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3 years ago

Please help if u can

Mathematics

2 answers:

Mashutka [201]3 years ago

5 0

Total area = area of parallelogram + area of triangle
= base*height + 1/2base*height
= 11*9 + 1/2*11*7
=99+38.5
=137.5 ...answer !!

laila [671]3 years ago

4 0

Area of parallelogram
A = bh
A = 11 x 9 = 99

area of triangle
A = 1/2bh
A = 1/2 (7)(11)
A = 38.5

area of figure

99 + 38.5 = 137.5

answer
137.5 cm^2

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Q # 1

Explanation

Given the parabola

$f\left(x\right)=\left(x-3\right)^2-1$

Openness

It OPENS UP, as 'a=1' is positive.

Finding Vertex

The vertex of an up-down facing parabola of the form

$y=ax^2+bx+c\:\mathrm{is}\:x_v=-\frac{b}{2a}$

$\mathrm{Rewrite}\:y=\left(x-3\right)^2-1\:\mathrm{in\:the\:form}\:y=ax^2+bx+c$

$y=x^2-6x+8$

$a=1,\:b=-6,\:c=8$

$x_v=-\frac{\left(-6\right)}{2\cdot \:1}$

$x_v=3$

Finding $y_v$

$y_v=3^2-6\cdot \:3+8$

$y_v=-1$

So vertex is:

$\left(3,\:-1\right)$

Horizontal Translation

$y=\left(x-3\right)^2$ moves the graph RIGHT 3 units.

Vertical Translation

$f\left(x\right)=\left(x-3\right)^2-1$ moves the graph DOWN 1 unit.

Stretch or Compress Vertically

As $a = 1$ , so it does not affect the stretchiness or compression.

Q # 2

Explanation:

$f\left(x\right)=-\left(x+1\right)^2-2$

Openness

It OPENS DOWN, as 'a=-1' is negative.

Vertex

$\mathrm{Rewrite}\:y=-\left(x+1\right)^2-2\:\mathrm{in\:the\:form}\:y=ax^2+bx+c$

$y=-x^2-2x-3$

$a=-1,\:b=-2,\:c=-3$

$x_v=-\frac{\left(-2\right)}{2\left(-1\right)}$

$x_v=-1$

$\mathrm{Plug\:in}\:\:x_v=-1\:\mathrm{to\:find\:the}\:y_v\:\mathrm{value}$

$y_v=-2$

So vertex is:

$\left(-1,\:-2\right)$

Horizontal Translation

$y=\left(x+1\right)^2$ moves the graph LEFT 1 unit.

Vertical Translation

$f\left(x\right)=\left(x+1\right)^2-2$ moves the graph DOWN 2 unit.

Stretch or Compress Vertically

As $a = -1$ < 0, so it is either stretched or compressed.

Q # 3

Explanation:

$f\left(x\right)=\frac{1}{3}\left(x-4\right)^2+6$

It OPENS UP, as 'a=1/3' is positive.

Vertex

$\mathrm{Rewrite}\:y=\frac{1}{3}\left(x-4\right)^2+6\:\mathrm{in\:the\:form}\:y=ax^2+bx+c$

$y=\frac{1\cdot \:x^2}{3}-\frac{8x}{3}+\frac{34}{3}$

$a=\frac{1}{3},\:b=-\frac{8}{3},\:c=\frac{34}{3}$

$x_v=-\frac{\left(-\frac{8}{3}\right)}{2\left(\frac{1}{3}\right)}$

$x_v=4$

Finding $y_v$

$y_v=\frac{1\cdot \:4^2}{3}-\frac{8\cdot \:4}{3}+\frac{34}{3}$

$y_v=6$

So vertex is:

$\left(4,\:6\right)$

Horizontal Translation

$f\left(x\right)=\left(x-4\right)^2$ moves the graph RIGHT 4 units.

Vertical Translation

$f\left(x\right)=}\left(x-4\right)^2+6$ moves the graph UP 6 unit.

Stretch or Compress Vertically

As $a=\frac{1}{3}$ , so it the graph is vertically compressed by a factor of 1/3.

Check the attached comparison graphs.

Q # 4

Explanation:

Given the function

$f\left(x\right)=-\left(x+3\right)^2$

It OPENS DOWN, as 'a=-1' is negative.

Vertex

The vertex of an up-down facing parabola of the form $y=a\left(x-m\right)\left(x-n\right)$

is the average of the zeros $x_v=\frac{m+n}{2}$

$y=-\left(x+3\right)^2$

$a=-1,\:m=-3,\:n=-3$

$x_v=\frac{m+n}{2}$

$x_v=\frac{\left(-3\right)+\left(-3\right)}{2}$

$x_v=-3$

Finding $y_v$

$y_v=-\left(-3+3\right)^2$

$y_v=0$

So vertex is:

$\left(-3,\:0\right)$

Horizontal Translation

$y=\left(x+3\right)^2$ moves the graph LEFT 3 units.

Vertical Translation

$y=\left(x+3\right)^2$ does not move the graph vertically.

Stretch or Compress Vertically

As $a=-1$ , so it the graph is either vertically stretched or compressed.

Q # 5

Explanation:

$f\left(x\right)=\left(x+5\right)^2-3$

Openness

It OPENS UP, as 'a=1' is positive.

Vertex

$\mathrm{Rewrite}\:y=\left(x+5\right)^2-3\:\mathrm{in\:the\:form}\:y=ax^2+bx+c$

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$y_v=\left(-5\right)^2+10\left(-5\right)+22$

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Horizontal Translation

$f\left(x\right)=\left(x+5\right)^2$ moves the graph LEFT 5 units.

Vertical Translation

$f\left(x\right)=\left(x+5\right)^2-3$ moves the graph DOWN 3 unit.

Stretch or Compress Vertically

As $a = 1$ , so it does not affect the stretchiness or compression.

Check the attached comparison graphs.

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