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2 years ago

Use the segment addition postulate to solve for BC when AB = 37 and AC

Mathematics

1 answer:

Ivan2 years ago

4 0

Answer:

BC = 56

Step-by-step explanation:

Segment Addition Postulate: AB + BC = AC

37 + BC = 93
Subtract 37 from both sides
BC = 56

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4a to 3rd power b to 4power times b to 3rd power times 2b to 3rd power divided by 4b

ddd [48]

(4a^3b^4(b^3)(2b^3))/4b

4a^3b^3(b^2)(2b^2)/4

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2 years ago

Dan earns a commission of 15% on his sales. Last month his sales were: $550.23, $793.23,

elixir [45]

9514 1404 393

Answer:

$533.86

Step-by-step explanation:

The sum of the sales amounts is $3,559.07. Dan's commission is ...

0.15 × $3,559.07 = $533.86

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2 years ago

F(x) = 2x + 3g(x) = 2x² + 6x + 13Find: (gºf)(x)

Butoxors [25]

Answer:

The function (gof)(x) is;

$(g\circ f)(x)=8x^2+36x+49$

Explanation:

Given the functions;

$\begin{gathered} f(x)=2x+3 \\ g(x)=2x^2+6x+13 \end{gathered}$

Solving for the function;

$(g\circ f)(x)=g(f(x))$

so, we have;

$\begin{gathered} g(f(x))=2(f(x))^2+6(f(x))+13 \\ g(f(x))=2(2x+3)^2+6(2x+3)+13 \\ g(f(x))=2(4x^2+12x+9)^{}+6(2x)+6(3)+13 \\ g(f(x))=8x^2+24x+18^{}+12x+18+13 \\ g(f(x))=8x^2+24x^{}+12x+18+18+13 \\ g(f(x))=8x^2+36x+49 \end{gathered}$

Therefore, the function (gof)(x) is;

$(g\circ f)(x)=8x^2+36x+49$

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3 months ago

What are the solutions to x2 – 10x = 39?

oee [108]

Answer: x=−3 or x=13

Step-by-step explanation:

Step 1: Subtract 39 from both sides.

x2−10x−39=39−39

x2−10x−39=0

Step 2: Factor left side of equation.

(x+3)(x−13)=0

Step 3: Set factors equal to 0.

x+3=0 or x−13=0

x=−3 or x=13

7 0

1 year ago

Find an equation of a line through √3,1 / 2) parallel to the line:

vagabundo [1.1K]

Answer:

(a) $x=\sqrt{3}$

(b) $y=\frac{1}{2}$

(c) Answer of part a is equation of line $x=\sqrt{3}$ which is parallel to y axis and in part by equation of line is $y=\frac{1}{2}$ which is parallel to x axis

Step-by-step explanation:

We have given points are $(\sqrt{3},\frac{1}{2})$

(a) Equation of line is given x = -9

We have to find the equation of line parallel to x= -9 and passing through $(\sqrt{3},\frac{1}{2})$

x = -9 line is parallel to y axis and so slope will be infinite

Equation of line is given by $y-y_1=m(x-x_1)$

So $y-\frac{1}{2}=\frac{1}{0}(x-\sqrt{3})$

$x=\sqrt{3}$

(b) Equation of line is given $y=-\sqrt{7}$

This line is parallel to x axis so slope will be zero

So equation of line will be $y-\frac{1}{2}=0(x-\sqrt{3})$

$y=\frac{1}{2}$

(c) Answer of part a is equation of line $x=\sqrt{3}$ which is parallel to y axis and in part by equation of line is $y=\frac{1}{2}$ which is parallel to x axis

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3 years ago