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2 years ago

Sean has to face a big problem but he becomes stronger as a result which saying best summarizes the change in Sean?

Mathematics

1 answer:

adell [148]2 years ago

5 0

Answer:

D.) Every cloud has a silver lining!

Step-by-step explanation:

Hope this helps!

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Find the value of csc theta, if cos =-3/5; 90 degrees < theta < 180 degrees

Sloan [31]

Answer: $\frac{5}{4}$

Step-by-step explanation:

Given

$\cos \theta =\frac{-3}{5}$

and $\theta$ lies between

$90^{\circ}$

and for this $\theta$ , $\sin$ and $\text{cosec}$ is Positive as they lie in 2 nd Quadrant

$\sin ^2\theta +\cos ^2\theta =1$

$\sin ^2\theta =1-(\frac{-3}{5})^2$

$\sin ^2\theta =1-\frac{9}{25}$

$\sin ^2\theta =\frac{16}{25}$

$\sin \theta =\frac{4}{5}$

$\therefore \text{cosec}\ \theta =\frac{5}{4}$

6 0

2 years ago

A cuboid has side lengths 3cm, 4cm and 7cm. A cylinder with a height double that of the longest side of the cuboid has the same

mart [117]

Step-by-step explanation:

Longest side of cuboid: 7cm

Height of cylinder:

2 × 7 = 14 cm

Volume of cuboid:

7x3x4=84 $cm^3$

Volume of cylinder:

$84=14\pi r^2 \\r=\sqrt{\frac{6}{\pi} } \ \ \ cm\\r=1.38 \ cm\\$

Surface area of the cylinder:

$\pi r^2 + 2 \pi r h$

$6+2\pi (\sqrt{\frac{6}{\pi} } ) (14)$

=128 cm^2

4 0

2 years ago

Explain why a set of points that defines a plane cannot be collinear

USPshnik [31]

Answer:

Three non-collinear points determine a plane.

Step-by-step explanation:

This statement means that if you have three points not on one line, then only one specific plane can go through those points. The plane is determined by the three points because the points show you exactly where the plane is

3 0

2 years ago

Use the following matrices, A, B, C and D to perform each operation.

Vinvika [58]

Step-by-step explanation:

$A=\left[\begin{array}{ccc}3&1\\5&7\end{array}\right]$

$B=\left[\begin{array}{ccc}4&1\\6&0\end{array}\right]$

$C=\left[\begin{array}{ccc}-2&3&1\\-1&0&4\end{array}\right]$

$D=\left[\begin{array}{ccc}-2&3&4\\0&-2&1\\3&4&-1\end{array}\right]$

$1.\\A+B=\left[\begin{array}{ccc}3&1\\5&7\end{array}\right]+\left[\begin{array}{ccc}4&1\\6&0\end{array}\right]=\left[\begin{array}{ccc}3+4&1+1\\5+6&7+0\end{array}\right]=\left[\begin{array}{ccc}7&2\\11&7\end{array}\right]$

$2.\\B-A=\left[\begin{array}{ccc}4&1\\6&0\end{array}\right]-\left[\begin{array}{ccc}3&1\\5&7\end{array}\right]=\left[\begin{array}{ccc}4-3&1-1\\6-5&0-7\end{array}\right]=\left[\begin{array}{ccc}1&0\\1&-7\end{array}\right]$

$3.\\3C=3\left[\begin{array}{ccc}-2&3&1\\-1&0&4\end{array}\right]=\left[\begin{array}{ccc}(3)(-2)&(3)(3)&(3)(1)\\(3)(-1)&(3)(0)&(3)(4)\end{array}\right]=\left[\begin{array}{ccc}-6&9&3\\-3&0&12\end{array}\right]$

$4.\\C\cdot D=\left[\begin{array}{ccc}-2&3&1\\-1&0&4\end{array}\right]\cdot\left[\begin{array}{ccc}-2&3&4\\0&-2&1\\3&4&-1\end{array}\right]\\\\=\left[\begin{array}{ccc}(-2)(-2)+(3)(0)+(1)(3)&(-2)(3)+(3)(-2)+(1)(4)&(-2)(4)+(3)(1)+(1)(-1)\\(-1)(-2)+(0)(0)+(4)(3)&(-1)(3)+(0)(-2)+(4)(4)&(-1)(4)+(0)(1)+(4)(-1)\end{array}\right]\\=\left[\begin{array}{ccc}7&-8&-6\\14&13&-8\end{array}\right]$