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3 years ago

What is the value of y? 2y y+10 50 all in degrees and in different angles of a triangle

Mathematics

2 answers:

bezimeni [28]3 years ago

5 0

Answer:

The value of y = 40

Step-by-step explanation:

Measure of one angle of the triangle is given to be 2y

Measure of second angle of the triangle is given to be y + 10

Measure of the third angle of the triangle is given to be 50

Now, By using the angles sum property of a triangle, We have :

Sum of all the three angles of a triangle is always supplementary

⇒ ∠1 + ∠2 + ∠3 = 180°

⇒ 2y° + (y + 10)° + 50 = 180°

⇒ 3y + 10 + 50 = 180°

⇒ 3y = 120

⇒ y = 40

Hence, The value of y = 40

s344n2d4d5 [400]3 years ago

3 0

You need to do 2y+y+10+50=180. Then you'd combine like terms to make 3y+60=180 then to get rid of the sixty subtract it from 180 to make 3y=120 then divide by three to get y by itself so 120 divided by three is 40 which makes y whisk forty. Hope this helped.

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How many terms of the arithmetic sequence {1,22,43,64,85,…} will give a sum of 2332? Show all steps including the formulas used

MA_775_DIABLO [31]

There's a slight problem with your question, but we'll get to that...

Consecutive terms of the sequence are separated by a fixed difference of 21 (22 = 1 + 21, 43 = 22 + 21, 64 = 43 + 21, and so on), so the n-th term of the sequence, a (n), is given recursively by

• a (1) = 1

• a (n) = a (n - 1) + 21 … … … for n > 1

We can find the explicit rule for the sequence by iterative substitution:

a (2) = a (1) + 21

a (3) = a (2) + 21 = (a (1) + 21) + 21 = a (1) + 2×21

a (4) = a (3) + 21 = (a (1) + 2×21) + 21 = a (1) + 3×21

and so on, with the general pattern

a (n) = a (1) + 21 (n - 1) = 21n - 20

Now, we're told that the sum of some number N of terms in this sequence is 2332. In other words, the N-th partial sum of the sequence is

a (1) + a (2) + a (3) + … + a (N - 1) + a (N) = 2332

or more compactly,

$\displaystyle\sum_{n=1}^N a(n) = 2332$

It's important to note that N must be some positive integer.

Replace a (n) by the explicit rule:

$\displaystyle\sum_{n=1}^N (21n-20) = 2332$

Expand the sum on the left as

$\displaystyle 21 \sum_{n=1}^N n-20\sum_{n=1}^N1 = 2332$

and recall the formulas,

$\displaystyle\sum_{k=1}^n1=\underbrace{1+1+\cdots+1}_{n\text{ times}}=n$

$\displaystyle\sum_{k=1}^nk=1+2+3+\cdots+n=\frac{n(n+1)}2$

So the sum of the first N terms of a (n) is such that

21 × N (N + 1)/2 - 20N = 2332

Solve for N :

21 (N ² + N) - 40N = 4664

21 N ² - 19 N - 4664 = 0

Now for the problem I mentioned at the start: this polynomial has no rational roots, and instead

N = (19 ± √392,137)/42 ≈ -14.45 or 15.36

so there is no positive integer N for which the first N terms of the sum add up to 2332.

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2 years ago

Apply the markup for each item. Then, find the retail price. Round to two decimals places when necessary.

Alborosie

Answer:

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Step-by-step explanation:

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3 years ago

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Step-by-step explanation:

using the following formula we can find a square root for a number which is a imperfect square

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so √38 = √36 + (38-36)/2√36

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3 years ago

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Answer:

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Step-by-step explanation:

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2 years ago