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garri49 [273]
3 years ago
13

What is the solution for x^2 +14x +15​

Mathematics
2 answers:
ololo11 [35]3 years ago
7 0
  • ...................................

Mademuasel [1]3 years ago
7 0

Step-by-step explanation:

the solution are

x = 1 or x = -15

x^2 + 14x - 15 =0

x^2 + 14x = 15

to write the left hand side as a perfect square we add 49 to both sides :

x^2 + 14x + 49 = 15 + 49

x^2 + 2.x.7 + 7^2 = 64

using the identity (a^2 + b^2) = a^2 + 2ab + b^2

(x + 7)^2 = 64

x + 7 = √64 or x + 7 = -√64

x = 8 - 7 or x = - 8 - 7

= 1 = - 15

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aalyn [17]

For this case we have the following expression:

(3x + 8) (3x-3)

We apply distributive property, which by definition establishes that:

(a + b) (c + d) = ac + ad + bc + bd

So:

(3x + 8) (3x-3) =\\9x ^ 2-9x + 24x-24 =\\9x ^ 2 + 15x-24

Answer:

9x ^ 2 + 15x-24

7 0
3 years ago
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What is the volume of the prism that can be constructed from this net?
Alecsey [184]
That would make a rectangular prism with measurements of:

Length = 9
Width = 4
Height = 10

V = lwh

V = 9 * 4 * 10

V = 360
7 0
3 years ago
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Choose whether it's always, sometimes, never 
Keith_Richards [23]

Answer: An integer added to an integer is an integer, this statement is always true. A polynomial subtracted from a polynomial is a polynomial, this statement is always true. A polynomial divided by a polynomial is a polynomial, this statement is sometimes true. A polynomial multiplied by a polynomial is a polynomial, this statement is always true.

Explanation:

1)

The closure property of integer states that the addition, subtraction and multiplication is integers is always an integer.

If a\in Z\text{ and }b\in Z, then a+b\in Z.

Therefore, an integer added to an integer is an integer, this statement is always true.

2)

A polynomial is in the form of,

p(x)=a_nx^n+a_{n-1}x^{x-1}+...+a_1x+a_0

Where a_n,a_{n-1},...,a_1,a_0 are constant coefficient.

When we subtract the two polynomial then the resultant is also a polynomial form.

Therefore, a polynomial subtracted from a polynomial is a polynomial, this statement is always true.

3)

If a polynomial divided by a polynomial  then it may or may not be a polynomial.

If the degree of numerator polynomial is higher than the degree of denominator polynomial then it may be a polynomial.

For example:

f(x)=x^2-2x+5x-10 \text{ and } g(x)=x-2

Then \frac{f(x)}{g(x)}=x^2+5, which a polynomial.

If the degree of numerator polynomial is less than the degree of denominator polynomial then it is a rational function.

For example:

f(x)=x^2-2x+5x-10 \text{ and } g(x)=x-2

Then \frac{g(x)}{f(x)}=\frac{1}{x^2+5}, which a not a polynomial.

Therefore, a polynomial divided by a polynomial is a polynomial, this statement is sometimes true.

4)

As we know a polynomial is in the form of,

p(x)=a_nx^n+a_{n-1}x^{x-1}+...+a_1x+a_0

Where a_n,a_{n-1},...,a_1,a_0 are constant coefficient.

When we multiply the two polynomial, the degree of the resultand function is addition of degree of both polyminals and the resultant is also a polynomial form.

Therefore, a polynomial subtracted from a polynomial is a polynomial, this statement is always true.

3 0
3 years ago
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Two types of defects are observed in the production of integrated circuit boards. It is estimated that 6% of the boards have sol
aalyn [17]

Answer:

(a) 0.09

(b) 0.06

(c) 0.0018

(d) 0.9118

(e) 0.03

Step-by-step explanation:

Let <em>A</em> = boards have solder defects and <em>B</em> = boards have surface defects.

The proportion of boards having solder defects is, P (A) = 0.06.

The proportion of boards having surface-finish defects is, P (B) = 0.03.

It is provided that the events A and B are independent, i.e.

P(A\cap B)=P(A)\times P(B)

(a)

Compute the probability that either a solder defect or a surface-finish defect or both are found as follows:

= P (A or B) + P (A and B)

=P(A)+P(B)-P(A\cap B)+P(A\cap B)\\=P(A)+P(B)\\=0.06+0.03\\=0.09

Thus, the probability that either a solder defect or a surface-finish defect or both are found is 0.09.

(b)

The probability that a solder defect is found is 0.06.

(c)

The probability that both defect are found is:

P(A\cap B)=P(A)\times P(B)\\=0.06\times0.03\\=0.0018

Thus, the probability that both defect are found is 0.0018.

(d)

The probability that none of the defect is found is:

P(A^{c}\cup B^{c})=1-P(A\cup B)\\=1-P(A)-P(B)+P(A\cap B)\\=1-P(A)-P(B)+[P(A)\times P(B)]\\=1-0.06-0.03+(0.06\times0.03)\\=0.9118

Thus, the probability that none of the defect is found is 0.9118.

(e)

The probability that the defect found is a surface finish is 0.03.

6 0
3 years ago
T a linen sale Mrs. Earle bought twice as many pillowcases for $2 each as sheets for $5 each. If she spent less than $40, not in
Anettt [7]
For this case, the first thing we must do is define variables:
 x: number of pillowcases
 y: number of sheets
 We now write the system of equations:
 2x + 5y = 40
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 Solving the system we have:
 x = 8.9
 y = 4.4
 Answer:
 
The maximum number of pillowcases she could have purchased is:
 
x = 8 (spent less than $ 40)
3 0
3 years ago
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