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2 years ago

Help me please ty ty

Mathematics

1 answer:

julsineya [31]2 years ago

3 0

Answer:

Step-by-step explanation:

(f - g)(x)

= f(x) - g(x)

= 3x² + 4x - 6 - (6x³ - 5x² - 2) ← distribute parenthesis by - 1

= 3x² + 4x - 6 - 6x³ + 5x² + 2 ← collect like terms

= - 6x³ + 8x² + 4x - 4

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X+4y=2v make x the subject

vfiekz [6]

X=-4y+2v is the answer, because you would put 4y on the other side do -4y to even zero out the y factor on the left side.

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3 years ago

A cube has an edge length of 9 meters. What is its volume, in cubic meters?

rjkz [21]

<h3>Answer: 729 cubic meters</h3>

Work Shown:

volume = side*side*side

volume = 9*9*9

volume = 729

A slight shortcut would be to say 9^3 = 729

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2 years ago

The points on a scatterplot lie close to the line whose equation is y = 21 - 4.7x. The slope of this line is

AlladinOne [14]

The slope is -4.7 because this equation is written in point slope form. This equation can be rewritten as y = -4.7x + 21. Point slope form is y = mx + b, with m being slope and b being y-intercept.

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3 years ago

points P and Q lie 240 m apart in line with and on opposite sides of a communications tower. the angles of elevation to the top

Leona [35]

Answer:

The height of the tower is 130.5m

Step-by-step explanation:

In the question, we are given the following values:

The angles of elevation to the top of the tower from P = 50°

The angles of elevation to the top of the tower from Q = 45°

The angles of elevation to the top of the tower from P = 50°

Hence,cot P = 1/ tan(50°)

The angles of elevation to the top of the tower from Q = 45°

Hence, tan 45° = 1

In the question,we are told Points P and Q lie 240 m apart in line with and on opposite sides of a communications tower.

Therefore,

PQ = height of the tower( tan Q + 1/tan P)

240m = height of the tower( tan 45° + 1/ tan 50°)

240m = h(1 + 1/tan 50°)

h = (240 m)/(1 + 1/tan (50°))

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3 years ago

I can't figure out how to do (i + j) x (i x j)for vector calc

Vinil7 [7]

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$\begin{gathered} \vec{A}=a_1\hat{i}+a_2\hat{j}+a_3\hat{k} \\ \vec{B}=b_1\hat{i}+b_2\hat{j}+b_3\hat{k} \\ \Rightarrow\vec{A}\times\vec{B}=\det (\begin{bmatrix}{\hat{i}} & {\hat{j}} & {\hat{k}} \\ {a_1} & {a_2} & {a_3} \\ {b_1} & {b_2} & {b_3}\end{bmatrix}) \end{gathered}$

Then, solving the determinant

$\Rightarrow\vec{A}\times\vec{B}=(a_2b_3-b_2a_3)\hat{i}+(b_1a_3+a_1b_3)\hat{j}+(a_1b_2-b_1a_2)\hat{k}$

In our case,

$\begin{gathered} (\hat{i}+\hat{j})=1\hat{i}+1\hat{j}+0\hat{k} \\ \text{and} \\ (\hat{i}\times\hat{j})=(1,0,0)\times(0,1,0)=(0)\hat{i}+(0)\hat{j}+(1-0)\hat{k}=\hat{k} \\ \Rightarrow(\hat{i}\times\hat{j})=\hat{k} \end{gathered}$

Where we used the formula for AxB to calculate ixj.

Finally,

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Thus, (i+j)x(ixj)=i-j

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1 year ago