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MAXImum [283]
2 years ago
6

PLEASE HELP!!!!!!!!!!!!!!!!!!!!!!!!

Mathematics
2 answers:
Karolina [17]2 years ago
8 0

Answer:

I think it is E 3000

because area of a triangle is

1\2 base * height

base = BC = 120

height = AH = 100

120\2 = 60

60 * 100 = 6000

6000\2 = 3000

charle [14.2K]2 years ago
3 0

Answer:

I think it is C.2920

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Which function is the inverse of f(x) = 2x + 3 ?<br> Help
bearhunter [10]

Answer:

The inverse is 1/2x -3/2

Step-by-step explanation:

y =2x+3

Exchange x and y

x = 2y+3

Solve for y, subtracting 3 from each side

x-3 = 2y+3-3

x-3 =2y

Divide each side by 2

(x-3)/2 = 2y/2

1/2x - 3/2 =y

The inverse is 1/2x -3/2

7 0
3 years ago
Read 2 more answers
What is the unknown fraction?
vaieri [72.5K]

Answer:

38/100

Step-by-step explanation:

.................................................

3 0
2 years ago
I need help asap :(:):)
zubka84 [21]
I think it’s the 3rd one
8 0
3 years ago
What is the area of the region bounded between the curves y=x and y=sqrt(x)?
erastova [34]

Answer:

\frac{1}{6}

Step-by-step explanation:

y = x     .....(1)

y=\sqrt{x}     .... (2)

By solving equation (1) and equation (2)

x = \sqrt{x}

\sqrt{x}\left ( \sqrt{x}-1 \right )=0

\sqrt{x}=0 or \left ( \sqrt{x}-1 \right )=0

x = 0, x = 1

y = 0, y = 1

A = \int_{0}^{1}ydx(curve)-\int_{0}^{1}ydx(line)

A = \int_{0}^{1}\sqrt{x}dx-\int_{0}^{1}xdx

A = \frac{2}{3}[x^\frac{3}{2}]_{0}^{1}-\frac{1}{2}[x^2]_{0}^{1}

A = \frac{2}{3}-\frac{1}{2}

A = \frac{1}{6}

8 0
3 years ago
Determine whether the equation defines y as a function of x.<br> x+y=9 and x^2+y^2=1 and x=y^2
Murrr4er [49]

Answer:

Function: x+y=9

Not Function: x^2+y^2=1    and     x=y^2

Step-by-step explanation:

Given

x+y=9

x^2+y^2=1

x=y^2

Required

Determine if y is a function of x

Solving x+y=9

x+y=9

Make y the subject of formula

y = 9 - x

<em>Hence; y is a function of x</em>

Solving x^2+y^2=1

x^2+y^2=1

Subtract x² from both sides

y^2=1 - x^2

Square root of both sides

y =\± \sqrt{1 - x^2}

This implies that

y =\sqrt{1 - x^2}     or     y =-\sqrt{1 - x^2}

<em>Because </em>y<em> can be any of those two expressions, it is not a function.</em>

Solving x=y^2

x=y^2

Reorder

y^2 = x

Take square roots

y = \±\sqrt{x}

This implies that

y = \sqrt{x}      or        y = -\sqrt{x}

<em>Because </em>y<em> can be any of those two expressions, it is not a function.</em>

4 0
3 years ago
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