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2 years ago

One challenge of federalism is that

Mathematics

1 answer:

padilas [110]2 years ago

7 0

Answer:

power and authority are concentrated in the central government and can take away people's right

Step-by-step explanation:

plsss mark brainliest

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The theater club spends $1,197 for 63 student

avanturin [10]

Answer:

$\$19\ per\ student$

Step-by-step explanation:

we know that

To find the the rate per studen, we need to divide the total amout spent by the number of students

$\frac{1,197}{63}=\$19\ per\ student$

6 0

3 years ago

Read 2 more answers

5 If the spinner landed on red 100 times

mario62 [17]

Answer:

50 I think. let me know if I'm wrong.

7 0

3 years ago

The 1st,12th and the last term of an AP are 4,31.5 and 376.5 respectively. Determine the number of terms in series

Nutka1998 [239]

Answer: 150

Step-by-step explanation:

To solve arithmetic progression, the nth term is calculated using the formula = a + (n - 1)d

From the question,

First term = a = 4

12th term = 31.5

We need to get the common difference which will be:

Since 12th term = 31.5

a + (n - 1)d = 31.5

a + (12 - 1)d = 31.5

a + 11d = 31.5

Recall that a = 4

4 + 11d = 31.5.

11d = 31.5 - 4

11d = 27.5

d = 27.5 / 11

d = 2.5

Since last term is 376.5, the number of terms in it will be calculated as:

a + (n - 1)d = 376.5

4 + (n - 1)2.5 = 376.5

4 + 2.5n - 2.5 = 376.5

2.5n = 376.5 - 4 + 2.5

2.5n = 375

n = 375 / 2.5

n = 150

The number of terms in the series is 150.

3 0

3 years ago

Find the area bounded by the given curves: y=2x−x2,y=2x−4

Andrej [43]

Answer:

$A = [\frac{32}{3}]$

Step-by-step explanation:

Given

$y_1 = 2x - x^2$

$y_2 = 2x - 4$

Required

Determine the area bounded by the curves

First, we need to determine their points of intersection

$2x - x^2 = 2x - 4$

Subtract 2x from both sides

$-x^2 = -4$

Multiply through by -1

$x^2 = 4$

Take square root of both sides

$x = 2$ or $x = -2$

This Area is then calculated as thus

$A = \int\limits^a_b {[y_1 - y_2]} \, dx$

Where a = 2 and b = -2

Substitute values for $y_1$ and $y_2$

$A = \int\limits^a_b {(2x - x^2) - (2x - 4)} \, dx$

Open Brackets

$A = \int\limits^a_b {2x - x^2 - 2x + 4} \, dx$

Collect Like Terms

$A = \int\limits^a_b {2x - 2x- x^2 + 4} \, dx$

$A = \int\limits^a_b {- x^2 + 4} \, dx$

Integrate

$A = [-\frac{x^{3}}{3} +4x](2,-2)$

$A = [-\frac{2^{3}}{3} +4(2)] - [-\frac{-2^{3}}{3} +4(-2)]$

$A = [-\frac{8}{3} +8] - [-\frac{-8}{3} -8]$

$A = [\frac{-8+ 24}{3}] - [\frac{8}{3} -8]$

$A = [\frac{-8+ 24}{3}] - [\frac{8-24}{3}]$

$A = [\frac{16}{3}] - [\frac{-16}{3}]$

$A = [\frac{16}{3}] + [\frac{16}{3}]$

$A = [\frac{16 + 16}{3}]$

$A = [\frac{32}{3}]$

Hence, the Area is:

$A = [\frac{32}{3}]$

7 0

4 years ago

PLS HELP WILL GIVE BRAIN THINGY

trasher [3.6K]

Answer:

Teachers are tough

Step-by-step explanation:

your welcome i hope you do good in ur test! :)

4 0

3 years ago