The angle which is four times its complement is a. 60 b. 30 C. 45 d. 72

Mathematics

1 answer:

grigory [225]2 years ago

5 0

Answer:

The answer is D.72°

Step-by-step explanation:

hope this helps.

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Find the angle between u = i+sqr of 7j and v = -i+9j. Round to the nearest tenth of a degree.

Zolol [24]

$\bf ~~~~~~~~~~~~\textit{angle between two vectors } \\\\ cos(\theta)=\cfrac{\stackrel{\textit{dot product}}{u \cdot v}}{\stackrel{\textit{magnitude product}}{||u||~||v||}} \implies \measuredangle \theta = cos^{-1}\left(\cfrac{u \cdot v}{||u||~||v||}\right) \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ \begin{cases} u=i+\sqrt{7}j\implies &\\\\ v=-i+9j\implies & \end{cases} \\\\[-0.35em] ~\dotfill$

$\bf u\cdot v\implies (1)(-1)~+~(\sqrt{7})(9)\implies -1+9\sqrt{7}\implies 9\sqrt{7}-1~\hfill dot~product \\\\[-0.35em] ~\dotfill\\\\ ||u||\implies \sqrt{1^2+(\sqrt{7})^2}\implies \sqrt{1+7}\implies \sqrt{8}~\hfill magnitudes \\\\\\ ||v||\implies \sqrt{(-1)^2+9^2}\implies \sqrt{1+81}\implies \sqrt{82} \\\\[-0.35em] ~\dotfill$

$\bf \theta =cos^{-1}\left( \cfrac{9\sqrt{7}-1}{\sqrt{8}\cdot \sqrt{82}} \right)\implies \theta =cos^{-1}\left( \cfrac{9\sqrt{7}-1}{\sqrt{656}} \right) \\\\\\ \theta \approx cos^{-1}(0.8906496638868531)\implies \theta \approx 27.05$

make sure your calculator is in Degree mode.

6 0

2 years ago

The equation a = (b1 + b2)h can be used to determine the area, a, of a trapezoid with height, h, and base lengths, b1 and b2. Wh

Arlecino [84]

Hb1+hb2=a
a/h=b1+b2
a/(b1+b2)=h

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3 years ago

Read 2 more answers

HELPPPPP PLEASEEEEE!!!

Pepsi [2]

Answer:

The height of right circular cone is h = 15.416 cm

Step-by-step explanation:

The formula used to calculate lateral surface area of right circular cone is: $s=\pi r\sqrt{r^2+h^2}$

where r is radius and h is height.

We are given:

Lateral surface area s = 236.64 cm²

Radius r = 4.75 cm

We need to find height of right circular cone.

Putting values in the formula and finding height:

$s=\pi r\sqrt{r^2+h^2}\\236.64=3.14(4.75)\sqrt{(3.75)^2+h^2} \\236.64=14.915\sqrt{(3.75)^2+h^2} \\\frac{236.64}{14.915}=\sqrt{14.0625+h^2} \\15.866=\sqrt{14.0625+h^2} \\Switching\:sides\:\\\sqrt{14.0625+h^2} =15.866\\Taking\:square\:on\:both\:sides\\(\sqrt{14.0625+h^2})^2 =(15.866)^2\\14.0625+h^2=251.729\\h^2=251.729-14.0625\\h^2=237.6665\\Taking\:square\:root\:on\:both\:sides\\\sqrt{h^2}=\sqrt{237.6665} \\h=15.416$