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madreJ [45]
3 years ago
7

An equilateral triangle has a perimeter of (42x^2+ 30y).What is the length of each side?

Mathematics
1 answer:
arsen [322]3 years ago
3 0

Answer:

which chapter is this? and page no:

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Can someone pls help
sweet-ann [11.9K]

z = 13/9?? i don't really know hope i helped

4 0
2 years ago
Read 2 more answers
Defining congruent triangles
seropon [69]

Answer:

E -->R

D-->K

O-->O

Step-by-step explanation:

Assuming you are asking for congruent parts, correspondence is the part which corresponds and is also equal.

Here Angle E corresponds and is congruent to Angle R in these congruent triangles.

Angle D is congruent and corresponds to Angle K for these congruent triangles.

Angle O is equal to Angle O as they are vertical angles across a vertex.

5 0
3 years ago
7.
Agata [3.3K]

Answers:

  • prime number = {2,3,5}
  • composite number = {4,6}
  • number less than four = {1,2,3}
  • number more than or equal to three = {3,4,5,6}

====================================================

Explanation:

The list of all possible outcomes on a single standard die is {1,2,3,4,5,6}. The curly braces indicate set notation.

A prime number is a number where its only factors are 1 and itself. The value 1 is not prime, and it's not composite either.

Something like 3 is prime because its only factors are 1 and 3. Something like 4 is composite because 4 = 2*2. A composite value has factors other than 1 and itself.

Parts (iii) and (iv) are fairly straight forward. You simply list items less than four for part (iii) and you list items that are three or greater for part (iv).

5 0
3 years ago
<img src="https://tex.z-dn.net/?f=%5Cfrac%7B3%7D%7B4%7D" id="TexFormula1" title="\frac{3}{4}" alt="\frac{3}{4}" align="absmiddle
oee [108]

\frac{3}{4}  +  \frac{5}{16}  \\  \\  \frac{12}{16}  +  \frac{5}{16} \\  \\  \frac{17}{16}
5 0
2 years ago
The population of bacteria in a culture grows at a rate proportional to the number of bacteria present at time t. After 3 hours
ella [17]

Answer:

137

Step-by-step explanation:

Use the model A = Pe^(kt), where P is the initial number and k is the growth constant.  Then, in case 1,

300 = Pe^(k*3)

and in case 2,

4000 = Pe^(k*10).  We need to find P and e^k.

300 = Pe^(k*3) can be rewritten as (300/P) = [e^k]^3, which in turn may be solved for e^k:

∛(300/P) = e^k

Now we go back to 4000 = Pe^(k*10) and rewrite it as (4000/P) = [e^k]^10.  Substituting ∛(300/P) for e^k, we obtain:              

(4000/P) =  (300/P )^(10/3)

which must now be solved for P.  Raising both sides by the power of 3, we get:

(4000/P)^3 = (300/P)^10.  Therefore,

4000^3       300^10

------------- = -------------

   P^3             P^10

which reduces to:

4000^3       300^10

------------- = -------------

       1             P^7

or

        1         P^7

------------  =  -----------

4000^3       300^10

or:

           300^10

P^7 = ---------------- = 9.226*10^13  = 92.26*10^12

           4000^3

Taking the 7th root of both sides results in P = (92.26*10^12)^(1/7), or

                                                                        P = 137.26

The initial number of bacteria was 137.

7 0
2 years ago
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