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3 years ago

Which shows all the names that apply to the figure?

Mathematics

2 answers:

podryga [215]3 years ago

8 0

B. This is classified as a parallelogram and a rhombus.

Lorico [155]3 years ago

5 0

<span>B. parallelogram, rhombus</span>

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Gerry ran 3/8 of a mile on Monday. He ran 5/8 of a mile on tuesday he ran 1/2 mile on Wednesday how many miles in all did Gerry

pickupchik [31]

1 and 1/2 miles in total

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2 years ago

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Use the remainder theorem to find the remainder for (x ^5 + 32) divided by (x+2) and state whether or not the binomial is a fact

marta [7]

$x+2$ is a factor of $x^5+32$ because (by the remainder theorem) the remainder upon dividing $x^5+32$ by $x+2$ is $(-2)^5+32=0$ .

There's also the sum of fifth powers formula,

$a^5+b^5=(a+b)(a^4-a^3b+a^2b^2-ab^3+b^4)$

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2 years ago

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40. Danielle's living room is a rectangle with the dimensions 16 feet

sammy [17]

Throw rug with diameter of 12 feet what is the approximate

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3 years ago

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The figure below shows a circular park. Its radius is 18 yd.

Naddika [18.5K]

Radius=r=18yd

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2πr
2(18)(3.14)
36(3.14)
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Area

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324π yd²

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2 years ago

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Use the given information to find the exact value of the trigonometric function

eimsori [14]

$\begin{gathered} \csc \theta=-\frac{6}{5} \\ \tan \theta>0 \\ \cos \frac{\theta}{2}=\text{?} \end{gathered}$

Half Angle Formula

$\cos \frac{\theta}{2}=\pm\sqrt[\square]{\frac{1+\cos\theta}{2}}$ $\tan \theta>0\text{ and csc}\theta\text{ is negative in the third quadrant}$ $\begin{gathered} \csc \theta=-\frac{6}{5}=\frac{r}{y} \\ x^2+y^2=r^2 \\ x=\pm\sqrt[\square]{r^2-y^2} \\ x=\pm\sqrt[\square]{6^2-(-5)^2} \\ x=\pm\sqrt[\square]{36-25} \\ x=\pm\sqrt[\square]{11} \\ \text{x is negative since the angle is on the 3rd quadrant} \end{gathered}$ $\begin{gathered} \cos \theta=\frac{x}{r}=\frac{-\sqrt[\square]{11}}{6} \\ \cos \frac{\theta}{2}=\pm\sqrt[\square]{\frac{1+\cos\theta}{2}} \\ \cos \frac{\theta}{2}is\text{ also negative in the 3rd quadrant} \\ \cos \frac{\theta}{2}=-\sqrt[\square]{\frac{1+\frac{-\sqrt[\square]{11}}{6}}{2}} \\ \cos \frac{\theta}{2}=-\sqrt[\square]{\frac{\frac{6-\sqrt[\square]{11}}{6}}{2}} \\ \cos \frac{\theta}{2}=-\sqrt[\square]{\frac{6-\sqrt[\square]{11}}{12}} \\ \\ \end{gathered}$

Answer:

$\cos \frac{\theta}{2}=-\sqrt[\square]{\frac{6-\sqrt[\square]{11}}{12}}$

Checking:

$\begin{gathered} \frac{\theta}{2}=\cos ^{-1}(-\sqrt[\square]{\frac{6-\sqrt[\square]{11}}{12}}) \\ \frac{\theta}{2}=118.22^{\circ} \\ \theta=236.44^{\circ}\text{ (3rd quadrant)} \end{gathered}$

Also,

$\csc \theta=\frac{1}{\sin\theta}=\frac{1}{\sin (236.44)}=-\frac{6}{5}\text{ QED}$

The answer is none of the choices

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10 months ago