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3 years ago

HELP PLEASE

Mathematics

1 answer:

postnew [5]3 years ago

5 0

x^2 - 14x - 32 = x^2 - 16x + 2x - 32 = x(x-16) + 2(x-16) = (x - 16)(x + 2)

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2n^2 - 7n - 15 =

= 2n^2 - 10n + 3n - 15 =

= 2n(n - 5) + 3(n - 5) =

= (n - 5)(2n + 3)

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You might be interested in

The vertices of ∆ABC are A(2, 8), B(16, 2), and C(6, 2). what is the perimeter and area in square units

ad-work [718]

Check the picture below.

the triangle has that base and that height, recall that A = 1/2 bh.

now as for the perimeter, you can pretty much count the units off the grid for the segment CB, so let's just find the lengths of AC and AB,

$\bf ~~~~~~~~~~~~\textit{distance between 2 points}\\\\
\begin{array}{ccccccccc}
&&x_1&&y_1&&x_2&&y_2\\
% (a,b)
&A&(~ 2 &,& 8~) 
% (c,d)
&C&(~ 6 &,& 2~)
\end{array}~~~ 
% distance value
d = \sqrt{( x_2- x_1)^2 + ( y_2- y_1)^2}
\\\\\\
AC=\sqrt{(6-2)^2+(2-8)^2}\implies AC=\sqrt{4^2+(-6)^2}
\\\\\\
AC=\sqrt{16+36}\implies AC=\sqrt{52}\implies AC=\sqrt{4\cdot 13}
\\\\\\
AC=\sqrt{2^2\cdot 13}\implies AC=2\sqrt{13}$

$\bf ~~~~~~~~~~~~\textit{distance between 2 points}\\\\
\begin{array}{ccccccccc}
&&x_1&&y_1&&x_2&&y_2\\
% (a,b)
&A&(~ 2 &,& 8~) 
% (c,d)
&B&(~ 16 &,& 2~)
\end{array}\\\\\\
AB=\sqrt{(16-2)^2+(2-8)^2}\implies AB=\sqrt{14^2+(-6)^2}
\\\\\\
AB=\sqrt{196+36}\implies AB=\sqrt{232}\implies AB=\sqrt{4\cdot 58}
\\\\\\
AB=\sqrt{2^2\cdot 58}\implies AB=2\sqrt{58}$

so, add AC + AB + CB, and that's the perimeter of the triangle.

8 0

3 years ago

HELP ME HELP ME HELP ME HELP ME HELP ME HELP ME HELP ME HELP ME HELP ME HELP ME HELP ME HELP ME HELP ME HELP ME HELP ME HELP ME

Kryger [21]

Answer:

Given the triangle RST with Coordinates R(2,1), S(2, -2) and T(-1 , -2).

A dilation is a transformation which produces an image that is the same shape as original one, but is different size.

Since, the scale factor $\frac{5}{3}$ is greater than 1, the image is enlargement or a stretch.

Now, draw the dilation image of the triangle RST with center (2,-2) and scale factor $\frac{5}{3}$

Since, the center of dilation at S(2,-2) is not at the origin, so the point S and its image $S{}'$ are same.

Now, the distances from the center of the dilation at point S to the other points R and T.

The dilation image will be $\frac{5}{3}$ of each of these distances,

$SR=3$ , so $S{}'R{}'$ =5 ;

$ST=3$ , so $S{}'T{}'=5$

Now, draw the image of RST i.e R'S'T'

Since, $RT=3\sqrt{2}$ [By using hypotenuse of right angle triangle] and $R{}'T{}'=5\sqrt{2}$ .

(a)

Disagree with the given statement.

Side Angle Side postulate (SAS) states that:

If two sides and the included angle of one triangle are congruent to two sides and the included angle of another triangle then these two triangles are congruent.

Given: B is the midpoint of $\overline{AC}$ i.e $\overline{AB}\cong \overline{BC}$

In the triangle ABD and triangle CBD, we have

$\overline{AB}\cong \overline{BC}$ (SIDE) [Given]

$\overline{BD}\cong \overline{BD}$ (SIDE) [Reflexive post]

Since, there is no included angle in these triangles.

∴ $\Delta ABD$ is not congruent to $\Delta CBD$ .

Therefore, these triangles does not follow the SAS congruence postulates.

(b)

SSS(SIDE-SIDE-SIDE) states that if three sides of one triangle are congruent to three sides of another triangle, then the triangles are congruent.

Since it is also given that $\overline{AD}\cong \overline{CD}$ .