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goblinko [34]
3 years ago
15

Can you help thanks.....

Mathematics
2 answers:
Jet001 [13]3 years ago
7 0

Answer:

32,27,42,47,52

Step-by-step explanation:

nirvana33 [79]3 years ago
5 0
32, 37, 42, 47, 52. Just add 5
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Answer:

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Step-by-step explanation:

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I need to know the improper fractions answers for:
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Answer:

Part 1) x=\frac{15}{2}\ units

Part 2) z=\frac{15\sqrt{3}}{2}\ units

Part 3) y= \frac{15\sqrt{3}}{4}\ units

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Step-by-step explanation:

step 1

Find the value of x

In the large right triangle

cos(60^o)=\frac{x}{15} ----> by CAH (adjacent side divided by the hypotenuse)

Remember that

cos(60^o)=\frac{1}{2}

substitute

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solve for x

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step 2

Find the value of z

In the large right triangle

Applying the Pythagorean Theorem

15^2=x^2+z^2

substitute the value of x

15^2=(\frac{15}{2})^2+z^2

solve for z

z^2=15^2-(\frac{15}{2})^2

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step 3

Find the value of y

In the right triangle of the right

sin(30^o)=\frac{y}{z} ---> by SOH (opposite side divided by the hypotenuse)

substitute the given values of y and z

Remember that

sin(30^o)=\frac{1}{2}

so

\frac{1}{2}=y:\frac{15\sqrt{3}}{2}

solve for y

\frac{1}{2}= \frac{2y}{15\sqrt{3}}

y= \frac{15\sqrt{3}}{4}\ units

step 4

Find the value of b

In the right triangle of the right

cos(30^o)=\frac{b}{z} ---> by CAH (adjacent side divided by the hypotenuse)

substitute the given values of y and z

Remember that

cos(30^o)=\frac{\sqrt{3}}{2}

so

\frac{\sqrt{3}}{2}=b:\frac{15\sqrt{3}}{2}

solve for y

\frac{\sqrt{3}}{2}= \frac{2b}{15\sqrt{3}}

b= \frac{45}{4}\ units

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Step-by-step explanation:

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