Fourteen decreased by twice a number equals -32

Answers

Answer 1

Answer: your equation is 14-2x=-32

Step 1- subtract 14 on both sides

then your equation is 2x=-46

Step 2- divide 2 on both sides

then you get x= 23

x=23

Answer 2

Answer:

Final answer:

To solve the equation, subtract 14 from both sides and divide by -2 to isolate the variable x, which equals 23.

Explanation:

To solve this equation, we can set up an equation using the given information: 14 - 2x = -32. To isolate the variable, we can first subtract 14 from both sides of the equation: -2x = -46. Then, we can divide both sides by -2 to solve for x: x = 23.

Learn more about Solving Equations here:

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Suppose the altitude to the hypotenuse of a right triangle bisects the hypotenuse. How does the length of the altitude compare with the lengths of the segments of the hypotenuse?a) The length of the altitude is equal to twice the length of one of the segments of the hypotenuse.
b) The length of the altitude is equal to half the length of one of the segments of the hypotenuse.
c) The length of the altitude is equal to the length of one of the segments of the hypotenuse.
d) The length of the altitude is equal to the sum of the lengths of the segments of the hypotenuse.

Answers

Answer:

Option: C is correct.

c) The length of the altitude is equal to the length of one of the segments of the hypotenuse.

Step-by-step explanation:

By the Right Triangle Altitude Theorem:

The measure of the altitude drawn from the vertex of the right angle of a right triangle to its hypotenuse is the geometric mean between the measures of the two segments of the hypotenuse.

From the figure we could say that:

$AD=√(CD\cdot DB)$

As the hypotenuse is divided into divided into two equal parts since the altitude bisects the hypotenuse of the right triangle.

This means that:

CD=DB

Hence,

$AD=\sqrt{CD^(2)}\n\nAD=CD$

Hence, we could say that:

c) The length of the altitude is equal to the length of one of the segments of the hypotenuse.

1. On a right triangle, how does the length of the median drawn to the ... lengths. D. C. B. A. Triangle ABC is a right triangle with is the median to the ... to the hypotenuse is one-half as long as the hypotenuse, ..... This segment is an altitude to both triangles, with bases. AD and DC. These two segments are equal in length.

You pick a card at random. Without putting the first card back, you pick a second card at random. [1] [2] [3]

What is P(3, greater than 2)? Write your answer as a percentage.

Answers

Probability of first getting a 3 = 4/52 = 1/13
The p(getting >2) = probability of getting one of the remaining 3's or 40 other cards ( Im taking the aces to be low value) = 43/51

Required probability = 1/13 * 43/51 = 43/663

If aces are counted high it will be 1/13 * 47/51 = 47/531

=

You have 60 marbles in a jar, two of which are red. (so 2 red, 58 not red)You pick 10 at random. What is the chance that you draw both red marbles?
Explanation please, not just the answer. Brainliest answer if you calculate it for 2 out of 251 total marbles as well.

Answers

Well, I agonized over this for a while, and I have something that we may want to consider.

The only probability formula I know how to use is

Probability = (number of possible successes) / (total number of possibilities).

Can we work with that ? Let's see . . .

The denominator of that fraction is the total number of ways to draw
10 marbles from a jar of 60 . . .

The first draw can be any one of 60 marbles. For each of those . . .
The second draw can be any one of the remaining 59. For each of those . . .
The third draw can be any one of the remaining 58.
.
.
etc.

So the total number of ways to draw 10 from 60 is

   (60 x 59 x 58 x 57 x 56 x 55 x 54 x 53 x 52 x 51)

That's a very big number. My calculator says something that rounds to
2.736 x 10¹⁷. But my calculator only shows 10 digits, so it can't show
all 18 digits in the number. Fortunately, we don't need to see the whole
number written out. We'll just write it in factorial notation, and go on
to do the numerator of the fraction, which is going to be much harder.

That number that we just found is equal to (60!) / (50!) . It's going to be
the denominator of the big fraction.

Now for the numerator. That's going to be the number of ways that the
two red marbles can be included among the ten marbles drawn.

One red marble can be any one of the 10 marbles pulled out.
   For each of those . . .
The other red marble can be any one of the other 9 that are picked.

So there are (10 x 9) = 90 ways for the selected 10 to include both red ones.

SO ! Now, the probability that the ten that are drawn will include the two
red ones should be (it might be, it could be) . . .

   90 divided by (60! / 50!)   .

Remember, that fraction on the right is just total number of ways
to pick 10 out of 60 . The probability of including both red ones
in the draw is 90 divided by that number. It's very small.

Again from my calculator, it's 3.29 x 10⁻¹⁴ percent.

I have no confidence in my answer, but I invite you to look it over, along with
all the real gurus out there.

If I'm wrong, then I've stolen only 5 points that I'm not entitled to, and at least
I did put some effort into it.

Here you have two proportions. 2/60 and 10/60. I would cross multiply.
Giving you 20/120.
Simplify that to get 1/6. So you have a 1 in 6 chance of getting both the red marbles.
(I'm not 100% sure since it's been a while so you might want to get a second opinion.)
Not sure what you mean by the second part, but hope I provided some help :)

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