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a_sh-v [17]
3 years ago
5

The school drama club had 40 students in it last year. This year, the club has 52 students.

Mathematics
1 answer:
anastassius [24]3 years ago
3 0

Answer:

30% increase

Step-by-step explanation:

You might be interested in
What are the solutions to x2 + 8x + 7 =.b?
loris [4]

Answer:

x = 7 and x = -1

Step-by-step explanation:

First, we need to factor this quadratic.

To do this, start by finding all the factors of C.

7 x 1, -7 x -1

Now, we need to pick out the factors that add up to the b number, which is 8. That would be 7 and 1.

Now, we can use the numbers 7 and 1 to create two binomials.

( x + 7 ) ( x + 1 ) = 0

The last thing we need to do is to set those binomials to 0 and solve each of them individually.

x + 7 = 0

x = - 7

Solving the first binomial gives us our first solution, -7.

x + 1 = 0

x = -1

Solving the second one gives us our second solution, -1.

The solutions are -7 and -1.

4 0
3 years ago
Solve 4|x + 5| + 8 = 24.
KIM [24]
Your answer would be C
6 0
2 years ago
Is it necessary to rename 4 1/4 if you subtract 3/4 from it
Nesterboy [21]
No because they both have the same denominator!
5 0
3 years ago
What is the solution to the equation 5 (x minus 6) = 2 (x + 3)?
Reika [66]
The answer is x=12 :))
8 0
3 years ago
Find an equation of a line that is tangent to the graph of f and parallel to the given line. Please see picture
ivanzaharov [21]

Answer:

y = 3x - 2 (smaller y-intercept)

y = 3x + 2 (larger y-intercept)

Step-by-step explanation:

First let's write the generic equation of a line:

y = ax + b

This line needs to be parallel to the line 3x - y + 5 = 0, so it needs to have the same slope of this line.

The line 3x - y + 5 = 0 has a slope of 3, so our line has a = 3:

y = 3x + b

Now we need to find the values of b that make this line tangent to the function f(x) = x^3

Let's first find the derivative of f(x) in relation to x:

df(x)/dx = 3x^2

This derivative is the slope of the tangent line to the function for any value of x. We need a slope of 3, so:

3x^2 = 3

x^2 = 1

x = ±1

Now, to find the y-values, we have:

f(1) = 1^3 = 1

f(-1) = (-1)^3 = -1

So, using the points (1,1) and (-1,-1) in our parallel line, we have:

first line using (1,1) : 1 = 3*1 + b

b = -2

second line using (-1,-1) : -1 = -3*1 + b

b = 2

The value of b is the y-intercept of the line, so the line with smaller y-intercept is y = 3x - 2, and the line with larger y-intercept is y = 3x + 2

8 0
3 years ago
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