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irinina [24]
3 years ago
14

How do I factor the quadratic equation

Mathematics
1 answer:
anastassius [24]3 years ago
4 0
Factor the following:
10 y^2 - 35 y + 30
Factor 5 out of 10 y^2 - 35 y + 30:
5 (2 y^2 - 7 y + 6)
Factor the quadratic 2 y^2 - 7 y + 6. 
The coefficient of y^2 is 2 and the constant term is 6. 
The product of 2 and 6 is 12. 
The factors of 12 which sum to -7 are -3 and -4. So 2 y^2 - 7 y + 6 = 2 y^2 - 4 y - 3 y + 6 = y (2 y - 3) - 2 (2 y - 3):
5 y (2 y - 3) - 2 (2 y - 3)
Factor 2 y - 3 from y (2 y - 3) - 2 (2 y - 3):
Answer:  5 (2 y - 3) (y - 2)
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Solve the system of 2x-5y+z=1 3 y+2z=5-24z=48
motikmotik
<span>A) 2x -5y +z = 1
B) 3 y + 2z = 5
C) -24 z = 48
Those 3 equations you typed were somewhat squished together.
Did I retype those correctly?
Let me know and I'll solve it.


</span>
8 0
3 years ago
Turn 2 2/5 into a percent
Naddika [18.5K]
2/5 = 0.4
2.4 x 100 = 240%
6 0
3 years ago
The word geometry comes from the greek words geo, which means "earth," and metron, which means "a measuring of." given this info
Leokris [45]

The geometry was developed by a Greek mathematician called Euclid and developed to meet some practical need in surveying, construction, astronomy, and various crafts.

In this question,

Early geometry was a collection of empirically discovered principles concerning lengths, angles, areas, and volumes, which were developed to meet some practical need in surveying, construction, astronomy, and various crafts.

It was developed by a Greek mathematician called Euclid. This branch of geometry deals with terms like points, lines, surfaces, dimensions of the solids, etc.,

In several ancient cultures there developed a form of geometry suited to the relationships between lengths, areas, and volumes of physical objects. This geometry was codified in Euclid's Elements about 300 bce on the basis of 10 axioms, or postulates, from which several hundred theorems were proved by deductive logic.

Hence we can conclude that the geometry was developed by a Greek mathematician called Euclid and developed to meet some practical need in surveying, construction, astronomy, and various crafts.

Learn more about geometry here

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6 0
2 years ago
(K3+2k2-82k-28)/(k+10)
Ne4ueva [31]

So for this, we will be using synthetic division. To set it up, have the equation so that the divisor is -10 (since that is the solution of k + 10 = 0) and the dividend are the coefficients. Our equation will look as such:

<em>(Note that synthetic division can only be used when the divisor is a 1st degree binomial)</em>

  • -10 | 1 + 2 - 82 - 28
  • ---------------------------

Now firstly, drop the 1:

  • -10 | 1 + 2 - 82 - 28
  •       ↓
  • -------------------------
  •        1

Next, you are going to multiply -10 and 1, and then combine the product with 2.

  • -10 | 1 + 2 - 82 - 28
  •       ↓ - 10
  • -------------------------
  •        1 - 8

Next, multiply -10 and -8, then combine the product with -82:

  • -10 | 1 + 2 - 82 - 28
  •       ↓ -10 + 80
  • -------------------------
  •        1 - 8 - 2

Next, multiply -10 and -2, then combine the product with -28:

  • -10 | 1 + 2 - 82 - 28
  •       ↓ -10 + 80 + 20
  • -------------------------
  •        1 - 8 - 2 - 8

Now, since we know that the degree of the dividend is 3, this means that the degree of the quotient is 2. Using this, the first 3 terms are k^2, k, and the constant, or in this case k² - 8k - 2. Now what about the last coefficient -8? Well this is our remainder, and will be written as -8/(k + 10).

<u>Putting it together, the quotient is k^2-8k-2-\frac{8}{k+10}</u>

8 0
3 years ago
6th grade math i mark as brainliest​
Alinara [238K]

Answer:

16

Step-by-step explanation:

4*4

6 0
3 years ago
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