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Advocard [28]
3 years ago
10

Would you classify the number 169 as a perfect square a perfect cube both neither

Mathematics
2 answers:
seropon [69]3 years ago
4 0
A perfect square because the square root of 169 is 13, but 169 can not be cube rooted perfectly. So 169 is a perfect square.
Alex_Xolod [135]3 years ago
3 0
Perfect square because the square root of 169 is 13 but does not have a cubic measurement.
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lisov135 [29]

In math, an isometry is a congruent transformation in which the distance (or length) and the angle is preserved or remains the same even after the transformation.

The transformation can be translation, rotation, reflection, etc.

Let us not use this definition of isometry to answer our question, one at a time.

(I) In here, as we can see the distances 10 and 5 and the angle 43 degrees has been preserved. So, <u>this is an isometry.</u>

(II) In here, distances have been halved, so this is<u> not an isometry</u>, even though the angles have been preserved.

(III) In here, the corresponding distances and the angles have been preserved. So, <u>this is an isometry.</u>

6 0
3 years ago
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ivanzaharov [21]

Answer:

Z=BY+R

Step-by-step explanation:

B=z-r/y

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Z =BY+R

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7 0
3 years ago
Use slope formula,m = StartFraction y 2 minus y 1 Over x 2 minus x 1 EndFraction, to find the slope of a line that passes throug
Leya [2.2K]

Answer:

a. Slope is -2/7

b. b = 1

c. y = 2x + 1

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Step-by-step explanation:

7 0
3 years ago
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kodGreya [7K]

Choice C, because

\frac{3}{4}  =  \frac{6}{8}

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6 0
3 years ago
M is inversely proportional to r if m= 9 when r= 4 find m when r=2
bazaltina [42]

m=18 when r = 2.

Step-by-step explanation:

Given,

m∝\frac{1}{r}

So,

m = k×\frac{1}{r},--------eq 1, here k is the constant.

To find the value of m when r = 2

At first we need to find the value of k

Solution

Now,

Putting the values of m=9 and r = 4 in eq 1 we get,

9 = \frac{k}{4}

or, k = 36

So, eq 1 can be written as m= \frac{36}{r}

Now, we put r =2

m = \frac{36}{2}

or, m= 18

Hence,

m=18 when r = 2.

4 0
3 years ago
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