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astraxan [27]
3 years ago
5

A total of 443 tickets were sold for the school play. They were either adult tickets or student tickets. There were 57 fewer stu

dent tickets sold than adult tickets. How many adult tickets were sold?
Mathematics
1 answer:
Sladkaya [172]3 years ago
3 0

Answer

386

Step-by-step explanation:

Do the math 443-57

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Identify the radicand in the radical below
Elena L [17]

Answer:

The radicand is whatever is under the the radical, so in this case, your answer would be D. 3x-5



4 0
3 years ago
A local school requires 3 teachers for every 93 students. Which equation represents this relationship, where t is the number of
bearhunter [10]
<span>t is the number of teachers and s is the number of students
</span>
<span>The school requires 3 teachers for every 93 students.
</span>
<span>So, 3t = 93 s    ⇒⇒⇒ divide by 3
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<span>∴ 3t/3 = 93s/3
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<span>∴ t = 31 s
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The <span>equation represents this relationship is ⇒⇒⇒ t = 31 s
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8 0
3 years ago
Describe lengths of three segments that could not be used to form a triangle. Segments with lengths of 5 in., 5 in., and ______
Vesna [10]

Answer:

c = 11

Step-by-step explanation:

Given

Let the three sides be a, b and c

Such that:

a = b= 5

Required

Find c such that a, b and c do not form a triangle

To do this, we make use of the following triangle inequality theorem

a + b > c

a + c > b

b + c > a

To get a valid triangle, the above inequalities must be true.

To get an invalid triangle, at least one must not be true.

Substitute: a = b= 5

5 + 5 > c ===> 10 > c

5 + c > 5  ===> c > 5 - 5 ===> c > 0

5 + c > 5  ===> c > 5 - 5 ===> c > 0

The results of the inequality is: 10 > c and c > 0

Rewrite as: c > 0 and c < 10

0 < c < 10

This means that, the values of c that make a valid triangle are 1 to 9 (inclusive)

Any value outside this range, cannot form a triangle

So, we can say:

c = 11, since no options are given

8 0
3 years ago
How many solutions does 3 (x-1)=x+7 have
Anit [1.1K]
Do you know how to solve algebra? If so just solve for X. If you get an invalid equation then 0 solutions, if you get x equaling 1 number then 1 solution, if you get x equaling 2 numbers then 2 solutions, and so on.
6 0
3 years ago
Read 2 more answers
Can the mean value theorum be applied to the function f(x)=1/x^2 on the interval [-2, 1]? Explain.
Temka [501]

Answer:

No, it can not be applied.

Step-by-step explanation:

f(x) = 1/x²

f(x) is a polynomial that is not continuous

As,

f(x) = 1/0 is undefines

Secondly, it is not differentiable (i.e. the derivative does not exists on the interval given)

Derivative of this function

f'(x) = (1)x^-2

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      =  -2/x³

f'(0) = -2/0 is undefined

Thus, mean value theorem can not be applied.

7 0
4 years ago
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