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pishuonlain [190]
3 years ago
5

Math geometry help please show work thanks

Mathematics
1 answer:
Tems11 [23]3 years ago
6 0

Part (a)

Refer to the diagram below to see the two triangles separated.

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

Part (b)

If the 3 meter portion is perpendicular to the horizontal portion, then the triangles are similar. We can prove this using the AA similarity theorem. One pair of angles would be the shared overlapping 25 degree angles. The other pair of congruent angles would be the two 90 degree angles. We only need two pairs of congruent angles to use AA. You could use three pairs (and the aptly named AAA theorem) but that's overkill in my book.

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

Part (c)

There are a few ways to approach this, but one way is shown below

y/3 = (8.4+6.6)/(6.6)

y/3 = 15/(6.6)

6.6y = 3*15 .......... cross multiply

6.6y = 45

y = 45/(6.6)

y = 6.81818 approximately

When rounding to one decimal place, we get y = 6.8

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                                      (112.5 m  -  40 m)  =  72.5 meters

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Since it was falling at the rate of  2.5 m/s, it took

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Assume y≠60 which expression is equivalent to (7sqrtx2)/(5sqrty3)
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The equivalent will be:

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Therefore, option 'a' is true.

Step-by-step explanation:

Given the expression

\frac{\sqrt[7]{x^2}}{\sqrt[5]{y^3}}

Let us solve the expression step by step to get the equivalent

\frac{\sqrt[7]{x^2}}{\sqrt[5]{y^3}}

as

\sqrt[7]{x^2}=\left(x^2\right)^{\frac{1}{7}}      ∵ \mathrm{Apply\:radical\:rule}:\quad \sqrt[n]{a}=a^{\frac{1}{n}}

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=x^{2\cdot \frac{1}{7}}

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also

\sqrt[5]{y^3}=\left(y^3\right)^{\frac{1}{5}}         ∵  \mathrm{Apply\:radical\:rule}:\quad \sqrt[n]{a}=a^{\frac{1}{n}}

\mathrm{Apply\:exponent\:rule:\:}\left(a^b\right)^c=a^{bc},\:\quad \mathrm{\:assuming\:}a\ge 0

=y^{3\cdot \frac{1}{5}}

=y^{\frac{3}{5}}

so the expression becomes

\frac{x^{\frac{2}{7}}}{y^{\frac{3}{5}}}

\mathrm{Apply\:exponent\:rule}:\quad \:a^{-b}=\frac{1}{a^b}

=\left(\:x^{\frac{2}{7}}\right)\left(y^{-\frac{3}{5}}\right)            ∵ \:\frac{1}{y^{\frac{3}{5}}}=y^{-\frac{3}{5}}

Thus, the equivalent will be:

\frac{\sqrt[7]{x^2}}{\sqrt[5]{y^3}}=\left(\:x^{\frac{2}{7}}\right)\left(y^{-\frac{3}{5}}\right)

Therefore, option 'a' is true.

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