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bogdanovich [222]
3 years ago
11

3 Ximena draws an X shape in the coordinate plane as shown. Draw the images of

Mathematics
1 answer:
VARVARA [1.3K]3 years ago
3 0

When a shape is rotated, it must be rotated around a point.

<em>See attachment for the image of each rotation.</em>

To do this, the top coordinates of the X shape will be transformed using the appropriate rotation rule; the same rule will then be applied to the other parts of the X shape.

The top coordinates of the X shape are:

A = (0,0)

B = (1,0)

C = (3,0)

D =(4,0)

For 90 degrees counterclockwise rotation, the rule is:

(x,y) \to (-y,x)

So, we have:

A' = (0,0)

B' = (0,1)

C' = (0,3)

D' = (0,4)

For 180 degrees rotation, the rule is:

(x,y) \to (-x,-y)

So, we have:

A' = (0,0)

B' = (-1,0)

C' = (-3,0)

D' = (-4,0)

For 270 degrees counter rotation, the rule is:

(x,y) \to (y,-x)

So, we have:

A' = (0,0)

B' = (0,-1)

C' = (0,-3)

D' = (0,-4)

See attachment for the image of each rotation

Read more about rotations at:

brainly.com/question/1571997

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4 years ago
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Solve for x : log(3x) + log(x + 4) = log(15).<br><br>Please explain it to me.​
zlopas [31]

Answer:

x = 1

Explanation:

\sf \rightarrow log(3x) + log(x + 4) = log(15)

Rule: log(a) + log(b) = log(ab)

\sf \rightarrow log(3x(x + 4)) =  log(15)

cancel out log on both sides

\sf \rightarrow 3x(x + 4) = 15

relocate constant variable

\sf \rightarrow 3x^2 + 12x -15 = 0

take 3 as a common factor

\sf \rightarrow 3(x^2 + 4x -5) = 0

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\sf \rightarrow x^2 + 5x -x-5 = 0

factor common terms

\sf \rightarrow x(x  + 5) -1(x+5)= 0

collect into groups

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set to zero

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2 years ago
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The top and bottom margins of a poster are each 12 cm and the side margins are each 8 cm. The area of printed material on the po
grin007 [14]

Answer:

Dimensions of printed poster are

length is 32 cm

width is 48 cm


Step-by-step explanation:

Let's assume

length of printed poster is x cm

width of printed poster is y cm

now, we can find area of printed poster

so, area of printed poster is

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we are given that area as 1536

so, we can set it to 1536

xy=1536

now, we can solve for y

y=\frac{1536}{x}

now, we are given

The top and bottom margins of a poster are each 12 cm and the side margins are each 8 cm

so, total area of poster is

A=(8+x+8)\times (12+y+12)

A=(x+16)\times (y+24)

now, we can plug back y

A=(x+16)\times (\frac{1536}{x}+24)

now, we have to minimize A

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A'=\frac{d}{dx}\left(\left(x+16\right)\left(\frac{1536}{x}+24\right)\right)

we can use product rule

A'=\frac{d}{dx}\left(x+16\right)\left(\frac{1536}{x}+24\right)+\frac{d}{dx}\left(\frac{1536}{x}+24\right)\left(x+16\right)

=\frac{d}{dx}\left(x+16\right)\left(\frac{1536}{x}+24\right)+\frac{d}{dx}\left(\frac{1536}{x}+24\right)\left(x+16\right)

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now, we can set it to 0

and then we can solve for x

A'=-\frac{24576}{x^2}+24=0

-\frac{24576}{x^2}x^2+24x^2=0\cdot \:x^2

-24576+24x^2=0

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Since, x is dimension

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so, we will only consider positive value

x=32

now, we can solve for y

y=\frac{1536}{32}

y=48

so, dimensions of printed poster are

length is 32 cm

width is 48 cm


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