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Gwar [14]
3 years ago
11

PLEASE HELP ILL GIVE BRAINLIEST!!!!

Mathematics
2 answers:
Free_Kalibri [48]3 years ago
5 0

Answer:

The first one is x=45

The second one is w=14

The third one is x=6

Step-by-step explanation:

konstantin123 [22]3 years ago
4 0
Which value of w below is the solution to this equation ANWSER - C. 14 I think
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What’s the length of the rectangle if the width is 9?
Vladimir [108]

Answer:

18

Step-by-step explanation:

9+9=18 both sides of the rectangle is 9 so the top is longer which means 18

3 0
3 years ago
1. The ratio of boys to girls in Mr. Okafor's after-school club is the same as the ratio of boys to girls in Ms. Williams' after
Serjik [45]
27 is correct I hope you pass !
8 0
3 years ago
Can someone help me with this math homework please!
statuscvo [17]

Answer:

1 = Input G

2 = Input F

3 = Input H

Step-by-step explanation:

5 0
3 years ago
Consider the following functions. f1(x) = x, f2(x) = x2, f3(x) = 2x − 4x2 g(x) = c1f1(x) + c2f2(x) + c3f3(x) Solve for c1, c2, a
bekas [8.4K]

9514 1404 393

Answer:

  • (c1, c2, c3) = (-2t, 4t, t) . . . . for any value of t
  • NOT linearly independent

Step-by-step explanation:

We want ...

  c1·f1(x) +c2·f2(x) +c3·f3(x) = g(x) ≡ 0

Substituting for the fn function values, we have ...

  c1·x +c2·x² +c3·(2x -4x²) ≡ 0

This resolves to two equations:

  x(c1 +2c3) = 0

  x²(c2 -4c3) = 0

These have an infinite set of solutions:

  c1 = -2c3

  c2 = 4c3

Then for any parameter t, including the "trivial" t=0, ...

  (c1, c2, c3) = (-2t, 4t, t)

__

f1, f2, f3 are NOT linearly independent. (If they were, there would be only one solution making g(x) ≡ 0.)

7 0
2 years ago
How many positive integers less than or equal to 100 have a prime factor that is greater than 4?
Aleks [24]

Answer:

97, 89, 83, 73, 71, 67, 61, 59, 53, 47,94,43,86, 41, 82, 37,74, 31,62,93, 29,58,87, 23,46,69, 92, 19, 38,57,76,95, 17,34,51,68,85, 13,26,39,52,65,78,91, 11,22,33,44,55,66,77,88,99, 7,14,21,28,35,42,49,56,63,70,77,84,91,98,

5,10,15,20,25,30,40,45,50,60,75

Step-by-step explanation:

We have to find the  positive integers less than or equal to 100 have a prime factor that is greater than 4

First let us find out the prime factors greater than 4 and less than 100

They are

5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89 and 97.

Let us start from the highest.

Greatest is 97, next 89, 83, 73, 71, 67, 61, 59, 53, 47,94, 43,86, 41, 82, 37,74, 31,62,93, 29,58,87, 23,46,69, 92, 19, 38,57,76,95, 17,34,51,68,85, 13,26,39,52,65,78,91, 11,22,33,44,55,66,77,88,99, 7,14,21,28,35,42,49,56,63,70,77,84,91,98, 5,10,15,20....100

i.e. we consider each prime number and write its multiples also below 100

Now let us remove the repititions.

The answer would be

97, 89, 83, 73, 71, 67, 61, 59, 53, 47,94,43,86, 41, 82, 37,74, 31,62,93, 29,58,87, 23,46,69, 92, 19, 38,57,76,95, 17,34,51,68,85, 13,26,39,52,65,78,91, 11,22,33,44,55,66,77,88,99, 7,14,21,28,35,42,49,56,63,70,77,84,91,98,

5,10,15,20,25,30,40,45,50,60,75

3 0
3 years ago
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