All categories

3 years ago

Will Mark Brainlest helpppp

Mathematics

1 answer:

Anvisha [2.4K]3 years ago

6 0

Answer:

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

Step-by-step explanation:

P - Q + R = I ( I is the identity matrix )

$\left[\begin{array}{ccc}2&4\\3&5\\\end{array}\right]$ - $\left[\begin{array}{ccc}-2&2\\4&1\\\end{array}\right]$ + R = $\left[\begin{array}{ccc}1&0\\0&1\\\end{array}\right]$ ( subtract corresponding elements )

$\left[\begin{array}{ccc}2-(-2)&4-2\\3-4&5-1\\\end{array}\right]$ + R = $\left[\begin{array}{ccc}1&0\\0&1\\\end{array}\right]$

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

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

You might be interested in

the volume v of a right circular cylinder of radius r and heigh h is V = pi r^2 h 1. how is dV/dt related to dr/dt if h is const

laiz [17]

In general, the volume

$V=\pi r^2h$

has total derivative

$\dfrac{\mathrm dV}{\mathrm dt}=\pi\left(2rh\dfrac{\mathrm dr}{\mathrm dt}+r^2\dfrac{\mathrm dh}{\mathrm dt}\right)$

If the cylinder's height is kept constant, then $\dfrac{\mathrm dh}{\mathrm dt}=0$ and we have

$\dfrac{\mathrm dV}{\mathrm dt}=2\pi rh\dfrac{\mathrm dt}{\mathrm dt}$

which is to say, $\dfrac{\mathrm dV}{\mathrm dt}$ and $\dfrac{\mathrm dr}{\mathrm dt}$ are directly proportional by a factor equivalent to the lateral surface area of the cylinder ( $2\pi r h$ ).

Meanwhile, if the cylinder's radius is kept fixed, then

$\dfrac{\mathrm dV}{\mathrm dt}=\pi r^2\dfrac{\mathrm dh}{\mathrm dt}$

since $\dfrac{\mathrm dr}{\mathrm dt}=0$ . In other words, $\dfrac{\mathrm dV}{\mathrm dt}$ and $\dfrac{\mathrm dh}{\mathrm dt}$ are directly proportional by a factor of the surface area of the cylinder's circular face ( $\pi r^2$ ).

Finally, the general case ( $r$ and $h$ not constant), you can see from the total derivative that $\dfrac{\mathrm dV}{\mathrm dt}$ is affected by both $\dfrac{\mathrm dh}{\mathrm dt}$ and $\dfrac{\mathrm dr}{\mathrm dt}$ in combination.

8 0

3 years ago

(-59)-(-18)/(+3)+(-2)(+4)+-27-(-3)

mojhsa [17]

-41/25 I believe, I'm not sure why you put some addition signs in the inside of parentheses

8 0

3 years ago

Read 2 more answers

Gail practices swimming for 1 and 1 fourth hours swimming every day how much time does she practice in a week

Lorico [155]

If Gail practises swimming for 1 and 1/4 hours a day, then in a week she'll do:

1 1/4 * 7 = 5/4 * 7 = 35/4 = 8 3/4 [hours]

In other words she'll practice 8 hours and 45 minutes a week (assuming that she's still practicing on weekends).

8 0

3 years ago

25 POINTS AND BRAINLIEST PLEASE HELP ASAP

dexar [7]

M is a midpoint of BC so:

$M=\left(\dfrac{\boxed{2}\boxed{a}+a}{\boxed{2}},\dfrac{\boxed{0}+b}{2}\right)=\left(\dfrac{\boxed{3}\boxed{a}}{\boxed{2}},\dfrac{\boxed{b}}{\boxed{2}}\right)$

Length of MA:

$MA=\sqrt{\left(\dfrac{\boxed{3}a}{2}\boxed{-}\boxed{0}\right)^2+\left(\dfrac{\boxed{b}}{2}\boxed{-}\boxed{0}\right)^2}=\\\\\\=
\sqrt{\left(\dfrac{\boxed{3}a}{\boxed{2}}\right)^2+\left(\dfrac{b}{2}\right)^2}=\sqrt{\dfrac{\boxed{9}a^2}{\boxed{4}}+\dfrac{\boxed{b}^2}{\boxed{4}}}$

Length of NB:

$NB=\sqrt{\left(\dfrac{a}{2}\boxed{-}\boxed{2}a\right)^2+\left(\dfrac{b}{2}\boxed{-}\boxed{0}\right)^2}=\\\\\\=\sqrt{\left(\dfrac{a}{2}\boxed{-}\dfrac{\boxed{4}\boxed{a}}{2}\right)^2+\left(\dfrac{b}{2}-\boxed{0}\right)^2}=\\\\\\
\sqrt{\left(\dfrac{-3a}{2}\right)^2+\left(\dfrac{b}{\boxed{2}}\right)^2}=\sqrt{\dfrac{\boxed{9}a^2}{\boxed{4}}+\dfrac{\boxed{b}^2}{\boxed{4}}}$

5 0

3 years ago

Read 2 more answers

7(3r - 1) - (r + 5) = -52<br><br> Please step by step help

MrMuchimi

7(3r-1)-(r+5)=-52

21r-7-r-5=-52

21r-r-5-7=-52

20r-12=-52

Add 12 to both sides

20r=-40

Divide by 20

r=-2

4 0

3 years ago

Read 2 more answers

Will Mark Brainlest helpppp​

Will Mark Brainlest helpppp