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2 years ago

What is the surface area of the shape?

Mathematics

2 answers:

Darya [45]2 years ago

5 0

Answer:

Step-by-step explanation:

U can find it easily it is so simple

Sladkaya [172]2 years ago

3 0

Answer:

Step-by-step explanation:

You might be interested in

Write two equivalent ratios 3to5??

vesna_86 [32]

Answer:

3 : 5

6 : 10

9 : 15

12 : 20

15 : 25

7 0

3 years ago

Which sequence of transformations produces R’S’T’ from RST?

inn [45]

Answer:

A translation 2 units right and then a reflection over the x-axis

Step-by-step explanation:

The given vertices of ΔRST are R(0, 0), S(-2, 3), and T(-3, 1)

The vertices of triangle ΔR'S'T' are (2, 0), (0, -3), (-1, -1)

The points are plotted with the aid of MS Excel, and by observation, we have that the image of ΔRST is located on the other side of the x-axis with each coordinate on ΔR'S'T' shifted 2 units to the right of ΔRST

A translation of ΔRST 2 units right gives;

(0 + 2, 0) = (2, 0), (-2 + 2, 3) = (0, 3), and (-3 + 2, 1) = (-1, 1), to give;

(2, 0), (0, 3), and (-1, 1)

A reflection of the point (x, y) across the x-axis gives (x, -y)

A reflection of the above points across the x-axis gives;

(2, 0) reflected about x-axis → (2, 0) reflected about x-axis → (0, -3), and (-1, 1) reflected about x-axis → (-1, -1), which are the points of ΔR'S'T'

Therefore, the sequence of transformations that produces R'S'T' from RST are;

A translation 2 units right and then a reflection over the x-axis

6 0

3 years ago

A six sided die is rolled 4 times. find the probability that a 2 or 3 is rolled at least once. round your answer to two decimal

goblinko [34]

The probability that a 2 or 3 is rolled at least once is 0.14.

<h3>Probability of not rolling a 2 or 3 in a single die roll of two dice</h3>

when a six die is rolled once there is 11/12 probability of not rolling 2 or 3.

rolling a six sided die two times, p(2 or 3)' = 11/12

rolling a six sided die four times, p(2 or 3)' = (11/12)² = 121/144

the probability that a 2 or 3 is rolled at least once is calculated as follows;

P = 1 - p(2 or 3)'

P = 1 - 121/144

P = 20/144

P = 0.14

Thus, the probability that a 2 or 3 is rolled at least once is 0.14.

Learn more about probability here: brainly.com/question/24756209

#SPJ4

4 0

1 year ago

Wich is bigger<br><br> 1.320. 1.302<br><br> <<br> > <br> =

elena-14-01-66 [18.8K]

1.320 >1.302 or 1.302<1.320

4 0

3 years ago

The students in Mr. Wilson's Physics class are making golf ball catapults. The

Mnenie [13.5K]

Answer:

Part a) About 48.6 feet

Part b) About 8.3 feet

Part c) The domain is $0 \leq x \leq 48.6\ ft$ and the range is $0 \leq y \leq 8.3\ ft$

Step-by-step explanation:

we have

$y=-0.014x^{2} +0.68x$

This is a vertical parabola open downward (the leading coefficient is negative)

The vertex represent a maximum

where

x is the ball's distance from the catapult in feet

y is the flight of the balls in feet

Part a) How far did the ball fly?

Find the x-intercepts or the roots of the quadratic equation

Remember that

The x-intercept is the value of x when the value of y is equal to zero

The formula to solve a quadratic equation of the form

$ax^{2} +bx+c=0$

is equal to

$x=\frac{-b(+/-)\sqrt{b^{2}-4ac}} {2a}$

in this problem we have

$-0.014x^{2} +0.68x=0$

$a=-0.014\\b=0.68\\c=0$

substitute in the formula

$x=\frac{-0.68(+/-)\sqrt{0.68^{2}-4(-0.014)(0)}} {2(-0.014)}$

$x=\frac{-0.68(+/-)0.68} {(-0.028)}$

$x=\frac{-0.68(+)0.68} {(-0.028)}=0$

$x=\frac{-0.68(-)0.68} {(-0.028)}=48.6\ ft$

therefore

The ball flew about 48.6 feet

Part b) How high above the ground did the ball fly?

Find the maximum (vertex)

$y=-0.014x^{2} +0.68x$

Find out the derivative and equate to zero

$0=-0.028x +0.68$

Solve for x

$0.028x=0.68$

$x=24.3$

<em>Alternative method</em>

To determine the x-coordinate of the vertex, find out the midpoint between the x-intercepts

$x=(0+48.6)/2=24.3\ ft$

To determine the y-coordinate of the vertex substitute the value of x in the quadratic equation and solve for y

$y=-0.014(24.3)^{2} +0.68(24.3)$

$y=8.3\ ft$

the vertex is the point (24.3,8.3)

therefore

The ball flew above the ground about 8.3 feet

Part c) What is a reasonable domain and range for this function?

we know that

A reasonable domain is the distance between the two x-intercepts

$0 \leq x \leq 48.6\ ft$

All real numbers greater than or equal to 0 feet and less than or equal to 48.6 feet

A reasonable range is all real numbers greater than or equal to zero and less than or equal to the y-coordinate of the vertex

we have the interval -----> [0,8.3]

$0 \leq y \leq 8.3\ ft$

All real numbers greater than or equal to 0 feet and less than or equal to 8.3 feet

8 0

2 years ago