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2 years ago

Evaluate P + Q if possible.

Mathematics

2 answers:

jenyasd209 [6]2 years ago

4 0

Since the matrices have different dimensions, it is not possible to calculate the expression...

Thus, Option C is correct!!~

Sunny_sXe [5.5K]2 years ago

3 0

Here we are given with two matrices as follows

${\bf{P=\begin{bmatrix}5& 0& 1 \\ 2& 8& 5\end{bmatrix}}}$

And

${\bf{Q=\begin{bmatrix}6& 5& 0 \\ 1& 2& 8 \\ 5& 4& 1 \end{bmatrix}}}$

We need to find P+Q . But , matrix addition is only possible iff the order of all matrices which are being added is same . Now , here order of P is 2 × 3 while that of Q is 3 × 3 . So , matrix addition isn't possible.

Hence , Option C) Not possible is correct

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A store is having a clearance sale where merchandise on the sales rack is reduced by 80% from the original price. if a jacket wa

miskamm [114]

First multiply 76 by 80/100
(76 x 80 )/100 = 60.8

Now because the price has to be reduced...
76 - 60.8 = $15.2

Hope this will help.

4 0

2 years ago

B) Let g(x) =x/2sqrt(36-x^2)+18sin^-1(x/6) Find g'(x) =

jolli1 [7]

I suppose you mean

$g(x) = \dfrac x{2\sqrt{36-x^2}} + 18\sin^{-1}\left(\dfrac x6\right)$

Differentiate one term at a time.

Rewrite the first term as

$\dfrac x{2\sqrt{36-x^2}} = \dfrac12 x(36-x^2)^{-1/2}$

Then the product rule says

$\left(\dfrac12 x(36-x^2)^{-1/2}\right)' = \dfrac12 x' (36-x^2)^{-1/2} + \dfrac12 x \left((36-x^2)^{-1/2}\right)'$

Then with the power and chain rules,

$\left(\dfrac12 x(36-x^2)^{-1/2}\right)' = \dfrac12 (36-x^2)^{-1/2} + \dfrac12\left(-\dfrac12\right) x (36-x^2)^{-3/2}(36-x^2)' \\\\ \left(\dfrac12 x(36-x^2)^{-1/2}\right)' = \dfrac12 (36-x^2)^{-1/2} - \dfrac14 x (36-x^2)^{-3/2} (-2x) \\\\ \left(\dfrac12 x(36-x^2)^{-1/2}\right)' = \dfrac12 (36-x^2)^{-1/2} + \dfrac12 x^2 (36-x^2)^{-3/2}$

Simplify this a bit by factoring out $\frac12 (36-x^2)^{-3/2}$ :

$\left(\dfrac12 x(36-x^2)^{-1/2}\right)' = \dfrac12 (36-x^2)^{-3/2} \left((36-x^2) + x^2\right) = 18 (36-x^2)^{-3/2}$

For the second term, recall that

$\left(\sin^{-1}(x)\right)' = \dfrac1{\sqrt{1-x^2}}$

Then by the chain rule,

$\left(18\sin^{-1}\left(\dfrac x6\right)\right)' = 18 \left(\sin^{-1}\left(\dfrac x6\right)\right)' \\\\ \left(18\sin^{-1}\left(\dfrac x6\right)\right)' = \dfrac{18\left(\frac x6\right)'}{\sqrt{1 - \left(\frac x6\right)^2}} \\\\ \left(18\sin^{-1}\left(\dfrac x6\right)\right)' = \dfrac{18\left(\frac16\right)}{\sqrt{1 - \frac{x^2}{36}}} \\\\ \left(18\sin^{-1}\left(\dfrac x6\right)\right)' = \dfrac{3}{\frac16\sqrt{36 - x^2}} \\\\ \left(18\sin^{-1}\left(\dfrac x6\right)\right)' = \dfrac{18}{\sqrt{36 - x^2}} = 18 (36-x^2)^{-1/2}$

So we have

$g'(x) = 18 (36-x^2)^{-3/2} + 18 (36-x^2)^{-1/2}$

and we can simplify this by factoring out $18(36-x^2)^{-3/2}$ to end up with

$g'(x) = 18(36-x^2)^{-3/2} \left(1 + (36-x^2)\right) = \boxed{18 (36 - x^2)^{-3/2} (37-x^2)}$

5 0

2 years ago

Which input value produces the sane output value for the two functions on the gragh f(x)=-2/3x+1 g(x)=1/3x-2

zimovet [89]

Answer:

X=3

Step-by-step explanation:

We have two linear functions which intersect at a point. This point is shown in the attached graph. Linear functions are lines which are made of points that satisfy the function or relationship. This means at the intersection, this point (3,-1), both functions have the same values. An input of x=3 produces y=-1 in both functions.

5 0

3 years ago

eric is making a dress. the design calls for a right triangular piece of material. if the second angle has to be 35% what is the

svet-max [94.6K]

U have a right triangle...so one angle is 90%......and the second angle is 35%...keep in mind, the angles of a triangle = 180%
so the measure of the third angle is : 180 - 90 - 35 = 55%

3 0

3 years ago

Please help I don't understand!

coldgirl [10]

Answer:

Step-by-step explanation:

I believe because on the other side it is AE over ED so I just repeated the pattern.

6 0

2 years ago