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2 years ago

What is the equation of the line that passes through (3,-1) and (4,7)

2 answers:

8 0

Gradient = 7-(-1)/4-3
= 8
when x = 3, y = -1,
y = 8x + c
-1 = 8(3) + c
c = -1 - 24
= -25
the equation of the line is y = 8x - 25.

cupoosta [38]2 years ago

4 0

Using y=mx+b the answer would be y=8(x-3)-1.

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For the figures below, assume they are made of semicircles, quarter circles and squares. For each shape, find the area and perim

ICE Princess25 [194]

Answer:

Part a) The area of the figure is $\frac{9}{2}(4+\pi )\ cm^{2}$

Part b) The perimeter of the figure is $3(2+2\sqrt{2}+ \pi)\ cm$

Step-by-step explanation:

Step 1

Find the area of the figure

In this problem we have that

The figure ABC is the half of a square and the other figure is a semicircle

<u>Find the area of the figure ABC</u>

we have

$AB=6\ cm, BC=6\ cm$

The area of the half square ABC is equal to find the area of triangle ABC

$A1=\frac{1}{2}*6*6=18\ cm^{2}$

<u>Find the area of the semicircle</u>

The area of the semicircle is equal to

$A2=\pi r^{2}/2$

we have that

$r=6/2=3\ cm$

substitute

$A2=\pi (3)^{2}/2$

$A2=(9/2) \pi\ cm^{2}$

The area of the figure is equal to

$18\ cm^{2}+(9/2) \pi\ cm^{2}= \frac{9}{2}(4+\pi )\ cm^{2}$

Step 2

Find the perimeter of the figure

The perimeter of the figure is equal to

$P=AB+AC+length\ CB$

we have

$AB=6\ cm$

Applying Pythagoras theorem

$AC=\sqrt{6^{2}+6^{2}}\\AC=6\sqrt{2}\ cm$

Remember that

the circumference of a semicircle is equal to

$C=\frac{1}{2}2\pi r=\pi r$

$r=6/2=3\ cm$

$C=\pi(3)$

$C=3 \pi\ cm$

The perimeter of the figure is equal to

$P=6\ cm+6\sqrt{2}\ cm+3 \pi\ cm$

Simplify

$P=3(2+2\sqrt{2}+ \pi)\ cm$

5 0

3 years ago

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3 years ago