All categories

2 years ago

36.05 rounded off to the nearest tens place

Mathematics

1 answer:

Doss [256]2 years ago

7 0

Answer:

36.1

Step-by-step explanation:

if the 100s place is 5 or above add 1 to the tens place. if it's 5 or below the tens will stay the same.

You might be interested in

Two pounds of cashews cost $19, and 8 pounds cost $76.

Artyom0805 [142]

Answer:

Yes there is a proportinal relationship because when I did 19 divided by 2 I got 9.5 and when I do 76 divided by 8 I got 9.5 so its proportinal. Yes I can find how much money 6 pounds is its 57. And yes I can do it without a table because all you have to do is divide the 19 and 2 and the other 2 numbers and if there the same then its proportinal.

Step-by-step explanation:

BRAINLIEST MAYBE???

3 0

3 years ago

Quick need the answer!

Nat2105 [25]

The answer is A because A shows a ray (it continues in one direction), but the figure shown above is a line (it continues in both directions)

3 0

3 years ago

Read 2 more answers

on a recent standardized test, Ahmed scored int he 90 percentile. If 2800 people took the test, then Ahmed's score was higher th

Readme [11.4K]

Answer:

Ahmed scored the same or better than 2,520 people that took the test

explanation:
percentile is the percentage of people you scored the same/better than on something
90% of 2800 is 2520
I hope this helps!

3 0

2 years ago

Question

krok68 [10]

Answer:

Acute angle between the two planes: approximately $43^\circ$ .

Step-by-step explanation:

Find the normal vector of each plane:

The normal vector of the plane $x - 2\, y + 5\, z = 3$ is $\displaystyle \begin{bmatrix}1 \\ -2 \\ 5\end{bmatrix}$ .
The normal vector of the plane $2\, x + y - 3\, z = 15$ is $\displaystyle \begin{bmatrix}2 \\ 1\\ -3\end{bmatrix}$ .

As the name suggests, there is a $90^\circ$ angle between a plane and its normal vector. The following four angles will correspond to the vertices of a quadrilateral:

The $90^\circ$ angle between the first plane and its normal vector.
The angle between the normal vector of each plane.
The $90^\circ$ angle between the second plane and its normal vector.
The smallest angle between these two planes.

The sum of these four angles should be $360^\circ$ . Two of these four angles were known to be $90^\circ$ . Once the third angle (the angle between the two normal vectors) is found, subtractions would give the measure of the other angle (the smallest angle between these two planes.)

Make use of the dot product to find the angle between these two normal vectors. Let $\theta$ denote the angle between these two vectors.

$\displaystyle \cos \theta = \frac{\begin{bmatrix}1 \\ -2 \\ 5\end{bmatrix} \cdot \begin{bmatrix}2 \\ 1 \\ -3\end{bmatrix}}{\sqrt{1^2 + (-2)^2 + 5^2} \cdot \sqrt{2^2 + 1^2 + (-3)^2}} \approx -0.73193$ .

Before continuing, notice that the smallest angle between the two planes would be $360^\circ - 90^\circ - 90^\circ - \theta = 180^\circ - \theta$ .

Consider the identity: $\cos\left(180^\circ - \theta\right) = -\cos \theta$ .

In other words, $\cos\left(180^\circ - \theta\right)$ , the cosine of the smallest angle between the two planes (which the question is asking for) will be the opposite of $\cos \theta$ , the cosine of the angle between the two normal vectors.

Therefore, the cosine of the smallest angle between the two planes will be $-(-0.73193) = 0.73193$ .

Apply the inverse cosine function to find the size of that angle:

$\arccos(0.73193) \approx 43^\circ$ .

8 0

3 years ago

If f(x) = x3 – x2 – 3, which of the following is equal to g(x) = f(2 – x)?

Andru [333]

Answer:

-x^3+5x^2-8x+1, which is choice A

======================================

Work Shown:

f(x) = x^3 - x^2 - 3

f(x) = (x)^3 - (x)^2 - 3

f(2-x) = (2-x)^3 - (2-x)^2 - 3 ................ see note 1 (below)

f(2-x) = (2-x)(2-x)^2 - (2-x)^2 - 3 ........... see note 2

f(2-x) = (2-x)(4-4x+x^2) - (4-4x+x^2) - 3 ..... see note 3

f(2-x) = -x^3+6x^2-12x+8 - (4-4x+x^2) - 3 ..... see note 4

f(2-x) = -x^3+6x^2-12x+8 - 4+4x-x^2 - 3 ....... see note 5

f(2-x) = -x^3+5x^2-8x+1

----------

note1: I replaced every copy of x with 2-x. Be careful to use parenthesis so that you go from x^3 to (2-x)^3, same for the x^2 term as well.

note2: The (2-x)^3 is like y^3 with y = 2-x. We can break up y^3 into y*y^2, so that means (2-x)^3 = (2-x)(2-x)^2

note3: (2-x)^2 expands out into 4-4x+x^2 as shown in figure 1 (attached image below). I used the box method for this and for note 4 as well. Each inner box or cell is the result of multiplying the outside terms. Example: in row1, column1 we have 2 times 2 = 4. You could use the FOIL rule or distribution property, but the box method is ideal so you don't lose track of terms.

note4: (2-x)(4-4x+x^2) turns into -x^3+6x^2-12x+8 when expanding everything out. See figure 2 (attached image below). Same story as note 3, but it's a bit more complicated.

note5: distribute the negative through to ALL the terms inside the parenthesis of (4-4x+x^2) to end up with -4+4x-x^2

4 0

4 years ago