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3 years ago

What are the solution(s) to the quadratic equation 50 - x² = 0?

Mathematics

1 answer:

grandymaker [24]3 years ago

7 0

Answer:

The answer is C.

Step-by-step explanation:

$50-x^2=0$

$x^2=50$

$x=\pm \sqrt{50} =\pm \sqrt{25*2}=\pm 5\sqrt{2}$

The answer is C (I am assuming that it isn't 5/2).

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PLS HELP ME! <br> Um doing venn digrams so u have to set one up for it to work... TWENTY POINTS

uysha [10]

Answer:

Creating a diagram..

Step-by-step explanation:

Alrighty, I've made a diagram. I've even uploaded it so everyone can see how good at drawing I am (not)

a. 7 do not own anything.

b. the probability that someone owns a bone or a stick is 50%.

c. the probability that someone owns a rock and a bone is 14%.

d. the probability that someone owns at least 2 items is 42%.

e. the probability that someone owns only 1 of these items is 51%.

Please tell me if im wrong! im only a student .m.

7 0

4 years ago

(X^2+x-6)(2x^2+4x) factor completely

MissTica

Answer:

$2x\left(x-2\right)\left(x+3\right)\left(x+2\right)$

Step-by-step explanation:

Lets go ahead and take a step by step approach to solving your problem.

First we must start by factoring $x^2+x-6$ .

To do this, we must break the expression into groups! Although I will also provide a definition to help you understand! For $ax^2+bx+c$ we need to find u, v > $u\cdot \:v=a\cdot \:c$ and $u+v=b$ which we will group into $\left(ax^2+ux\right)+\left(vx+c\right)$ .

The values we have are... $a=1,\:b=1,\:c=-6$ and $u\cdot v=-6,\:u+v=1$ .

Now we need the factors of 6 which are 1, 2, 3 and 6. We also need the negative factors which you get simply by multiplying the positives by -1 or just reversing them.

Now we need to check for every two factors if u + v = 1.

$\mathrm{Check}\:u=1,\:v=-6: \:u\cdot v=-6,\:u+v=-5 \Rightarrow \mathrm{False}$

$\mathrm{Check}\:u=2,\:v=-3: \:u\cdot v=-6,\:u+v=-1 \Rightarrow \mathrm{False}$

$\mathrm{Check}\:u=3,\:v=-2: \:u\cdot v=-6,\:u+v=1 \Rightarrow \mathrm{True}$

$\mathrm{Check}\:u=6,\:v=-1: \:u\cdot v=-6,\:u+v=5 \Rightarrow \mathrm{False}$

Therefore $u=3,\:v=-2$ .

Now we want to $\mathrm{Group\:into\:}\left(ax^2+ux\right)+\left(vx+c\right)$ . Which is $\left(x^2-2x\right)+\left(3x-6\right)$ .

Now we must factor x from $x^2-2x$ .

Lets apply the exponent rule of $a^{b+c}=a^ba^c$ which means $x^2=xx$ .

Therefore we now have xx - 2x. Lets factor out the common term of x to get x(x - 2).

Now factor 3 out of $3x-6$ .

Rewrite 6 as 3 * 2. $3x-3\cdot \:2$ .

Now factor out the common term of 3 > $3\left(x-2\right)$ .

$x\left(x-2\right)+3\left(x-2\right)$ > Factor out the common term of x - 2 > $\left(x-2\right)\left(x+3\right)$ .

Now lets factor $2x^2+4x$ .

Apply the previous exponent rule of $a^{b+c}=a^ba^c$ > $x^2=xx$ > $2xx+4x$ .

Now rewrite 4 as 2 * 2 > $2xx+2\cdot \:2x$ .

Now factor out the common term of 2x to get 2x(x + 2).

Combine it all and we get $2x\left(x-2\right)\left(x+3\right)\left(x+2\right)$ .

Hope this helps!

7 0

3 years ago

|3k-2|=2|k+2| how do I do this when the 2 is on the other side??

grin007 [14]

|3k-2|=2|k+2l
Do Distributive Property and take 2 times by everything in the parenthesis. That will leave you with |3k-2|=2k+4. Then add like terms. 3k+2k=5k. Then -2+4=2. So you end up with 5k+2. I believe I'm right. If not please tell me.

3 0

3 years ago

What are the coordinates of point Q?

Ganezh [65]

Answer:

(0,-3) because its the coordinate

3 0

3 years ago

Suppose that there were a strong correlation between the variables g and h

Elan Coil [88]

A. g must cause h........

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3 years ago