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garik1379 [7]
4 years ago
10

What is the value of the expression 6z2+2z+10 when z=3?

Mathematics
1 answer:
grigory [225]4 years ago
5 0

Hi there! Hope this helps, and if it does, please mark brainliest!

Answer:

346

Step-by-step explanation:

6z2+2z+10

(6 x 3)^2 + 2 x 3 + 10

18^2 + 12 + 10

324 + 22

346

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In circle V, a diameter and one chord intersect at point W. Find tha value of x. A. 4.9 B. 6.3 C. 8 D. 20
solmaris [256]
V - the center of a circle;
r = 10
In the triangle VWS :
WS = x
VS = 10 ( hypotenuse )
VW = 10 - 4 = 6
We will use the Pythagorean theorem:
x² + 6² = 10²
x² + 36 = 100
x² = 100 - 36
x² = 64
x = √ 64 = 8
Answer:
C ) 8
6 0
4 years ago
Hi. I need help with this. If anyone could explain how to solve this problem it would be greatly appreciated!
vaieri [72.5K]

Answer: Just simply solve the problem, i'm sorry but i cant see the problem do you think you could just copy the question down?

Step-by-step explanation:

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4 years ago
An urn contains n white balls andm black balls. (m and n are both positive numbers.) (a) If two balls are drawn without replacem
Genrish500 [490]

DISCLAIMER: Please let me rename b and w the number of black and white balls, for the sake of readability. You can switch the variable names at any time and the ideas won't change a bit!

<h2>(a)</h2>

Case 1: both balls are white.

At the beginning we have b+w balls. We want to pick a white one, so we have a probability of \frac{w}{b+w} of picking a white one.

If this happens, we're left with w-1 white balls and still b black balls, for a total of b+w-1 balls. So, now, the probability of picking a white ball is

\dfrac{w-1}{b+w-1}

The probability of the two events happening one after the other is the product of the probabilities, so you pick two whites with probability

\dfrac{w}{b+w}\cdot \dfrac{w-1}{b+w-1}=\dfrac{w(w-1)}{(b+w)(b+w-1)}

Case 2: both balls are black

The exact same logic leads to a probability of

\dfrac{b}{b+w}\cdot \dfrac{b-1}{b+w-1}=\dfrac{b(b-1)}{(b+w)(b+w-1)}

These two events are mutually exclusive (we either pick two whites or two blacks!), so the total probability of picking two balls of the same colour is

\dfrac{w(w-1)}{(b+w)(b+w-1)}+\dfrac{b(b-1)}{(b+w)(b+w-1)}=\dfrac{w(w-1)+b(b-1)}{(b+w)(b+w-1)}

<h2>(b)</h2>

Case 1: both balls are white.

In this case, nothing changes between the two picks. So, you have a probability of \frac{w}{b+w} of picking a white ball with the first pick, and the same probability of picking a white ball with the second pick. Similarly, you have a probability \frac{b}{b+w} of picking a black ball with both picks.

This leads to an overall probability of

\left(\dfrac{w}{b+w}\right)^2+\left(\dfrac{b}{b+w}\right)^2 = \dfrac{w^2+b^2}{(b+w)^2}

Of picking two balls of the same colour.

<h2>(c)</h2>

We want to prove that

\dfrac{w^2+b^2}{(b+w)^2}\geq \dfrac{w(w-1)+b(b-1)}{(b+w)(b+w-1)}

Expading all squares and products, this translates to

\dfrac{w^2+b^2}{b^2+2bw+w^2}\geq \dfrac{w^2+b^2-b-w}{b^2+2bw+w^2-b-w}

As you can see, this inequality comes in the form

\dfrac{x}{y}\geq \dfrac{x-k}{y-k}

With x and y greater than k. This inequality is true whenever the numerator is smaller than the denominator:

\dfrac{x}{y}\geq \dfrac{x-k}{y-k} \iff xy-kx \geq xy-ky \iff -kx\geq -ky \iff x\leq y

And this is our case, because in our case we have

  1. x=b^2+w^2
  2. y=b^2+w^2+2bw so, y has an extra piece and it is larger
  3. k=b+w which ensures that k<x (and thus k<y), because b and w are integers, and so b<b^2 and w<w^2

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3 years ago
Which graph represents the funcion f(x) = - x ^ 2 + 5?
loris [4]

Step-by-step explanation:

you did not provide any graphs but here's what I got.

Hope this helps!

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3 years ago
What is y to the power of x
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X would be the exponent and y would be the number
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