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AlexFokin [52]
2 years ago
7

Identify the term that completes the equation. AC^2 = (DC)(?) BC AD BD AB

Mathematics
1 answer:
fiasKO [112]2 years ago
3 0

Answer:

BC

Step-by-step explanation:

the height of a right-angled triangle to the Hypotenuse is the square root of the product of the lengths of the 2 segments of the Hypotenuse.

so the square of the height is then the product of the 2 line segments of the Hypotenuse.

therefore

AC² = DC × BC

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Simplify -4x - x. Plzzzz Help
Monica [59]

-4x - x

Combine like terms

= -5x

Hope it helps

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3 years ago
Stacey and Michele both babysit in the evening.staceys hours rate is $10 for one child plus $2per additional child Michael hourl
lara31 [8.8K]
They have two children...
8 + 3c = 10 + 2c
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7 0
3 years ago
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Paige borrows $700 from her parents for a new computer. She repays $100 the first month, then begins to repay her parents $75 pe
GrogVix [38]
The recursive formula that models the amount owed is
an = a(n-1) -75; a1 = 600. 
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7 0
3 years ago
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A hiker is hiking in a valley. The height of the valley is h(x,y)=4x2+y2 where x and y are the east-west and north-south distanc
Ainat [17]

Answer:

A. \frac{\partial{h}}{\partial{t}}=0

Step-by-step explanation:

A. The problems asked for 2 ways to solve it, expanding the equation with the substitution  x(t)=2 cos(t) and y(t)=4 sin(t) to differentiate it . The other way is by chain rule.

Expanding and differentiating:

We start by substituting x(t)=2 cos(t) and y(t)=4 sin(t) in h(x,y)=4x2+y2:

h(x,y)=4x^{2}+y^{2}= 4(2cos(t))^{2}+(4sin(t))^{2}\\h(x,y)=4(4cos^{2}(t))+(16sen^{2}(t))\\h(x,y)=16cos^{2}(t)+16sen^{2}(t)=16(sen^{2}(t)+cos^{2}(t))\\h(x,y)=16

So, in the path that the hiker chose:

\frac{\partial{h}}{\partial{t}}=0

Chain rule:

We start differentiating h(x,y) using chain rule as follows:

\frac{\partial{h}}{\partial{t}}= \frac{\partial{h}}{\partial{x}}\frac{\partial{x}}{\partial{t}}+\frac{\partial{h}}{\partial{y}}\frac{\partial{y}}{\partial{t}}

Now, it´s easy to find all these derivatives:

\frac{\partial{h}}{\partial{x}}=8x\\\frac{\partial{x}}{\partial{t}}=-2sin(t)\\\frac{\partial{h}}{\partial{y}}=2y\\\frac{\partial{y}}{\partial{t}}=4cos(t)

Now we replace them in the chain rule, with the replacement x=2cos(t) and y=4sin(t) in the x,y that are left and we operate everything:

\frac{\partial{h}}{\partial{t}}= 8x(-2sin(t))+2y(4cos(t)

\frac{\partial{h}}{\partial{t}}= 8(2cos(t))(-2sin(t))+2(4sin(t))(4cos(t)

\frac{\partial{h}}{\partial{t}}= -32cos(t)sin(t)+32sin(t)cos(t)

\frac{\partial{h}}{\partial{t}}= 0

This will be our answer

6 0
3 years ago
The roots of x2 - 2x - 2 = 0 are
WINSTONCH [101]

Answer:

-2X=O

X=0

THE ANSWER IS X=0

Step-by-step explanation:

Hopes it helps

7 0
3 years ago
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