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1 year ago

6. rule (x,y) ---> is the image similar to the original pre-image?

Mathematics

1 answer:

Artyom0805 [142]1 year ago

3 0

Here, we want to check for the relationship between the image and its pre-image

The pre-image is (x,y)

The image is (3x,3y+5)

As we can see, the pre-image is not similar

This is because the transformation applied to the two values are not same

Thus, we have that;

No, the image is not similar to the pre-image as the translations applied to both coordinates are not same

The pre-image was transformed by dilating the x-coordinate of the pre-image by 3 units while the y-coordinate was transformed by dilating the y-coordinate of the pre-image by 3 and translating it upward by 5 units

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There were 60 ducks living in a pond. If the number living at the pond increased by 30%, how many ducks now live at the pond?

timama [110]

Answer:

Step-by-step explanation:

You take the original total: 60, and multiply it by 30 to get 1,800. You then divide 1,800 by 100 to get 18. 18 is your 30%. Now, you add the 30% to 60 getting your total number of ducks: 78.

The equation version:

$\frac{a}{60} = \frac{30}{100}$

It is just proportions.

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3 years ago

What is the mass of a 15cm cube of lron if density of iron is 78grams/cm

amm1812

Answer:

M = 1170 g

Step-by-step explanation:

Mass = density × volume

M = p x v

M = 78 x 15

M = 1170 g

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3 years ago

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marishachu [46]

Answer:

$M = \log(10000)$

Step-by-step explanation:

Given

$M = \log(\frac{I}{I_o})$

$I = 10000I_o$ ---- intensity is 10000 times reference earthquake

Required

The resulting equation

We have:

$M = \log(\frac{I}{I_o})$

Substitute the right values

$M = \log(\frac{10000I_o}{I_o})$

$M = \log(10000)$

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3 years ago

What does this equal <br>32p = 160

Assoli18 [71]

You should divide each side by 32. Then when you divide the 160 from 32 you should get 5. Because 32x5 =160

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4 years ago

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Help with this math question

tresset_1 [31]

The given inequality is:

$|-8x+24| \leq 16$

This inequality can be divided in two parts as:

a) $-16 \leq -8x +24$
b) $-8x + 24 \leq 16$

Solving part a:

$-16 \leq -8x+24 \\ \\ 
-40 \leq -8x \\ \\ 
5 \geq x \\ 
or \\ 
x \leq 5$

Solving part b:

$-8x+24 \leq 16 \\ \\ 
-8x \leq -8 \\ \\ 
x \geq 1$

Therefore, the solution to the given inequality is $x \leq 5$ and $x \geq 1$ . Combining both the ranges we get the solution: $1 \leq x \leq 5$ .

In interval notation, this solution can be expressed as [1,5]

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4 years ago

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