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Elenna [48]
3 years ago
8

Does — бр + 7 = 7 – бр have No Solution, One Solution, or Infinitely

Mathematics
2 answers:
julsineya [31]3 years ago
4 0

Answer:

Infinitely

Step-by-step explanation:

Hi,

Both sides of the equation are equal to each other, this means they're the exact same line and overlap. This is infinitely many solutions.

Hope this helps :)

Serhud [2]3 years ago
4 0

Answer:

Infinitely

Step-by-step explanation:

-6p+7=7-6p can be reshuffled to read: -6p+7=-6+7. Both sides of the equation are the same, so there are infinite solutions.

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1.5x - 3y = 6 (solve for y)
d1i1m1o1n [39]
Hello : 
<span>1.5x - 3y = 6
3y = 1.5x - 6
y = (1.5x - 6)/3</span>
7 0
3 years ago
Layla arked a group of students to chooue their favorite type of exerche from the choices of jogging, biling, and wwimming. The
Inessa [10]

Answer:

50 students

Step-by-step explanation:

20 + 30 = 50

7 0
2 years ago
What is 40t^2 - 400t + 500 = 0 factored
Dmitry_Shevchenko [17]
= 40(t^2 - 10t + 12.5)  answer
6 0
3 years ago
Given that −4i is a zero, factor the following polynomial function completely. Use the Conjugate Roots Theorem, if applicable. f
GalinKa [24]

Answer:

\large \boxed{\sf \bf \ \ f(x)=(x-4i)(x+4i)(x+3)(x-5) \ \ }

Step-by-step explanation:

Hello, the Conjugate Roots Theorem states that if a complex number is a zero of real polynomial its conjugate is a zero too. It means that (x-4i)(x+4i) are factors of f(x).

\text{Meaning that } (x-4i)(x+4i) =x^2-(4i)^2=x^2+16 \text{ is a factor of f(x).}

The coefficient of the leading term is 1 and the constant term is -240 = 16 * (-15), so we a re looking for a real number such that.

f(x)=x^4-2x^3+x^2-32x-240\\\\ =(x^2+16)(x^2+ax-15)\\\\ =x^4+ax^3-15x^2+16x^2+16ax-240

We identify the coefficients for the like terms, it comes

a = -2 and 16a = -32 (which is equivalent). So, we can write in \mathbb{R}.

\\f(x)=(x^2+16)(x^2-2x-15)

The sum of the zeroes is 2=5-3 and their product is -15=-3*5, so we can factorise by (x-5)(x+3), which gives.

f(x)=(x^2+16)(x^2-2x-15)\\\\=(x^2+16)(x^2+3x-5x-15)\\\\=(x^2+16)(x(x+3)-5(x+3))\\\\=\boxed{(x^2+16)(x+3)(x-5)}

And we can write in \mathbb{C}

f(x)=\boxed{(x-4i)(x+4i)(x+3)(x-5)}

Hope this helps.

Do not hesitate if you need further explanation.

Thank you

7 0
3 years ago
Which fraction can be replaced with 1/2 when estimating with fractions
White raven [17]

Answer: 7/16

Step-by-step explanation:

i got it right

5 0
2 years ago
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