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3 years ago

The two men measure the angle of elevation to the highest point on the rock to be 22.6°. They

Mathematics

1 answer:

IgorC [24]3 years ago

3 0

Answer:

15 m

Step-by-step explanation:

Refer to attached picture

The height = h
Angles of elevation 22.6° and 38.3°
Closer distance = d

Use tangent to solve

tan 22.6° = h/(d + 17) ⇒ h = (d + 17) tan 22.6° ⇒ h = 0.42(d + 17)
tan 38.3° = h/d ⇒ h = d tan 38.3° ⇒ h = 0.79d

Compare the equations above and find the value of d:

0.42(d + 17) = 0.79d
0.42d + 7.14 = 0.79d
0.79d - 0.42d = 7.14
0.37 d = 7.14
d = 7.14/0.37
d = 19.3 m

Now find the value of h:

h = 0.79d
h = 0.79*19.3
h = 15

Wave Rock is 15 m tall

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3 years ago

What is the quotient when 4x3 + 2x + 7 is divided by x + 3?

Arte-miy333 [17]

Answer:

The quotient of this division is $(4x^2 -12x + 38)$ . The remainder here would be $-26$ .

Step-by-step explanation:

The numerator $4x^3 + 2x + 7$ is a polynomial about $x$ with degree $3$ .

The divisor $x + 3$ is a polynomial, also about $x$ , but with degree $1$ .

By the division algorithm, the quotient should be of degree $3 - 1 = 2$ , while the remainder shall be of degree $1 - 1 = 0$ (i.e., the remainder would be a constant.) Let the quotient be $a\,x^2 + b\, x + c$ with coefficients $a$ , $b$ , and $c$ .

$4x^3 + 2x + 7 = \left(a\,x^2 + b\, x + c\right)(x + 3)$ .

Start by finding the first coefficient of the quotient.

The degree-three term on the left-hand side is $4 x^3$ . On the right-hand side, that would be $a\, x^3$ . Hence $a = 4$ .

Now, given that $a = 4$ , rewrite the right-hand side:

$\begin{aligned}&\left(4\,x^2 + b\, x + c\right)(x + 3) \cr =& \left(4x^2 + (b\, x + c)\right)(x + 3) \cr =& 4x^2(x + 3) + (bx + c)(x + 3) \cr =& 4x^3 + 12x^2 + (bx + c)(x + 3)\end{aligned}$ .

Hence:

$4x^3 + 2x + 7 = 4x^3 + 12x^2 + (b\,x + c)(x + 3)$

Subtract $\left(4x^3 + 12x^2\right$ from both sides of the equation:

$-12x^2 + 2x + 7 = (b\,x + c)(x + 3)$ .

The term with a degree of two on the left-hand side has coefficient $(-12)$ . Since the only term on the right hand side with degree two would have coefficient $b$ , $b = -12$ .

Again, rewrite the right-hand side:

$\begin{aligned}&\left(-12 x + c\right)(x + 3) \cr =& \left(-12 x+ c\right)(x + 3) \cr =& (-12x)(x + 3) + c(x + 3) \cr =& -12x^2 -36x + (bx + c)(x + 3)\end{aligned}$ .

Subtract $-12x^2 -36x$ from both sides of the equation:

$38x + 7 = c(x + 3)$ .

By the same logic, $c = 38$ .

Hence the quotient would be $(4x^2 - 12x + 38)$ .

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3 years ago

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2 years ago

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3 years ago

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