All categories

4 years ago

If a right hexagonal prism became an oblique hexagonal prism, would that have any effect on the volume and height of the

Mathematics

1 answer:

Nitella [24]4 years ago

3 0

Answer:

Even if a right solid became oblique, the volume would not change, because the cross sectional area and height would remain the same. Therefore, by Cavalieri's principle, the volume is the same as the original solid.

Step-by-step explanation:

sample answer

You might be interested in

Match each equation with a diagram:

denis23 [38]

Answers:
Equation 1: B
Equation 2: D
Equation 3: A
Equation 4: C

*I am pretty sure are right

3 0

3 years ago

Read 2 more answers

Find a cubic polynomial that goes through points (4, – 22) and (3, - 26) and has tangents with slopes

Salsk061 [2.6K]

Let f(x) = ax ³ + bx ² + cx + d.

The graph of f(x) passes through (4, -22) and (3, -26), which means f (4) = -22 and f (3) = -26, so that

64a + 16b + 4c + d = -22

27a + 9b + 3c + d = -26

When the question says it has tangents at some point, I take that to mean the slope of the tangent line at that point is the given number. So f ' (4) = 11 and f ' (3) = -2. We have

f '(x) = 3ax ²+ 2bx + c

so that

48a + 8b + c = 11

27a + 6b + c = -2

Solve the system:

• Eliminate d :

(64a + 16b + 4c + d) - (27a + 9b + 3c + d) = -22 - (-26)

→ 37a + 7b + c = 4

• Eliminate c :

(48a + 8b + c) - (27a + 6b + c) = 11 - (-2)

→ 21a + 2b = 13

(48a + 8b + c) - (37a + 7b + c) = 11 - 4

→ 11a + b = 7

• Eliminate b, then solve for a and the other variables:

(21a + 2b) - 2 (11a + b) = 13 - 2 (7)

-a = -1

a = 1 → b = -4 → c = -5 → d = -2

Then

f(x) = x ³ - 4x ² - 5x - 2

7 0

4 years ago

When Julia is writing a first draft, there is 0.7 probability that there will be no spelling mistakes on a page. One day, Julia

tester [92]

Answer:

The required probability is given by, 0.9919.

Step-by-step explanation:

Let, X be the random variable denoting the no. of pages among those 4 pages which Julia writes where she makes no spelling mistake.

clearly,

X $\sim$ Binomial (4, 0.7)

So, P(X = x) = $^4C_{x} \times (0.7)^{x} \times (0.3)^{(4 - x)}$

[when x = 0, 1, 2, 3, 4]

= 0 otherwise

According to the question, we are to find out P(X ≥ 1) .

Now, P(X ≥ 1)

= 1 - P(X = 0)

= $1 - (^4C_{0} \times (0.7)^{0} \times (0.3)^{4})$

= $1 - 0.0081$

= 0.9919

So, the required probability is given by, 0.9919

8 0

4 years ago

Read 2 more answers

How do you do coordinates?

suter [353]

Answer:

Start at (0, 0), or the origin. Just go to (0, 0), which is the intersection of the x and y axes, right in the center of the coordinate plane. 2. Move over x units to the right or left. Let's say you're working with the set of coordinates (5, -4). Your x coordinate is 5.

Step-by-step explanation:

4 0

3 years ago

(a) By inspection, find a particular solution of y'' + 2y = 14. yp(x) = (b) By inspection, find a particular solution of y'' + 2

SOVA2 [1]

Answer:

(a) The particular solution, y_p is 7

(b) y_p is -4x

(d) y_p is 8x + (7/2)

Step-by-step explanation:

To find a particular solution to a differential equation by inspection - is to assume a trial function that looks like the nonhomogeneous part of the differential equation.

(a) Given y'' + 2y = 14.

Because the nonhomogeneus part of the differential equation, 14 is a constant, our trial function will be a constant too.

Let A be our trial function:

We need our trial differential equation y''_p + 2y_p = 14

Now, we differentiate y_p = A twice, to obtain y'_p and y''_p that will be substituted into the differential equation.

y'_p = 0

y''_p = 0

Substitution into the trial differential equation, we have.

0 + 2A = 14

A = 6/2 = 7

Therefore, the particular solution, y_p = A is 7

(b) y'' + 2y = −8x

Let y_p = Ax + B

y'_p = A

y''_p = 0

0 + 2(Ax + B) = -8x

2Ax + 2B = -8x

By inspection,

2B = 0 => B = 0

2A = -8 => A = -8/2 = -4

The particular solution y_p = Ax + B

is -4x

Let y_p = Ax + B

y'_p = A

y''_p = 0

0 + 2(Ax + B) = -8x + 14

2Ax + 2B = -8x + 14

By inspection,

2B = 14 => B = 14/2 = 7

2A = -8 => A = -8/2 = -4

The particular solution y_p = Ax + B

is -4x + 7

(d) Find a particular solution of y'' + 2y = 16x + 7

Let y_p = Ax + B

y'_p = A

y''_p = 0

0 + 2(Ax + B) = 16x + 7

2Ax + 2B = 16x + 7

By inspection,

2B = 7 => B = 7/2

2A = 16 => A = 16/2 = 8

The particular solution y_p = Ax + B

is 8x + (7/2)

8 0

3 years ago