearest ten thousand – 2,620,000

nearest hundred thousand – Two million, six hundred thousand

Step-by-step explanation:

hope this helps.

5 0

3 years ago