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4 years ago

Find the area of the triangle. [? ] ft2 Don't round

Mathematics

2 answers:

Cerrena [4.2K]4 years ago

8 0

Step-by-step explanation:

$Area = \: \frac{bh}{2} \\ : b = 16.9 \: h = 10.4 \\$

$Area = \frac{16.9 \times 10.4}{2} = \frac{175.76}{2} = 87.88 {ft}^{2}$

Base = 16.9

height = 10.4

Area = ½b×h

A = ½(16.9×10.4)

A = ½(175.76)

Area = 175.76/2

A = 87.88ft²

sashaice [31]4 years ago

7 0

Answer:

$\Large \boxed{\mathrm{87.88 \ ft^2 }}$

Step-by-step explanation:

$\displaystyle area \ of \ triangle \ = \ \frac{base \times height }{2}$

$\displaystyle area \ of \ triangle \ = \ \frac{16.9 \times 10.4 }{2}$

$\displaystyle area \ of \ triangle \ = \ \frac{175.76}{2}$

$\displaystyle area \ of \ triangle \ = \ 87.88$

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Yanka [14]

Answer:

$x=\frac{12}{7} \\y=\frac{12}{5} \\z=-12$

Step-by-step explanation:

Let's re-write the equations in order to get the variables as separated in independent terms as possible \:

First equation:

$\frac{xy}{x+y} =1\\xy=x+y\\1=\frac{x+y}{xy} \\1=\frac{1}{y} +\frac{1}{x}$

Second equation:

$\frac{xz}{x+z} =2\\xz=2\,(x+z)\\\frac{1}{2} =\frac{x+z}{xz} \\\frac{1}{2} =\frac{1}{z} +\frac{1}{x}$

Third equation:

$\frac{yz}{y+z} =3\\yz=3\,(y+z)\\\frac{1}{3} =\frac{y+z}{yz} \\\frac{1}{3}=\frac{1}{z} +\frac{1}{y}$

Now let's subtract term by term the reduced equation 3 from the reduced equation 1 in order to eliminate the term that contains "y":

$1=\frac{1}{y} +\frac{1}{x} \\-\\\frac{1}{3} =\frac{1}{z} +\frac{1}{y}\\\frac{2}{3} =\frac{1}{x} -\frac{1}{z}$

Combine this last expression term by term with the reduced equation 2, and solve for "x" :

$\frac{2}{3} =\frac{1}{x} -\frac{1}{z} \\+\\\frac{1}{2} =\frac{1}{z} +\frac{1}{x} \\ \\\frac{7}{6} =\frac{2}{x}\\ \\x=\frac{12}{7}$

Now we use this value for "x" back in equation 1 to solve for "y":

$1=\frac{1}{y} +\frac{1}{x} \\1=\frac{1}{y} +\frac{7}{12}\\1-\frac{7}{12}=\frac{1}{y} \\ \\\frac{1}{y} =\frac{5}{12} \\y=\frac{12}{5}$

And finally we solve for the third unknown "z":

$\frac{1}{2} =\frac{1}{z} +\frac{1}{x} \\\\\frac{1}{2} =\frac{1}{z} +\frac{7}{12} \\\\\frac{1}{z} =\frac{1}{2}-\frac{7}{12} \\\\\frac{1}{z} =-\frac{1}{12}\\z=-12$

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3 years ago

The point (1, −1) is on the terminal side of angle θ, in standard position. What are the values of sine, cosine, and tangent of

Lilit [14]

Check the picture below.

$\bf (\stackrel{a}{1}~,~\stackrel{b}{-1})\qquad \impliedby \textit{let's find the hypotenuse} \\\\\\ \textit{using the pythagorean theorem} \\\\ c^2=a^2+b^2\implies c = \sqrt{a^2+b^2} \qquad \begin{cases} c=hypotenuse\\ a=adjacent\\ b=opposite\\ \end{cases} \\\\\\ c=\sqrt{1^2+(-1)^2}\implies c=\sqrt{2} \\\\[-0.35em] ~\dotfill$

$\bf sin(\theta ) \implies \cfrac{\stackrel{opposite}{-1}}{\stackrel{hypotenuse}{\sqrt{2}}}\implies \cfrac{-1}{\sqrt{2}}\cdot \cfrac{\sqrt{2}}{\sqrt{2}}\implies -\cfrac{\sqrt{2}}{(\sqrt{2})^2}\implies -\cfrac{\sqrt{2}}{2}$

$\bf cos(\theta ) \implies \cfrac{\stackrel{adjacent}{1}}{\stackrel{hypotenuse}{\sqrt{2}}}\implies \cfrac{1}{\sqrt{2}}\cdot \cfrac{\sqrt{2}}{\sqrt{2}}\implies \cfrac{\sqrt{2}}{(\sqrt{2})^2}\implies \cfrac{\sqrt{2}}{2} \\\\\\ tan(\theta ) = \cfrac{\stackrel{opposite}{-1}}{\stackrel{adjacent}{1}}\implies tan(\theta ) = -1$

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3 years ago

Read 2 more answers

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Brilliant_brown [7]

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2<br> 1 point<br> The point, (-3, 7), is on the line y = -3x+7.<br> True<br> O False

sladkih [1.3K]

Step-by-step explanation:

in order to know if the the point is on the line plug in the x coordinate to check if it gives you the y coordinate as an output.
If you plug in the x coordinate and it gives you the given y coordinate then the point is on the given line
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Andrew [12]

Answer:
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Explanation:
y = 1/2x + 14 is perpendicular to y = -x = 18 and passes through the slope (-6, 11)

Tip:
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3 years ago