All categories

3 years ago

The radius of a sphere is 6 units.

Mathematics

2 answers:

Travka [436]3 years ago

8 0

Answer:

C. or 3

Step-by-step explanation:

Yanka [14]3 years ago

7 0

Answer:

The answer is Three-Fourthst(12)2

You might be interested in

A rectangular box has length x and width 3. The volume of the box is given by y = 3x(8 – x). The greatest x-intercept of the gra

CaHeK987 [17]

Consider rectangular box with

length x units (x≥0);
width 3 units;
height (8-x) units (8-x≥0, then x≤8).

The volume of the rectangular box can be calculated as

$V_{box}=\text{length}\cdot \text{width}\cdot \text{height}.$

In your case,

$V_{box}=3\cdot x\cdot (8-x).$

Note that maximal possible value of the height can be 8 units (when x=0 - minimal possible length) and the minimal possible height can be 0 units (when x=8 - maximal possible length).

From the attached graph you can see that the greatest x-intercept is x=8, then the height will be minimal and lenght will be maximal.

Then the volume will be V=0 (minimal).

Answer: correct choices are B (the maximum possible length), C (the minimum possible height)

7 0

3 years ago

Which statement best describes the relationship between the two variables? There is an association because the relative frequenc

grandymaker [24]

Answer:

I think it is B if it is your welcome if not sorry its been a while since I learned about these.

Step-by-step explanation:

5 0

3 years ago

PLEASE HELP! Given the function f(x) = 3x2 − 4, what is f(-2)?

EleoNora [17]

Answer: Variant C

Step-by-step explanation:

f(x) = 3x^2 − 4

For finding f(-2) just replace x with - 2 like this:

f(-2)=3*((-2))^2-4=3*4-4=12-4=8

5 0

3 years ago

State the property. 4(2+3)=(2+3)4

Komok [63]

Hmm, it looks like it could be commutative property of multiplication.

The order of the expression is different.

4(2+3)=

4(5)=20

(2+3)4=

(5)4= 20

8 0

3 years ago

Find x so that the points (x,x+1), (x+2,x+3) and (x+3,2x+4) form a right-angled triangle.

azamat

Let a, b, and c be vectors each starting at the origin and terminating at the points (x, x + 1), (x + 2, x + 3), and (x + 3, 2x + 4), respectively.

Then the vectors a - b, a - c, and b - c are vectors that point in directions parallel to each of the legs formed by the triangle with these points as its vertices.

If this triangle is to contain a right angle, then exactly one of these pairs of vectors must be orthogonal. In other words, one of the following must be true:

(a - b) • (a - c) = 0

or

(a - b) • (b - c) = 0

or

(a - c) • (b - c) = 0

We have

a - b = (x, x + 1) - (x + 2, x + 3) = (-2, -2)

a - c = (x, x + 1) - (x + 3, 2x + 4) = (-3, -x - 3)

b - c = (x + 2, x + 3) - (x + 3, 2x + 4) = (-1, -x - 1)

Case 1: If (a - b) • (a - c) = 0, then

(-2, -2) • (-3, -x - 3) = (-2)×(-3) + (-2)×(-x - 3) = 2x + 12 = 0 ==> x = -6

which would make a - c = (-3, 3) and b - c = (-1, 5), and their dot product is not zero. Then the triangles vertices are at the points (-6, -5), (-4, -3), and (-3, -8).

Case 2: If (a - b) • (b - c) = 0, then

(-2, -2) • (-1, -x - 1) = (-2)×(-1) + (-2)×(-x - 1) = 2x + 4 = 0 ==> x = -2

which would make a - c = (-3, -1) and b = (-1, 1), and their dot product is also not zero. The vertices are the points (-2, -1), (0, 1), and (1, 0).

Case 3: If (a - c) • (b - c) = 0, then

(-3, -x - 3) • (-1, -x - 1) = (-3)×(-1) + (-x - 3)×(-x - 1) = x ² + 4x + 6 = 0

but the solutions to x here are non-real, so we throw out this case.

So there are two possible values of x that make a right triangle, x = -6 and x = -2.

3 0

3 years ago