Completely Factor 2n^2-23n+45=0

Alla [95]

Answer:

y=5x is the correct answer

4 0

3 years ago

Adaline is designing a garden in the shape of a 45º-45º-90º triangle, with an area of 32 ft2. To the nearest foot, how much fenc

wel

Answer:

27 ft

Step-by-step explanation:

Given: A garden in the shape of a 45º-45º-90º triangle.

Area of triangle= 32 ft²

As given dimension of triangle, we understand that it is a right angle and triangle and since two angles are equal, therefore two sides of triangle are also equal.

∴ Base and height will be equal.

Assuming base is also as "h"

Now, finding the length of base and height by using area of triangle formula.

Area of triangle= $\frac{1}{2} \times base\times height$

⇒ $32= \frac{1}{2} \times h\times h$

Multiplying both side by 2

⇒ $64= h^{2}$

Square rooting both side, remember; √a²= a

⇒ $\sqrt{64} = h$

∴ h= 8 feet

Hence, both base and height of right angle triangle is 8 feet.

Next, finding total fencing required to enclose the garden.

Perimeter of garden is equal to the amount of fencing required by Adaline.

Therefore, finding the length of hypotenous of triangle too.

Using pythogorean theoram.

$h^{2} = a^{2} +b^{2}$ , where a is the height and b is the base of triangle.

⇒ $h^{2}= 8^{2} +8^{2}$

⇒ $h^{2} = 128$

Square rooting both side

⇒ $h= \sqrt{128}$

∴ h= 11.3 feet

Perimeter of garden= $base+height+hypotenous$

⇒ Perimeter of garden= $8+8+11.3= 27.3\ feet$ $\approx 27\ feet$

Hence, 27 feet fencing is required to enclose the garden.

3 0

4 years ago

Read 2 more answers

Function rule and y intercept

stiks02 [169]

to get the equation for any straight line, all we need is two points from it, so hmmm say from this table let's use (2,0) and (6,-8)

$\bf \begin{array}{|r|r|r|r|r|r|} \cline{1-6} days&x&2&4&6&8\\ \cline{1-6} dollars&y&0&-4&-8&-12\\ \cline{1-6} \end{array}~\hfill (\stackrel{x_1}{2}~,~\stackrel{y_1}{0})\qquad (\stackrel{x_2}{6}~,~\stackrel{y_2}{-8}) \\\\\\ \stackrel{slope}{m}\implies \cfrac{\stackrel{rise} {\stackrel{y_2}{-8}-\stackrel{y1}{0}}}{\underset{run} {\underset{x_2}{6}-\underset{x_1}{2}}}\implies \cfrac{-8}{4}\implies \cfrac{\stackrel{rise}{-2}}{\underset{run}{1}}\implies \stackrel{slope}{-2}$

$\bf \begin{array}{|c|ll} \cline{1-1} \textit{point-slope form}\\ \cline{1-1} \\ y-y_1=m(x-x_1) \\\\ \cline{1-1} \end{array}\implies y-\stackrel{y_1}{0}=\stackrel{m}{-2}(x-\stackrel{x_1}{2}) \\\\\\ \stackrel{~\hfill y-intercept}{y=-2x\stackrel{\downarrow }{+4}}\qquad \impliedby \begin{array}{|c|ll} \cline{1-1} slope-intercept~form\\ \cline{1-1} \\ y=\underset{y-intercept}{\stackrel{slope\qquad }{\stackrel{\downarrow }{m}x+\underset{\uparrow }{b}}} \\\\ \cline{1-1} \end{array}$