Simplify the expression <br><br> c+2+c+c+4

Find the factorization of the polynomial

makvit [3.9K]

There is only one real root, at x=-2, so the polynomial describing this parabola has factors of (x+2) with multiplicity 2. The y-intercept tells you the vertical stretch is 1.

The factorization is y = (x +2)².

6 0

3 years ago

There are 3 islands A,B,C. Island B is east of island A, 8 miles away. Island C is northeast of A, 5 miles away and northwest of

Nostrana [21]

Answer:

The bearing needed to navigate from island B to island C is approximately 38.213º.

Step-by-step explanation:

The geometrical diagram representing the statement is introduced below as attachment, and from Trigonometry we determine that bearing needed to navigate from island B to C by the Cosine Law:

$AC^{2} = AB^{2}+BC^{2}-2\cdot AB\cdot BC\cdot \cos \theta$ (1)

Where:

$AC$ - The distance from A to C, measured in miles.

$AB$ - The distance from A to B, measured in miles.

$BC$ - The distance from B to C, measured in miles.

$\theta$ - Bearing from island B to island C, measured in sexagesimal degrees.

Then, we clear the bearing angle within the equation:

$AC^{2}-AB^{2}-BC^{2}=-2\cdot AB\cdot BC\cdot \cos \theta$

$\cos \theta = \frac{BC^{2}+AB^{2}-AC^{2}}{2\cdot AB\cdot BC}$

$\theta = \cos^{-1}\left(\frac{BC^{2}+AB^{2}-AC^{2}}{2\cdot AB\cdot BC} \right)$ (2)

If we know that $BC = 7\,mi$ , $AB = 8\,mi$ , $AC = 5\,mi$ , then the bearing from island B to island C:

$\theta = \cos^{-1}\left[\frac{(7\mi)^{2}+(8\,mi)^{2}-(5\,mi)^{2}}{2\cdot (8\,mi)\cdot (7\,mi)} \right]$

$\theta \approx 38.213^{\circ}$

The bearing needed to navigate from island B to island C is approximately 38.213º.

8 0

3 years ago

A toy that originally sold for 12 dollars went of sale for 8 dollars what was the percent discount? Remember to show your work!

harkovskaia [24]

12———-100%
8 ———— x
X=(8*100)/12=800/12=66.66%
Then discount percent is 100-66.66=33.33%

3 0

3 years ago

30 POINTS,NEED HELP ASAP !!!

zaharov [31]

Answer: OPTION A.

Step-by-step explanation:

You can observe that in the figure CDEF the vertices are:

$C(-2,-1),\ D(-2,0),\ E(2,2)\ and\ F(2,1)$

And in the figure C'D'E'F' the vertices are:

$C'(-8,-4),\ D'(-8,0),\ E'(8,8)\ and\ F'(8,4)$

For this case, you can divide any coordinate of any vertex of the figure C'D'E'F' by any coordinate of any vertex of the figure CDEF:

For C'(-8,-4) and C(-2,-1):

$\frac{-8}{-2}=4\\\\\frac{-4}{-1}=4$

Let's choose another vertex. For E'(8,8) and E(2,2):

$\frac{8}{2}=4\\\\\frac{8}{2}=4$

You can observe that the coordinates of C' are obtained by multiplying each coordinate of C by 4 and the the coordinates of E' are also obtained by multiplying each coordinate of E by 4.

Therefore, the rule that yields the dilation of the figure CDEF centered at the origin is:

$(x, y)$ → $(4x, 4y)$

7 0

3 years ago

Read 2 more answers

In the figure below, segment CD is parallel to segment EF and point H bisects segment DE :

Grace [21]

Answer:

See explanation

Step-by-step explanation:

In the figure below, segment CD is parallel to segment EF, DE is a transversal, then angles DIH and HGI are congruent as alternate interior angles when two parallel lines are cut by a transversal.

Consider triangles DIH and EGH. In these triangles,

$\angle DIH\cong \angle EGH$ as alternate interior angles;
$\angle DHI\cong \angle GHE$ as vertical angles;
$DH\cong HE$ because point H bisects segment DE (given).

Thus,

$\triangle DIH\cong \triangle EGH$ by AAS postulate

4 0

3 years ago

Simplify the expression c+2+c+c+4