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expeople1 [14]
3 years ago
5

Which of the following constants can be added to x^2+15x to form a perfect square trinomial?

Mathematics
2 answers:
ycow [4]3 years ago
3 0

Answer:

Answer b. \frac{225}{4}

Step-by-step explanation:

In order to get a perfect square trinomial from a binomial of the form: x^2 +bx. the term to add should be the square of half of the coeficcient "b". That is: (\frac{b}{2} )^2. Such will give you a trinomial that comes from the perfect square of the binomial: (x+\frac{b}{2})^2=x^2+bx+(\frac{b}{2}  )^2

In your case, b=15 therefore (\frac{b}{2} )^2=(\frac{15}{2} )^2=\frac{225}{4}

MatroZZZ [7]3 years ago
3 0

<u>Answer:</u>

Option C. The constant is \frac{225}{4}

<u>Solution: </u>

Let us assume that the constant is c.

Now to the equation will be x^{2}+15 x+c-------- (i)

We know the square formula  (a+b)^{2}=a^{2}+2 a b+b^{2}

As per the formula we can write the equation as,

x^{2}+15 x+c

=x^{2}+2 \times x \times\left(\frac{15}{2}\right)+c

Now if we need make the equation perfect square,

Then as per the formula c should be \left(\frac{15}{2}\right)^{2}=\frac{225}{4}

And the equation will be the perfect square as \left(x+\frac{15}{2}\right)^{2}

So, the constant is \frac{225}{4}

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16x⁴-4x²-4x-1 and 8x³-1<br>HCF​
IgorLugansk [536]

Answer:

We have expanded formula of (-4x-1)² = a²+2ab+b².

So, we write the formula in square form as (a+b)².

Since we have a²-b² in step 4. We further write this as (a+b)(a-b). This is the factor formula of a²-b².

As we had two terms in place of in (a+b)(a-b), we multiply the term 'b' with '+' and '-' sign respectively.  

Write the second expression given in the question.

Write the terms in the form of cube.

Write the factor formula of a³-b³) in the form of (a-b)(a²+ab+b²).

Write the H.C.F. (Highest Common Factor) of the given expressions by analysing the factors you generated in each expressions. Here, (4x²+2x+1) are the common factors.

4 0
2 years ago
I need help and have to show my work
Oksi-84 [34.3K]
Okay 
5.
m+4=17
   -4   -4
    m=13
6.
12=24-y
24-12
y=12
7.
15-b=12
-12   -12
b=3
8.
10t=90
90/10
t=9
9.
22/y=2
22/2
y=11
10.
54=6b
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7 0
3 years ago
Can someone please help me.
Anika [276]

Answer:

Radius= 8.99 mm (9 mm)

Step-by-step explanation:

We use the formula for the volume of a cone which is

1/3 pi ×(radius)^2 *height

932.58 = (pi) × (11) ×(r^2/3)

Radius^2 = (932.58×3) / (11pi)

= 80.96m

Radius= 8.99mm

5 0
2 years ago
Read 2 more answers
given examples of relations that have the following properties 1) relexive in some set A and symmetric but not transitive 2) equ
rodikova [14]

Answer: 1) R = {(a, a), (а,b), (b, a), (b, b), (с, с), (b, с), (с, b)}.

It is clearly not transitive since (a, b) ∈ R and (b, c) ∈ R whilst (a, c) ¢ R. On the other hand, it is reflexive since (x, x) ∈ R for all cases of x: x = a, x = b, and x = c. Likewise, it is symmetric since (а, b) ∈ R and (b, а) ∈ R and (b, с) ∈ R and (c, b) ∈ R.

2) Let S=Z and define R = {(x,y) |x and y have the same parity}

i.e., x and y are either both even or both odd.

The parity relation is an equivalence relation.

a. For any x ∈ Z, x has the same parity as itself, so (x,x) ∈ R.

b. If (x,y) ∈ R, x and y have the same parity, so (y,x) ∈ R.

c. If (x.y) ∈ R, and (y,z) ∈ R, then x and z have the same parity as y, so they have the same parity as each other (if y is odd, both x and z are odd; if y is even, both x and z are even), thus (x,z)∈ R.

3) A reflexive relation is a serial relation but the converse is not true. So, for number 3, a relation that is reflexive but not transitive would also be serial but not transitive, so the relation provided in (1) satisfies this condition.

Step-by-step explanation:

1) By definition,

a) R, a relation in a set X, is reflexive if and only if ∀x∈X, xRx ---> xRx.

That is, x works at the same place of x.

b) R is symmetric if and only if ∀x,y ∈ X, xRy ---> yRx

That is if x works at the same place y, then y works at the same place for x.

c) R is transitive if and only if ∀x,y,z ∈ X, xRy∧yRz ---> xRz

That is, if x works at the same place for y and y works at the same place for z, then x works at the same place for z.

2) An equivalence relation on a set S, is a relation on S which is reflexive, symmetric and transitive.

3) A reflexive relation is a serial relation but the converse is not true. So, for number 3, a relation that is reflexive but not transitive would also be serial and not transitive.

QED!

6 0
3 years ago
Find the values of a and b such that x^2-2x+2=(x-a)^2+b
vekshin1

Answer:

a = 1, b = 1

Step-by-step explanation:

Expand the right side and compare the coefficients of like terms on both sides, that is

right side

(x - a)² + b ← expand factor using FOIL

= x² - 2ax + a² + b

Compare to left side x² - 2x + 2

Compare the coefficients of the x- term

- 2a = - 2 ( divide both sides by - 2 )

a = 1

Compare the constant terms

a² + b = 2 ( substitute a = 1 )

1² + b = 2

1 + b = 2 ( subtract 1 from both sides )

b = 1

Thus a = 1, b = 1

6 0
2 years ago
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