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2 years ago

7

How many different 5-digit numbers can be written using only numbers 0, 1, 2, 3, and 4, if number should contain each of these n

umbers: only once?

1 answer:

gizmo_the_mogwai [7]2 years ago

8 0

Answer:

3125

Step-by-step explanation:

5* 5* 5* 5 = 3125

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Use the Change of Base Formula to evaluate log5 92. Then convert log5 92 to a logarithm in base 3. Round to the nearest thousand

ANTONII [103]

I was thinking that the answer could be

2.810; log3 21.903

OR

<span> 2.810; log3 55.2 </span>

So,<span> log5 of 92 is between 2.5 and 3 Correct answer is A</span>

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2 years ago

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Someone help need this done right now I’ll mark u brainliest

Ad libitum [116K]

Answer:

89 meters

Step-by-step explanation:

We know it's 9 + "the height of the triangle".

Fortunately, the height of the triangle (let's call it y) can be computed from:

y^2 + 18^2 = 82^2 (Pythagoras theorem).

y^2 = 82^2 - 18^2 = 6724 - 324 = 6400

y = 80 (or y = -80 but it doesn't make sense in geometry).

so x = y + 9 = 89;

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2 years ago

An exam has two probability problems, 1 and 2. If 37% of the students solved problem 1 and 12% of the students solved both probl

Zanzabum

Answer:

about 32%

Step-by-step explanation:

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2 years ago

Suppose the tree diagram below represents all the students in a high school

Lunna [17]

Answer:

Ok to find the probity we add the amount of boys and the amount of girls together to get 500 and out of these 100 are left handed so 1/5 also know as 20%

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8 0

2 years ago

Match each set of vertices with the type of triangle they form.

Andrew [12]

Answer: The calculations are done below.

Step-by-step explanation:

(i) Let the vertices be A(2,0), B(3,2) and C(5,1). Then,

$AB=\sqrt{(2-3)^2+(0-2)^2}=\sqrt{5},\\\\BC=\sqrt{(3-5)^2+(2-1)^2}=\sqrt{5},\\\\CA=\sqrt{(5-2)^2+(1-0)^2}=\sqrt{10}.$

Since, AB = BC and AB² + BC² = CA², so triangle ABC here will be an isosceles right-angled triangle.

(ii) Let the vertices be A(4,2), B(6,2) and C(5,3.73). Then,

$AB=\sqrt{(4-6)^2+(2-2)^2}=\sqrt{4}=2,\\\\BC=\sqrt{(6-5)^2+(2-3.73)^2}=\sqrt{14.3729},\\\\CA=\sqrt{(5-4)^2+(3.73-2)^2}=\sqrt{14.3729}.$

Since, BC = CA, so the triangle ABC will be an isosceles triangle.

(iii) Let the vertices be A(-5,2), B(-4,4) and C(-2,2). Then,

$AB=\sqrt{(-5+4)^2+(2-4)^2}=\sqrt{5},\\\\BC=\sqrt{(-4+2)^2+(4-2)^2}=\sqrt{8},\\\\CA=\sqrt{(-2+5)^2+(2-2)^2}=\sqrt{9}.$

Since, AB ≠ BC ≠ CA, so this will be an acute scalene triangle, because all the angles are acute.

(iv) Let the vertices be A(-3,1), B(-3,4) and C(-1,1). Then,

$AB=\sqrt{(-3+3)^2+(1-4)^2}=\sqrt{9}=3,\\\\BC=\sqrt{(-3+1)^2+(4-1)^2}=\sqrt{13},\\\\CA=\sqrt{(-1+3)^2+(1-1)^2}=\sqrt 4.$

Since AB² + CA² = BC², so this will be a right angled triangle.

(v) Let the vertices be A(-4,2), B(-2,4) and C(-1,4). Then,

$AB=\sqrt{(-4+2)^2+(2-4)^2}=\sqrt{8},\\\\BC=\sqrt{(-2+1)^2+(4-4)^2}=\sqrt{1}=1,\\\\CA=\sqrt{(-1+4)^2+(4-2)^2}=\sqrt{13}.$

Since AB ≠ BC ≠ CA, and so this will be an obtuse scalene triangle, because one angle that is opposite to CA will be obtuse.

Thus, the match is done.

4 0

3 years ago

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