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2 years ago

What is the length x of the right triangle, rounded to the nearest tenth?

Mathematics

1 answer:

Zarrin [17]2 years ago

6 0

Answer:

109.5; B

Step-by-step explanation:

From your identity,

CosA = adjacent/ hypothenus

A represent an arbitrary angle between the sides in question.

In the question above, A=64

Hypothenus is the longest side and adjacent is the side just below the angle .

In the above case,

Hypothenus= X

adjacent =48

This means;

Cos64 = 48 /X

X = 48 / cos64; [ from cross multiplication and diving through by cos64]

X = 48 /0.4383 [ cos64 in radian = 0.4383]

= 109.51

= 109.5 to the nearest tenth.

Note( do your calculation of angle in radian or else, you won't get the answer)

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Find the standard form of the equation of the parabola with a vertex at the origin and a focus at (0, -7).

creativ13 [48]

So hmm check the picture below

we know the vertex is at the origin, and the focus point is below it, that means two things, the parabola is vertical and it's opening downwards

notice the distance "p", from the vertex to the focus point, is just 7 units, however, since the parabola is opening downwards, the "p" value will be negative, so p = -7

$\bf \textit{parabola vertex form with focus point distance}\\\\
\begin{array}{llll}
(y-{{ k}})^2=4{{ p}}(x-{{ h}}) \\\\
\boxed{4{{ p}}(y-{{ k}})=(x-{{ h}})^2 }\\
\end{array}
\qquad 
\begin{array}{llll}
vertex\ ({{ h}},{{ k}})\\\\
{{ p}}=\textit{distance from vertex to }\\
\qquad \textit{ focus or directrix}
\end{array}\\\\
-------------------------------\\\\
4{{ p}}(y-{{ k}})=(x-{{ h}})^2\quad 
\begin{cases}
h=0\\
k=0\\
p=-7
\end{cases}\implies 4(-7)(y-0)=(x-0)^2
\\\\\\
-28y=x^2\implies y=-\cfrac{1}{28}x^2$

8 0

3 years ago

Want an equation in point slope form of the line through the given point and with the given slope

coldgirl [10]

Answer:

<h3>The answer is option D</h3>

Step-by-step explanation:

To find an equation of a line in point slope form when given the slope and a point we use the formula

$y - y_1 = m(x - x_1)$

where

m is the slope

( x1 , y1) is the point

From the question the point is ( - 2 , -5) and slope 0 is

The equation of the line is

$y + 5 = 0(x + 2) \\$

We have the final answer as

Hope this helps you

3 0

3 years ago

If 'f(x)=x^2-18' and 'g(x)=(x-9)^(-1)(x+9)', find 'g(x)*f(x)

mezya [45]

The answer I would say would be B.

4 0

3 years ago

Simplify: <br>{(12)^1 + (13)^-1}/[(1/5)^-2 × {(1/5)^-1 + (1/8)^-1}^-1]

CaHeK987 [17]

Step-by-step explanation:

$\underline{\underline{\sf{➤\:\:Solution}}}$

$\sf \dashrightarrow \: \dfrac{ \left(\left(12 \right)^{ - 1} + \left(13 \right)^{ - 1} \right) }{\left( \dfrac{1}{5}\right) ^{ - 2} \times\left( \left( \dfrac{1}{5} \right) ^{ - 1} +\left( \dfrac{1}{8} \right) ^{ - 1} \right) ^{ - 1}}$

$\sf \dashrightarrow \: \dfrac{ \left(\dfrac{1}{12} + \dfrac{1}{13} \right) }{\left( \dfrac{5}{1}\right) ^{ 2} \times\left( \dfrac{5}{1} + \dfrac{8}{1} \right) ^{ - 1}}$

LCM of 12 and 13 is 156

$\sf \dashrightarrow \: \dfrac{ \left(\dfrac{1 \times 13 = 13}{12 \times 13 = 156} + \dfrac{1 \times 12 = 12}{13 \times 12 = 156} \right) }{\ \dfrac{25}{1} \times\left( \dfrac{5 + 8}{1} \right) ^{ - 1}}$

$\sf \dashrightarrow \: \dfrac{ \left(\dfrac{13}{156} + \dfrac{12}{156} \right) }{\ \dfrac{25}{1} \times\left( \dfrac{13}{1} \right) ^{ - 1}}$

$\sf \dashrightarrow \: \dfrac{ \left(\dfrac{13 + 12}{156} \right) }{\ \dfrac{25}{1} \times\dfrac{1}{13} }$

$\sf \dashrightarrow \: \dfrac{25}{156} \div \dfrac{25}{13}$

$\sf \dashrightarrow \: \dfrac{ \cancel{25}}{156} \times \dfrac{13}{ \cancel{25} }$

$\sf \dashrightarrow \: \dfrac{13}{156}$

$\sf \dashrightarrow \: \dfrac{1}{12}$

$\sf \dashrightarrow \: Answer = \underline{\boxed{ \sf{ \dfrac{1}{12} }}}$

━━━━━━━━━━━━━━━━━━━━━━━━

$\underline{\underline{\sf{★\:\:Laws\:of\: Exponents :}}}$

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$\sf Case : (i) \: if \: a > b \: then, \bigg( \dfrac{m}{n}\bigg) ^{a} \div \bigg( \dfrac{m}{n}\bigg)^{b} = \bigg( \dfrac{m}{n}\bigg)^{a - b}$

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