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3 years ago

The ΔPQR is right-angled at P, and PN is an altitude. If QN = 12 in and NR = 6 in, find PN, PQ, PR.

Mathematics

1 answer:

zaharov [31]3 years ago

4 0

Answer:

PR = 6√2, PN = 6, PQ = 12√2

Step-by-step explanation:

Find the unknowns through Sin, Cos, Tan formulae.

The explanation is attached in the pictures. !!

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The ratio of cats to dogs is 8:9. If there are 27 dogs, how many cats are there (Help)

Katyanochek1 [597]

Answer:

24 yw :)

Step-by-step explanation:

3 0

2 years ago

Subtract (3 + 2i) from (–9 – 8i). –17 – 5i –6 – 6i –12 – 10i 12 + 10

kupik [55]

Answer:

-12 - 10i

Step-by-step explanation:

We are subtracting 3 + 2i from -9 - 8i. Rewrite the left side as -3 - 2i and then ADD this result to -9 - 8i:

-9 - 8i

-3 -2i

-----------

-12 - 10i

8 0

3 years ago

Read 2 more answers

Algebraic Reasoning please help!

stiv31 [10]

Answer:

$\left[\begin{array}{ccc}7\\4\\2\end{array}\right]$

The answer is a single-column matrix (7,4,2)

Step-by-step explanation:

In such multiplication of matrices, you have to proceed by multiplying each ROW of the first matrix by the COLUMN of the second matrix. So,

$\left[\begin{array}{ccc}3&6&1\end{array}\right] * \left[\begin{array}{ccc}2\\0\\1\end{array}\right] = (3 * 2) + (6 * 0) + (1 * 1) = 6 + 0 + 1 = 7$

then...

$\left[\begin{array}{ccc}2&4&0\end{array}\right] * \left[\begin{array}{ccc}2\\0\\1\end{array}\right] = (2 * 2) + (4 * 0) + (0 * 1) = 4 + 0 + 0 = 4$

and

$\left[\begin{array}{ccc}0&6&2\end{array}\right] * \left[\begin{array}{ccc}2\\0\\1\end{array}\right] = (0 * 2) + (6 * 0) + (2 * 1) = 0 + 0 + 2= 2$

I hope it helps.

5 0

3 years ago

Integrate G(x,y,z)equalsz over the parabolic cylinder yequalszsquared, 0less than or equalsxless than or equals2, 0less than

yKpoI14uk [10]

I gather you're supposed to compute the integral of $G(x,y,z)=z$ over a surface $S$ that is the part of the parabolic cylinder $y=z^2$ with $0\le x\le2$ and $0\le z\le\frac{\sqrt{15}}2$ .

We can parameterize $S$ by

$\vec s(x,z)=x\,\vec\imath+z^2\,\vec\jmath+z\,\vec k$

with the given constraints on $x$ and $z$ . Take the normal vector to $S$ to be

$\vec s_x\times\vec s_z=-\vec\jmath+2z\,\vec k$

so that the surface element is

$\mathrm dS=\|\vec s_x\times\vec s_z\|\,\mathrm dx\,\mathrm dz=\sqrt{1+4z^2}\,\mathrm dx\,\mathrm dz$

Then in the integral, we have

$\displaystyle\iint_Sz\,\mathrm dS=\int_0^2\int_0^{\sqrt{15}/2}z\sqrt{1+4z^2}\,\mathrm dz\,\mathrm dx=\boxed{\frac{21}2}$

5 0

3 years ago

Can you guys please help me

VARVARA [1.3K]

Answer:

Number 1 I think sorry if I am incorrect.

8 0

3 years ago

Read 2 more answers