Solve x^2+20x+100=50

Mathematics

1 answer:

uysha [10]2 years ago

6 0

1. You con solve the quadratic equation x^2+20x+100=50 by following the proccedure below:

2. Pass the number 50 from the right member to the left member. Then you obtain:

x^2+20x+100-50=0
 x^2+20x+50=0

3. Then, you must apply the quadratic equation, which is:

x=(-b±√(b^2-4ac))/2a

x^2+20x+50=0

a=1
b=20
c=50

4. Therefore, when you substitute the values into the quadratic equation and simplify ir, you obtain that the result is:

-10±5√2 (It is the last option).

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Triangle XYZ was dilated by a scale factor of 2 to create triangle ACB and sin ∠X = 5 over 5 and 59 hundredths.

Vikentia [17]

The relationship between Traingle XYZ and ACB is that they are similar triangles. tanX = tanA = 5 over 2 and 5 tenths, where

AC = 2 x XY
CB = 2 x YZ

<h3>What is the dilation about?</h3>

Triangle XYZ was dilated by a scale factor of 2 to create triangle ACB, hence XYZ and ACB are similar triangles.

Angles Y and C are said to measure 90 degrees, and angles A and X are known to be congruent. Thus:

tanX = tanA = 5 over 2 and 5 tenths

AC = 2 x XY

CB = 2 x YZ

Another way to solve for it is by:

Note that Dilation of the triangle ΔXYZ was by a factor of "2".

Since m∠Y = m∠C = 90º - given

ΔXYZ and ΔACB are said to be right triangles.

Since ∠X ≅ ∠A - given

Then ∠X and ∠Z are said to becomplementary angles

Since m∠Z = 90° - m∠X -- given

Then ∠A and ∠B are said to be complementary angles

Since m∠B = 90° - m∠A --- given

Then, ∠Z ≅ ∠B

Therefore ΔXYZ ∼ ΔACB are similar triangles because it has its corresponding sides to be proportional and its corresponding angles to be congruent.

Note:

sin ∠X = 5/5.59

sin ∠X = YZ/XY

YZ = 5 seen in ΔXYZ

XZ = 5.59 seen in hypotenuse of ΔXYZ

Then one need to Calculate the length of one aspect of XY:

XY = √((5.59)2 - 52)

= 2.4996

Note CB/YZ = 2

CD = 2*YZ = 2 x 5

= 10

AC/XY = 2