4y+(-2e)+6xx=98e=36y=14Please help me

Mathematics

1 answer:

Sidana [21]3 years ago

8 0

Answer:

572

Step-by-step explanation:

4y + (-2e) + 6x

4(14) + (-2(36)) + 6(98)

56 - 72 + 588

-16 + 588

572

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if a 42.9ft tall flagpole casts a 253.1ft long shadow then how long is the shadow that a 6.2ft tall woman casts?

vovangra [49]

Set up a proportion.
flagpole/shadow = woman/her shadow.

f/s1 = w/s2
f = 42.9
s1 = 253
w = 6.2
s2 = ??? = x

42.9/253 = 6.2 / x
42.9 x = 253*6.2
42.9 x = 1568.6
x = 1568.6 / 42.9
x = 36.56 The womans shadow is 36.56 feet long <<<< ===== answer.

3 0

3 years ago

Can anybody show me how to do this half angle identity problem STEP BY STEP? I always seem to be missing something

klio [65]

Answer: $\bold{-\dfrac{7\sqrt2}{10}}$

<u>Step-by-step explanation:</u>

It is given that θ is between 270° and 360°, which means that θ is located in Quadrant IV ⇒ (x > 0, y < 0). Furthermore, the half-angle will be between 135° and 180°, which means the half-angle is in Quadrant II ⇒ $cos\ \dfrac{\theta}{2}$

It is given that sin θ = $-\dfrac{7}{25}$ ⇒ y = -7 & hyp = 25

Use Pythagorean Theorem to find "x":

x² + y² = hyp²

x² + (-7)² = 25²

x² + 49 = 625

x² = 576

x = 24

Use the "x" and "hyp" values to find cos θ:

$cos\ \theta=\dfrac{x}{hyp}=\dfrac{24}{25}$

Lastly, input cos θ into the half angle formula:

$cos\bigg(\dfrac{\theta}{2}\bigg)=\pm \sqrt{\dfrac{1+cos\ \theta}{2}}\\\\\\.\qquad \quad =\pm \sqrt{\dfrac{1+\dfrac{24}{25}}{2}}\\\\\\.\qquad \quad =\pm \sqrt{\dfrac{\dfrac{25}{25}+\dfrac{24}{25}}{2}}\\\\\\.\qquad \quad =\pm \sqrt{\dfrac{\dfrac{49}{25}}{2}}\\\\\\.\qquad \quad =\pm \sqrt{\dfrac{49}{50}}\\\\\\.\qquad \quad =\pm \dfrac{7}{5\sqrt2}}\\\\\\.\qquad \quad =\pm \dfrac{7}{5\sqrt2}}\bigg(\dfrac{\sqrt2}{\sqrt2}\bigg)\\\\\\.\qquad \quad =\pm \dfrac{7\sqrt2}{10}$