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S_A_V [24]
2 years ago
9

Joan went to her local zoo that featured 18 whale exhibits. If the zoo features 45 exhibits in total then what percent of the zo

o”s exhibits are not feature whales
Mathematics
1 answer:
Alex787 [66]2 years ago
6 0
18/45 turns into 40%
You might be interested in
(x-8)*2/5=2<br>помогите пожалуйста
sweet-ann [11.9K]
(x-8)*2/5=2
0.4x - 3.2=2
0.2x-1.6=1
0.2x=2.6
x=2.6*5
x=13

^_^

7 0
3 years ago
An octagonal pyramid ... how many faces are there, how many vertices and how many edges? A triangular prism ... how many faces a
ahrayia [7]

1: 8 faces and 9 with the base 9 vertices and 16 edges

2: 3 faces and 5 with the bases 6 vertices and 9 edges

3: 3 faces and 4 with the base 4 vertices and 6 edges

4 0
4 years ago
Boris chooses three different numbers the sum of the three numbers is 36 one of the numbers is a cube number the other two numbe
OlgaM077 [116]

Answer:

3,4,5

Step-by-step explanation:

Here, we need to asume that the values are integers, as if not, there will be infinite solutions.

Lets go by parts:

The 1st number is a cubic, lets call it A

The second is a factor of 20, lets call it B. The same for the 3rd, call it C.

So, A + B+ C = 36

There are no many factors of 20, those are the numbers that when multiplied gives as the value of 20. Those are 1, 2, 4, 5, 10 and 20. So, B and C are some of these numbers.

Then, we know A if less than 36, other way the whole sum will be greater than 36. How many cubic number with integer cubic roots are less than 36? Well, lets guess:

1^3 = 1 -> valid, as it is less than 36

2^3 = 8 -> valid, as it is less than 36

3^2 = 27 -> valid, less than 36

4^3 = 64 -> not valid, as it is greater than 36

So, A ir 1, 2 or 3.

If A is 1, then B+C needs to be 35. But, from the factors of 20 that we listed, there are no combinations of 2 numbers that sum 35. So, A CAN'T be 1.

If A = 2, then we have:

8 + B + C = 36

B + C = 28

And again, there are not combinations of two factors of 20 that sum 28 (try yourself).

If A=3:

27 + B + C = 36

B +C = 9

And we have a winner!!! If B=4 and C=5 (or viceversa C=4 and B=5)

27 + 4 + 5 = 36

So, A=3, B=4 and C=5 or A=3, C=4 and B=5 are solutions.

3 0
4 years ago
1/4(x-2)+4=12 <br> 1/4(x-2)=8<br> x-2=2<br> x=4 <br> correct the error
konstantin123 [22]

Step-by-step explanation:

(i) x + 3 = 0

L.H.S. = x + 3

By putting x = 3,

L.H.S. = 3 + 3 = 6 ≠ R.H.S.

∴ No, the equation is not satisfied.

(ii) x + 3 = 0

L.H.S. = x + 3

By putting x = 0,

L.H.S. = 0 + 3 = 3 ≠ R.H.S.

∴ No, the equation is not satisfied.

(iii) x + 3 = 0

L.H.S. = x + 3

By putting x = −3,

L.H.S. = − 3 + 3 = 0 = R.H.S.

∴ Yes, the equation is satisfied.

(iv) x − 7 = 1

L.H.S. = x − 7

By putting x = 7,

L.H.S. = 7 − 7 = 0 ≠ R.H.S.

∴ No, the equation is not satisfied.

(v) x − 7 = 1

L.H.S. = x − 7

By putting x = 8,

L.H.S. = 8 − 7 = 1 = R.H.S.

∴ Yes, the equation is satisfied.

(vi) 5x = 25

L.H.S. = 5x

By putting x = 0,

L.H.S. = 5 × 0 = 0 ≠ R.H.S.

∴ No, the equation is not satisfied.

(vii) 5x = 25

L.H.S. = 5x

By putting x = 5,

L.H.S. = 5 × 5 = 25 = R.H.S.

∴ Yes, the equation is satisfied.

(viii) 5x = 25

L.H.S. = 5x

By putting x = −5,

L.H.S. = 5 × (−5) = −25 ≠ R.H.S.

∴ No, the equation is not satisfied.

(ix) = 2

L.H.S. =

By putting m = −6,

L. H. S. = ≠ R.H.S.

∴No, the equation is not satisfied.

(x) = 2

L.H.S. =

By putting m = 0,

L.H.S. = ≠ R.H.S.

∴No, the equation is not satisfied.

(xi) = 2

L.H.S. =

By putting m = 6,

L.H.S. = = R.H.S.

∴ Yes, the equation is satisfied.

8 0
3 years ago
Read 2 more answers
Need help please help me
ANTONII [103]
Megan:
x to the one third power =  x ^{1/3}
<span>x to the one twelfth power = </span>x ^{1/12}

<span>The quantity of x to the one third power, over x to the one twelfth power is:
</span>\frac{x ^{1/3}}{x ^{1/12}}
<span>
Since </span>\frac{ x^{a} }{ x^{b} } = x ^{a-b}
then \frac{x ^{1/3}}{x ^{1/12}} = x^{1/3-1/12}

Now, just subtract exponents:
1/3 - 1/12 = 4/12 - 1/12 = 3/12 = 1/4

\frac{x ^{1/3}}{x ^{1/12}} = x^{1/3-1/12} = x^{1/4}


Julie:
x times x to the second times x to the fifth = x * x² * x⁵

<span>The thirty second root of the quantity of x times x to the second times x to the fifth is
</span>\sqrt[32]{x* x^{2} * x^{5} }
<span>
Since </span>x^{a}* x^{b}= x^{a+b}
Then \sqrt[32]{x* x^{2} * x^{5} }= \sqrt[32]{ x^{1+2+5} } =\sqrt[32]{ x^{8} }

Since \sqrt[n]{x^{m}} = x^{m/n} }
Then \sqrt[32]{ x^{8} }= x^{8/32} = x^{1/4}

Since both Megan and Julie got the same result, it can be concluded that their expressions are equivalent.
3 0
3 years ago
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