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2 years ago

(a-2)^2=64 Please help solve?

Mathematics

2 answers:

Ann [662]2 years ago

8 0

Answer:

Step-by-step explanation:

a=10

AysviL [449]2 years ago

8 0

Answer:

a=10

Step-by-step explanation:

$(a-2)^{2}$ =64

$(a-2)^{2}$ = $8^{2}$

cancel the squares,

a-2=8

a=10

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The net change for a certain stock is the dollar value change in the stock's closing price from the previous day's closing price

Natali [406]

-3 has the greatest absolute value.

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3 years ago

What is the length of BC

elena55 [62]

Answer:

BC = 2

Step-by-step explanation:

We first need to use triangle BAD to find BD,

BD is the hypotenuse, so use pythagorean theorem...

(√23)² + (√13)² = (BD)²

23 + 13 = (BD)²

36 = (BD)²

6 = BD

Now BD is the hypotenuse of triangle BCD, so use pythagorean theorem agian to find BC

(4√2)² + (BC)² = 6²

32 + (BC)² = 36

(BC)² = 4

BC = 2

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3 years ago

What square root is larger that the square root of 17

Ymorist [56]

Answer:

i would have to say the square root of 26 since its 5.099

Step-by-step explanation:

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3 years ago

Find the price-demand equation for a particular brand television when the demand is 20 TVs per week at $150 per TV, given that t

Alina [70]

Answer:

~ $150=50e^{-0.01()20}$

~ $p(100)=\$127.7$

Step-by-step explanation:

From the question we are told that:

Price of 20TVs per week $P_{20}=\$150$

Marginal price-demand function $p'(x)=-0.5e-0.01x$

Generally the The Marginal price function is mathematically given by

$p'(x)=-0.5e^{-0.01x}$

$p(x)=\int-0.5e^{-0.01x}$

$p(x)=50e^{-0.001x}+C$

Therefore the equation when the demand is 20 TVs per week at $150 per TV

$150=50e^{-0.01()20}$

Giving

$p(x)=50e^{-0.01x}+150-50e^{-0.01(20)}$

Therefore the Price when the demand is 100 TVs per week

$p(100)=50e^{-0.01(100)}+150-50e^{-0.01(20)}$

$p(100)=\$127.7$

7 0

2 years ago

I really really really need help!!!!

Yakvenalex [24]

For $f$ to be continuous at $x=1$ , you need to have the limit from either side as $x\to1$ to be the same.

$\displaystyle\lim_{x\to1^-}f(x)=\lim_{x\to1^-}(|x-1|+2)=2$
$\displaystyle\lim_{x\to1^+}f(x)=\lim_{x\to1^+}(ax^2+bx)=a+b$

If $a=2$ and $b=3$ , then the limit from the right would be $2+3=5\neq2$ , so the answer to part (1) is no, the function would not be continuous under those conditions.

This basically answers part (2). For the function to be continuous, you need to satisfy the relation $a+b=2$ .

Part (c) is done similarly to part (1). This time, you need to limits from either side as $x\to2$ to match. You have

$\displaystyle\lim_{x\to2^-}f(x)=\lim_{x\to2^-}(ax^2+bx)=4a+2b$
$\displaystyle\lim_{x\to2^+}f(x)=\lim_{x\to2^+}(5x-10)=0$

So, $a$ and $b$ have to satisfy the relation $4a+2b=0$ , or $2a+b=0$ .

Part (4) is done by solving the system of equations above for $a$ and $b$ . I'll leave that to you, as well as part (5) since that's just drawing your findings.

8 0

2 years ago