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Drupady [299]
2 years ago
14

\sqrt{2x^2+x+1}+\sqrt{x^2-x+1}=3x

Mathematics
1 answer:
djyliett [7]2 years ago
7 0
X=1


Mark brainliest please


Hope this helps you
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Do the following lengths satisfy the Pythagorean Theorem? Lengths =14, 50, 48.<br> Yes<br> No
seraphim [82]

Answer:

  yes

Step-by-step explanation:

  14² +48² = 196 +2304 = 2500 = 50²

The three lengths (14, 48, 50) are a Pythagorean triple.

__

<em>Additional comment</em>

The two smaller numbers correspond to side lengths. The larger one corresponds to the hypotenuse length.

We typically write the numbers of a Pythagorean triple in increasing order, but the order doesn't really matter. As sides of a triangle, the triangle formed is a right triangle, regardless of the order.

6 0
2 years ago
The polygons below are similar. Find the value of y.
elena-14-01-66 [18.8K]

Answer:

Im sorry bro but there's no picture

Step-by-step explanation:

7 0
3 years ago
HELP54465745653475vccbxdsztgfghdcdx
vredina [299]

Answer:

Solution is invalid

Step-by-step explanation:

Let the time when they will have the same water content be x hours

At x hours;

For the first tank, the content will be;

50 + 10x

For the second;

29 + 3x

Equating both, will give us x

29 + 3x = 50 + 10x

We will get x as a negative number here

Since number of hours can not be negative, the solution is invalid

4 0
3 years ago
PEF and triangle GHI are similar right triangles.
LekaFEV [45]

Answer: it’s the first 1

Step-by-step explanation:

6 0
2 years ago
Suppose that 40 percent of the drivers stopped at State Police checkpoints in Storrs on Spring Weekend show evidence of driving
lesantik [10]

Answer:

a) 0.778

b) 0.9222

c) 0.6826

d) 0.3174

e) 2 drivers

Step-by-step explanation:

Given:

Sample size, n = 5

P = 40% = 0.4

a) Probability that none of the drivers shows evidence of intoxication.

P(x=0) = ^nC_x P^x (1-P)^n^-^x

P(x=0) = ^5C_0  (0.4)^0 (1-0.4)^5^-^0

P(x=0) = ^5C_0 (0.4)^0 (0.60)^5

P(x=0) = 0.778

b) Probability that at least one of the drivers shows evidence of intoxication would be:

P(X ≥ 1) = 1 - P(X < 1)

= 1 - P(X = 0)

= 1 - ^5C_0 (0.4)^0 * (0.6)^5

= 1 - 0.0778

= 0.9222

c) The probability that at most two of the drivers show evidence of intoxication.

P(x≤2) = P(X = 0) + P(X = 1) + P(X = 2)

^5C_0  (0.4)^0  (0.6)^5 + ^5C_1  (0.4)^1  (0.6)^4 + ^5C_2  (0.4)^2  (0.6)^3

= 0.6826

d) Probability that more than two of the drivers show evidence of intoxication.

P(x>2) = 1 - P(X ≤ 2)

= 1 - [^5C_0  (0.4)^0  (0.6)^5 + ^5C_1  (0.4)^1  (0.6)^4 + ^5C_2 * (0.4)^2  (0.6)^3]

= 1 - 0.6826

= 0.3174

e) Expected number of intoxicated drivers.

To find this, use:

Sample size multiplied by sample proportion

n * p

= 5 * 0.40

= 2

Expected number of intoxicated drivers would be 2

7 0
3 years ago
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