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melamori03 [73]
3 years ago
5

="\sqrt{x^{2} +5x+3} +2x^{2} +10x-15=0" alt="\sqrt{x^{2} +5x+3} +2x^{2} +10x-15=0" align="absmiddle" class="latex-formula">
Mathematics
1 answer:
Anettt [7]3 years ago
8 0
<h3>Factor</h3><h3>√x²+5x+3+2x²+10x-15=0</h3><h3>√x²+2x²+5x+10x+3-15=0</h3><h3>√3x²+15x-12=0</h3><h3>√3x²+(3(5x)+3×-4=0</h3><h3>√3(x²+5x)+3×-4=0</h3><h3>√3(x²+5x-4)=0</h3><h3> </h3>

please mark this answer as brainlist

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Find the range of the following set of data:12,2,14,80,100
Leviafan [203]

Answer: 98

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Step-by-step explanation:

The parts listed in the congruence statements don't correspond, so they aren't necessarily congruent.

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2 years ago
Suppose we have 3 cards identical in form except that both sides of the first card are colored red, both sides of the second car
Nikolay [14]

Answer:

probability that the other side is colored black if the upper side of the chosen card is colored red = 1/3

Step-by-step explanation:

First of all;

Let B1 be the event that the card with two red sides is selected

Let B2 be the event that the

card with two black sides is selected

Let B3 be the event that the card with one red side and one black side is

selected

Let A be the event that the upper side of the selected card (when put down on the ground)

is red.

Now, from the question;

P(B3) = ⅓

P(A|B3) = ½

P(B1) = ⅓

P(A|B1) = 1

P(B2) = ⅓

P(A|B2)) = 0

(P(B3) = ⅓

P(A|B3) = ½

Now, we want to find the probability that the other side is colored black if the upper side of the chosen card is colored red. This probability is; P(B3|A). Thus, from the Bayes’ formula, it follows that;

P(B3|A) = [P(B3)•P(A|B3)]/[(P(B1)•P(A|B1)) + (P(B2)•P(A|B2)) + (P(B3)•P(A|B3))]

Thus;

P(B3|A) = [⅓×½]/[(⅓×1) + (⅓•0) + (⅓×½)]

P(B3|A) = (1/6)/(⅓ + 0 + 1/6)

P(B3|A) = (1/6)/(1/2)

P(B3|A) = 1/3

5 0
3 years ago
Help? i'll mark brainliest if correct
Strike441 [17]

Answer:

36,686

Step-by-step explanation:

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240 x 26 = 36,686

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3 years ago
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Ludmilka [50]

Step-by-step explanation:

a. 3⁴

b. 7⁵

c. 10²

d. 5⁷

Hope this helps?

8 0
3 years ago
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