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guajiro [1.7K]
2 years ago
12

What is the factored form of 36x4 – 25?

Mathematics
2 answers:
gladu [14]2 years ago
8 0
36x^4 - 25=(6x^2)^2-5^2=(6x^2-5)(6x^2+5)
babunello [35]2 years ago
8 0

its A (6x2-5)6x2+5)

Step-by-step explanation:

That is the factored form if 36x4-25

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Which statement is an example of the symmetric property of congruence?
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The symmetric property of congruence states that if a quantity, say, A is congruent to a quantity, say B, then, the quantity B is congruent to the quantity A.

Mathematically, this statement of the symmetric property of congruence can be written as:

If A\cong B, then B\cong A.

Applying this symmetric property of congruence to the question that we are given, we see that only Option D fulfills the conditions of the definition of the property of congruence as Option D clearly states that:

If \Delta EFG\cong \Delta HJK, then \Delta HJK\cong \Delta EFG which is the exact statement of the symmetric property of congruence when applied to triangles.

Thus, Option D is the correct option.

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1 year ago
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PLEASE HELP ME!
dezoksy [38]
Thicknesses at different point are: <span>41, 38, 36, 29, 34, 44, 46, 43, 35, 40


In increasing order: 29, 34, 35, 36, 38, 40, 41, 43, 44, 46

Median = (38+40)/2 = 39m</span>

Median thickness is 39m
5 0
2 years ago
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An experiment is carried out 400 times.
MissTica

Answer:

You would expect the outcome to be void 240 times

Step-by-step explanation:

1000 x 0.24

6 0
2 years ago
A team of 10 players is to be selected from a class of 6 girls and 7 boys. Match each scenario to its probability. You have to d
tankabanditka [31]
The selection of r objects out of n is done in

C(n, r)= \frac{n!}{r!(n-r)!} many ways.

The total number of selections 10 that we can make from 6+7=13 students is 

C(13,10)= \frac{13!}{3!(10)!}= \frac{13*12*11*10!}{3*2*1*10!}= \frac{13*12*11}{3*2}=  286
thus, the sample space of the experiment is 286

A. 
<span>"The probability that a randomly chosen team includes all 6 girls in the class."

total number of group of 10 which include all girls is C(7, 4), because the girls are fixed, and the remaining 4 is to be completed from the 7 boys, which can be done in C(7, 4) many ways.


</span>C(7, 4)= \frac{7!}{4!3!}= \frac{7*6*5*4!}{4!*3*2*1}= \frac{7*6*5}{3*2}=35
<span>
P(all 6 girls chosen)=35/286=0.12

B.
"</span>The probability that a randomly chosen team has 3 girls and 7 boys.<span>"

with the same logic as in A, the number of groups were all 7 boys are in, is 

</span>C(6, 3)= \frac{6!}{3!3!}= \frac{6*5*4*3!}{3!3!}= \frac{6*5*4}{3*2*1}=20
<span>
so the probability is 20/286=0.07

C.
"</span>The probability that a randomly chosen team has either 4 or 6 boys.<span>"

case 1: the team has 4 boys and 6 girls

this was already calculated in part A, it is </span>0.12.
<span>
case 2, the team has 6 boys and 4 girls.

there C(7, 6)*C(6, 4) ,many ways of doing this, because any selection of the boys which can be done in C(7, 6) ways, can be combined with any selection of the girls. 

</span>C(7, 6)*C(6, 4)= \frac{7!}{6!1}* \frac{6!}{4!2!} =7*15= 105
<span>
the probability is 105/286=0.367

since  case 1 and case 2 are disjoint, that is either one or the other happen, then we add the probabilities:

0.12+0.367=0.487 (approximately = 0.49)

D.
"</span><span>The probability that a randomly chosen team has 5 girls and 5 boys.</span><span>"

selecting 5 boys and 5 girls can be done in 

</span>C(7, 5)*C(6,5)= \frac{7!}{5!2} * \frac{6!}{5!1}=21*6=126

many ways,

so the probability is 126/286=0.44
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2 years ago
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PART A: Marvin has a coupon the discount is the rental of a full-size car by $25 they decide to buy insurance for each day if th
natita [175]

Answer:

i dont know about the other parts but part A is 19

Step-by-step explanation:

6 0
1 year ago
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