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Kaylis [27]
2 years ago
8

What is the equation of the line passing through the points (-25, 50) and (25, 50) in slope-intercept form?

Mathematics
1 answer:
babunello [35]2 years ago
5 0

Answer:

Y = 50x

Step-by-step explanation:

Remember that slope intercept form is Y = mx + b indicating that m is the slope

To solve this equation use the formula

Y2 - y1 , x2 - x1:

50 - (- 50) = 100

25 - (-25) = 50

100/50 = 50/1

So the answer is y = 50x

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Use the function f(x) = -3x^2+13x+3 to find x when f(x) = -7
Bond [772]

Answer: - 235

<u>Step-by-step explanation:</u>

f(x) = -3x² + 13x + 3

f(-7) = -3(-7)² + 13(-7) + 3

      = -3(49) - 91 + 3

      = -147 - 88

      = -235  

7 0
3 years ago
Read 2 more answers
What is the standard form of the equation<br><br>Center (-14,5), circumference: 22π
IRISSAK [1]

The standard equation for a circle with center at (h,k) and radius r is

(x-h)^2 + (y-k)^2 = r^2, where (h,k) is the center of the circle.

The formula for the circumference of a circle is C = 2pi*r.  In this particular problem, we need to determine the radius of the circle.  That radius is:  r = C/[2pi].  Here, C = 22pi, so we get r = 22pi/[2pi], and so r^2 = 11^2.

Putting to use the given info, we have:

(x+14)^2 + (y-5)^2 = 11^2

6 0
3 years ago
An item is regularly priced at $80 . It is on sale for 20% off the regular price. How much (in dollars) is discounted from the r
Mazyrski [523]

80 * .20=16

It is 16 dollars off the regular price


3 0
3 years ago
Find a factorization of x² + 2x³ + 7x² - 6x + 44, given that<br> −2+i√√7 and 1 - i√/3 are roots.
Levart [38]

A factorization of x^4+2x^3+7x^2-6x+44 is (x^2+4x+11)(x^2-2x+4).

<h3>What are the properties of roots of a polynomial?</h3>
  • The maximum number of roots of a polynomial of degree n is n.
  • For a polynomial with real coefficients, the roots can be real or complex.
  • The complex roots of a polynomial with real coefficients always exist in a pair of conjugate numbers i.e., if a+ib is a root, then a-ib is also a root.

If the roots of the polynomial p(x)=ax^4+bx^3+cx^2+dx+e are r_1,r_2,r_3,r_4, then it can be factorized as p(x)=(x-r_1)(x-r_2)(x-r_3)(x-r_4).

Here, we are to find a factorization of p(x)=x^4+2x^3+7x^2-6x+44. Also, given that -2+i\sqrt{7} and 1-i\sqrt{3} are roots of the polynomial.

Since p(x)=x^4+2x^3+7x^2-6x+44 is a polynomial with real coefficients, so each complex root exists in a pair of conjugates.

Hence, -2-i\sqrt{7} and 1+i\sqrt{3} are also roots of the given polynomial.

Thus, all the four roots of the polynomial p(x)=x^4+2x^3+7x^2-6x+44, are: r_1=-2+i\sqrt{7}, r_2=-2-i\sqrt{7}, r_3=1-i\sqrt{3}, r_4=1+i\sqrt{3}.

So, the polynomial p(x)=x^4+2x^3+7x^2-6x+44 can be factorized as follows:

\{x-(-2+i\sqrt{7})\}\{x-(-2-i\sqrt{7})\}\{x-(1-i\sqrt{3})\}\{x-(1+i\sqrt{3})\}\\=(x+2-i\sqrt{7})(x+2+i\sqrt{7})(x-1+i\sqrt{3})(x-1-i\sqrt{3})\\=\{(x+2)^2+7\}\{(x-1)^2+3\}\hspace{1cm} [\because (a+b)(a-b)=a^2-b^2]\\=(x^2+4x+4+7)(x^2-2x+1+3)\\=(x^2+4x+11)(x^2-2x+4)

Therefore, a factorization of x^4+2x^3+7x^2-6x+44 is (x^2+4x+11)(x^2-2x+4).

To know more about factorization, refer: brainly.com/question/25829061

#SPJ9

3 0
1 year ago
Read 2 more answers
3. Which of the following is the surface area of this right triangular prism? (
Ostrovityanka [42]
160in 5x8=40x4=160 is the answer to ur question
8 0
2 years ago
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