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erma4kov [3.2K]
3 years ago
12

Fill in the blank to make the two fractions equivalent.

Mathematics
2 answers:
NNADVOKAT [17]3 years ago
8 0
7*5=35 ,so you must do the same with the numerator to make it equal it would be 4*5=20. The Answer is 20/35
vovikov84 [41]3 years ago
6 0
The answer is 20. Its there but erased
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In triangle abc angle a = 40◦, side b = 11 and side c = 5. use the cosine rule to find the length of side a, to two decimal plac
Sonbull [250]
A^2 = b^2 + c^2 - 2bc cos a

    = 11^2 + 5^2 - 2*5*11 cos 40

   =  7.86 to 2 DP's

to find the remaining angles use the sine rule:-
a / sin A = b / sin B so

7.857/ sin 40 =  11 / sin B
sin B = 11 sin40 / 7.857 = 0.8999

<B = 64 degrees 

so <C = 180 - 64-40  =  76 degrees
5 0
3 years ago
PLZ HELP PLZ! TY! WILL GIVE BRAINLIEST TO WHOEVER ANSWERS
Marysya12 [62]
Take the y and x and 4 as a common factor and you will get 4xy(2yz+3x) the second one is the answer
7 0
4 years ago
Factor 26r3s + 52r5 – 39r2s4.
algol [13]

Answer:

13r²(2rs + 4r³ - 3s⁴)

Step-by-step explanation:

In equation 26r³s + 52r⁵ - 39r²s⁴;

The GCF of 26, 52, and 39 = 13

The GCF of r³, r⁵ and r² = r²

The GCF of s, (no "s"), and s⁴ = no "s" (Since one of the number doesn't have "s")

Now we can factor out 13r² from all three expressions;

26r³s + 52r⁵ - 39r²s⁴

=> <em>13r²(2rs) + 13r²(4r³) - 13r²(3s⁴)</em>

To factor it all together;

<u>13r²(2rs + 4r³ - 3s⁴)</u>

Hope this helps!

7 0
3 years ago
I run a book club with n people, not including myself. Every day, for 365 days, I invite three members in the club to review a b
Bezzdna [24]
<h3>Answer:   15</h3>

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

Explanation:

The order doesn't matter. A group like {A,B,C} is the same as {B,A,C}.

All that matters is the overall group rather than the positioning of the members.

We'll use the nCr combination formula since order doesn't matter.

The value of n is unknown, but we know that r = 3 members are to be selected.

Let's pick a value for n at random. Let's say n = 10.

Plug n = 10 and r = 3 into the nCr formula below.

n C r = \frac{n!}{r!(n-r)!}\\\\10 C 3 = \frac{10!}{3!*(10-3)!}\\\\10 C 3 = \frac{10!}{3!*7!}\\\\10 C 3 = \frac{10*9*8*7!}{3!*7!}\\\\ 10 C 3 = \frac{10*9*8}{3!}\\\\ 10 C 3 = \frac{10*9*8}{3*2*1}\\\\ 10 C 3 = \frac{720}{6}\\\\ 10 C 3 = 120\\\\

Unfortunately we don't reach 365 or larger.

Let's try n = 11

n C r = \frac{n!}{r!(n-r)!}\\\\11 C 3 = \frac{11!}{3!*(11-3)!}\\\\11 C 3 = \frac{11!}{3!*8!}\\\\11 C 3 = \frac{11*10*9*8!}{3!*8!}\\\\ 11 C 3 = \frac{11*10*9}{3!}\\\\ 11 C 3 = \frac{11*10*9}{3*2*1}\\\\ 11 C 3 = \frac{990}{6}\\\\ 11 C 3 = 165\\\\

We're still under our target. The good news is that the nCr value is increasing.

So the idea is to do trial and error with various values of n. Keep incrementing n until nCr = nC3 is equal to 365 or larger.

Here's a table of values where r = 3 the entire time

\begin{array}{|c|c|} \cline{1-2}\text{n} & \text{nCr}\\\cline{1-2}10 & 120\\\cline{1-2}11 & 165\\\cline{1-2}12 & 220\\\cline{1-2}13 & 286\\\cline{1-2}14 & 364\\\cline{1-2}15 & 455\\\cline{1-2}\end{array}

The nCr values are also found in Pascal's Triangle. Each of those values are the fourth entry of each row.

When n = 14, we have nCr = 364 which is very close. We're one short unfortunately.

So we have to go for <u>n = 15</u> instead. This makes the nCr value well over 365 of course, but it guarantees that you'll have plenty of trios to choose from such that no group of three is repeated. Unfortunately some trios will be left out.

7 0
2 years ago
Which equation shown below shows a linear relationship between x and y ?
KiRa [710]

Answer:

Its probably a because the ones with squared results in a u shape

5 0
3 years ago
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