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Nikolay [14]
3 years ago
5

I have 12 coins in nickels and dimes. if the total value of these coins is 85 cents, how many nickels are in this collection

Mathematics
1 answer:
tiny-mole [99]3 years ago
8 0
You would have a equal amount of both dimes and nickels. Therefore 6 dimes and 6 nickels is the answer.

please vote my answer brainliest. thanks!
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Which function describes this graph a. y=(x+5)(x-4) b. y=(x-3)(x-6) c. y=x^2-2x+4 d. y=x^2*9x+18
lara [203]

Answer:

I believe that the answer is B

6 0
3 years ago
The Greatest Problems on Campus
PSYCHO15rus [73]

Answer: D.) This is an example of inductive reasoning because a general conclusion is reached based on a specific example,

Step-by-step explanation: Inductive reasoning simply refers to making conclusion about a specific subject or topic from patterns or insights derived from related examples. In the scenario above, the conclusion reached encompasses the overall full time 4 years college student. However, this conclusion was inferred based on a specific example comprising of only a randomized sample of 1200 full time 4 years college students in 100 campuses. random. The example failed to incorporate every student, Hence, the conclusion is induced as the choice of a sample of students may not convey the choice or decision of all.

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8 0
3 years ago
Multiply.
Nady [450]

\boxed{3\sqrt{22}\sqrt{58}\sqrt{18}=18\sqrt{638}}

<h2>Explanation:</h2>

Here we have the following expression:

3\sqrt{22}\sqrt{58}\sqrt{18}

So we need to simplify that radical expression. By property of radicals we know that:

\sqrt[n]{a}\sqrt[n]{b}=\sqrt[n]{ab}

So:

3\sqrt{22}\sqrt{58}\sqrt{18}=3\sqrt{22\times 58 \times 18}=3\sqrt{22968}

The prime factorization of 22968 is:

22968=2^3\cdot 3^2\cdot11\cdot 29

Hence:

3\sqrt{22968}=3\sqrt{2^3\cdot 3^2\cdot11\cdot 29}=3\sqrt{2^2\cdot 3^2\cdot 2\cdot 11\cdot 29}

By property:

\sqrt[n]{a^n}=a

So:

3\sqrt{2^2\cdot 3^2\cdot 2\cdot 11\cdot 29} \\ \\ 3(2)(3)\sqrt{2\cdot 11\cdot 29}=18\sqrt{638}

Finally:

\boxed{3\sqrt{22}\sqrt{58}\sqrt{18}=18\sqrt{638}}

<h2>Learn more:</h2>

Radical expressions: brainly.com/question/13452541

#LearnWithBrainly

3 0
3 years ago
A circle has a radius of 6 inches. Which is the closest to the area, in square inches, of the circle?
Gelneren [198K]

Answer:

131.1 in^2

Step-by-step explanation:

a=πr^2

7 0
3 years ago
Suppose a, b denotes of the quadratic polynomial x² + 20x - 2022 &amp; c, d are roots of x² - 20x + 2022 then the value of ac(a
Alja [10]
<h3><u>Correct Question :- </u></h3>

\sf\:a,b \: are \: the \: roots \: of \:  {x}^{2} + 20x - 2020 = 0 \: and \:  \\  \sf \: c,d \: are \: the \: roots \: of \:  {x}^{2}  -  20x  + 2020 = 0 \: then \:

\sf \: ac(a - c) + ad(a - d) + bc(b - c) + bd(b - d) =

(a) 0

(b) 8000

(c) 8080

(d) 16000

\large\underline{\sf{Solution-}}

Given that

\red{\rm :\longmapsto\:a,b \: are \: the \: roots \: of \:  {x}^{2} + 20x - 2020 = 0}

We know

\boxed{\red{\sf Product\ of\ the\ zeroes=\frac{Constant}{coefficient\ of\ x^{2}}}}

\rm \implies\:ab = \dfrac{ - 2020}{1}  =  - 2020

And

\boxed{\red{\sf Sum\ of\ the\ zeroes=\frac{-coefficient\ of\ x}{coefficient\ of\ x^{2}}}}

\rm \implies\:a + b = -  \dfrac{20}{1}  =  - 20

Also, given that

\red{\rm :\longmapsto\:c,d \: are \: the \: roots \: of \:  {x}^{2}  -  20x  + 2020 = 0}

\rm \implies\:c + d = -  \dfrac{( - 20)}{1}  =  20

and

\rm \implies\:cd = \dfrac{2020}{1}  = 2020

Now, Consider

\sf \: ac(a - c) + ad(a - d) + bc(b - c) + bd(b - d)

\sf \:  =  {ca}^{2} -  {ac}^{2} +  {da}^{2} -  {ad}^{2} +  {cb}^{2} -  {bc}^{2} +  {db}^{2} -  {bd}^{2}

\sf \:  =  {a}^{2}(c + d) +  {b}^{2}(c + d) -  {c}^{2}(a + b) -  {d}^{2}(a + b)

\sf \:  = (c + d)( {a}^{2} +  {b}^{2}) - (a + b)( {c}^{2} +  {d}^{2})

\sf \:  = 20( {a}^{2} +  {b}^{2}) + 20( {c}^{2} +  {d}^{2})

\sf \:  = 20\bigg[ {a}^{2} +  {b}^{2} + {c}^{2} +  {d}^{2}\bigg]

We know,

\boxed{\tt{  { \alpha }^{2}  +  { \beta }^{2}  =  {( \alpha   + \beta) }^{2}  - 2 \alpha  \beta  \: }}

So, using this, we get

\sf \:  = 20\bigg[ {(a + b)}^{2} - 2ab +  {(c + d)}^{2} - 2cd\bigg]

\sf \:  = 20\bigg[ {( - 20)}^{2} +  2(2020) +  {(20)}^{2} - 2(2020)\bigg]

\sf \:  = 20\bigg[ 400 + 400\bigg]

\sf \:  = 20\bigg[ 800\bigg]

\sf \:  = 16000

Hence,

\boxed{\tt{ \sf \: ac(a - c) + ad(a - d) + bc(b - c) + bd(b - d) = 16000}}

<em>So, option (d) is correct.</em>

4 0
2 years ago
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