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jenyasd209 [6]
4 years ago
15

Can someone help me please solving these maths problems thanks.

Mathematics
2 answers:
blondinia [14]4 years ago
8 0
2/4 on the 0 to 1 and 5/4 on the 1 to 2
Zielflug [23.3K]4 years ago
4 0
2/4 goes in the first box and 5/4 goes in the second
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Plzz help and no wrong answer need to be done asap
Ann [662]

Answer:

1/500

Step-by-step explanation:

7 0
2 years ago
Solve sec2theta + tantheta - 3 =0 for 0
Serhud [2]

Answer:

Ф = \frac{\pi }{4} ,\frac{5\pi}{4}

Step-by-step explanation:

It is a bit difficult to input the work here, so I uploaded an image

  • First we can use the trig identities to change sec²(Ф) to tan²(Ф) + 1
  • Then we can combine like terms
  • Then we can factor this as a polynomial function
  • Then we can set each term equal to zero and solve for Ф
  • The first term tan(Ф) - 2 = 0 has no solution because tan(Ф) ≠ -2 anywhere
  • The second term tan(Ф) - 1 = 0 has two solutions of \frac{\pi}{4} and \frac{5\pi}{4} so these are the solutions to the problem

7 0
3 years ago
Must Just one match to fix the equation ?<br> 6 + 4 = 4
zhannawk [14.2K]

Answer:

6 + 4 × -2/4 = 4

Step-by-step explanation:

hi hope u understand

3 0
3 years ago
Read 2 more answers
A jeweler orders necklaces from a website that offers $6 shipping for any-size order. Each necklace costs $7. The jeweler wants
Furkat [3]

Answer:

c = $7n + $6

Step-by-step explanation:

$6 shipping for any size

$7 per necklace

let n equal the quantity of necklaces and c equal the total cost.

you buy n amount of necklaces and add the shipping cost to find your cost.

7 0
3 years ago
Read 2 more answers
If $x$ is the average of $13$, $-16$, and $6$ and if $y$ is the cube root of $8$, find $x^2 + y^3$. Please Help
Alekssandra [29.7K]

Answer:

If we have a set of N elements:

{x₁, x₂, ..., xₙ}

The mean value (also called the average value) Is calculated as:

Average = (x₁ + x₂ + ... + xₙ)/n

So if x is the average of 13, -16 and 6 ( a total of 3 values)

x will be equal to:

x = (13 + (-16) + 6)/3 = (19 - 16)/3 = 3/3 = 1

x = 1

And we know that:

y = ∛8

Remember that:

2*2 = 4

and

4*2 = 8

then

2*2*2 = 2^3 = 8

then ∛8 = 2.

So we have:

y = ∛8 = 2

Now we can replace these values in the equation:

x^2 + y^3

replacing:

x = 1

y = 2

we get:

1^2 + 2^3 = 1 + 8 = 9

5 0
3 years ago
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