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4 years ago

10

There are 10 fifth-grade students attending a class field trip to the city’s art museum. If each admission ticket costs $6, what

is the total cost of admission for the 10 students

1 answer:

pshichka [43]4 years ago

4 0

Answer:

$60

Step-by-step explanation:

10x6=60. 6+6+6+6+6+6+6+6+6+6= 60

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The following shape is only based on squares, semicircles and quarter circles. Find the Area and Perimeter of the shaded part. A

charle [14.2K]

The perimeter of the shaded region is 8π cm and the area of the shaded region is 32(π - 2) cm² or 36.53 square cm.

<h3>What is a circle?</h3>

It is described as a set of points, where each point is at the same distance from a fixed point (called the center of a circle)

We have shown a square in which a one-fourth circle is shown and making an ellipse.

The perimeter of the circle = 2πr = 2π(8) = 16π cm

The perimeter of the one-fourth is circle = (16π)/4 = 4π cm

The perimeter of 2 one-fourth circle = 2(4π) = 8π cm

The area of the one-fourth circle = (1/4)πr² = (1/4)π(8)² = 16π square cm

The area of the right-angle triangle = (1/2)(8)(8) = 32 square cm

The area of the shaded region = 2(16π - 32) = 36.53 square cm

or

= 32(π - 2)cm²

Thus, the perimeter of the shaded region is 8π cm and the area of the shaded region is 32(π - 2) cm² or 36.53 square cm.

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6 0

2 years ago

R=5(CA−0.3) solve for c

Maurinko [17]

R = 5(CA - 0.3)

<span>Multiply everything in the parenthesis by 5.</span>

R = 5CA - 1.5

Add 1.5 to both sides

1.5 + R = 5CA

Divide 5A on both sides.

C = 1.5 + R / 5A

Hope this helps!

4 0

3 years ago

Read 2 more answers

If you roll a fair six-sided die and a fair four-sided die, what is the probability that the dice add to 6 or higher

Lena [83]

Okay so you have to figure out all the possible ways to get six or higher. between a 6 sided dice and a 4 sided dice. We do 6*24 to find all the possibilities.

The probability of getting a 6 or higher is 14/24

4 0

3 years ago

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HELP WITH THESE SERIES OF QUESTIONS!!

Virty [35]

1. Simplified already
2. 5x^2 x (3x+5)
3. 3x^3 x (3x^4 +11)

5 0

3 years ago

Y= 2x^3 - 15x^2 + 36x - 20 find turning points and show working out

Murljashka [212]

Given:

The function is:

$y=2x^3-15x^2+36x-20$

To find:

The turning points.

Solution:

We have,

$y=2x^3-15x^2+36x-20$

Differentiate the given function with respect to x.

$y'=2(3x^2)-15(2x)+36(1)-(0)$

$y'=6x^2-30x+36$

$y'=6(x^2-5x+6)$

For turning point, $y'=0$ .

$6(x^2-5x+6)=0$

$x^2-3x-2x+6=0$

$x(x-3)-2(x-3)=0$

$(x-3)(x-2)=0$

Using zero product property, we get

$x-3=0$ and $x-2=0$

$x=3$ and $x=2$

Therefore, the turning points of the given function are at $x=2$ and $x=3$ .

8 0

3 years ago