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3 years ago

Multiply.

Mathematics

1 answer:

KATRIN_1 [288]3 years ago

7 0

The answer could be
-1287
im not sure

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Factor completely. n^2 + 7n - 44

4vir4ik [10]

(n+11)(n-4) is your answer

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3 years ago

Read 2 more answers

The angles of a quadrilateral measure 80, 100, 100, 80 in this order what kind of quadrilateral has this shape what attributes h

Harrizon [31]

Answer:

Isosceles trapezoid

Step-by-step explanation:

-An isosceles trapezoid is also sometimes called a convex quadrilateral.

-It properties include:

A line of symmetry can often bisects a pair of opposite sides.

It has to obtuse alternating with each other after which a pair of acute angles alternate with each other i.e 80°, 80°,100°,100°
It has a trapezoidal shape which by definition has a two pair of parallel sides.
The angles on it's base or ceiling are equal hence the name Isosceles Trapezoid.

8 0

3 years ago

6 - 2x = 6x - 10x + 4

yarga [219]

$6 - 2x = 6x - 10x + 4$

$6 - 2x = - 4x + 4$

$6 = - 4x + 4 + 2x$

$6 = - 2x + 4$

$- 2x + 4 = 6$

$- 2x = 6 - 4$

$- 2x = 2$

$x = \frac{ \: \: \: 2}{ - 2}$

$x = - 1$

Therefore , the value of x = -1

6 0

3 years ago

A polynomial has a leading coefficient of 1 and the

Hoochie [10]

Answer:

(A) $[x-(2+i)][x-(2-i)][x-\sqrt{2}][x+\sqrt{2}]$

Step-by-step explanation:

A polynomial has a leading coefficient of 1 and the following factors with multiplicity 1:

$x-(2+i)\\x-\sqrt{2}$

We apply the following to find the factored form of the polynomial.

If a complex number is a root of a polynomial with real coefficients, its complex conjugate is also a root of that polynomial.

If the polynomial has an irrational root $a+\sqrt{b}$ , where a and b are rational and b is not a perfect square, then it has also a conjugate root $a-\sqrt{b}$ .