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3 years ago

Number 3 and 4 please ASAP

Mathematics

1 answer:

Tpy6a [65]3 years ago

3 0

Step-by-step explanation:

Question 3) there's 60 minutes in a hour

3 times 60 = 180 books in 1 hour

3 times 180 = 540 books in 3 hours

540 + 180 + 180 = 900 books in 5 hours

The difference between those numbers : 900-540= 360 and 540-180= 360 so every 2 hours 360 books is done each time so just add 360 books each time for every 2 hours added

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In △ABC, DG=32 cm . What is the length of AD¯¯¯¯¯?

Sladkaya [172]

I think it's 96 cm. good luck

6 0

3 years ago

A species of phytoplankton measures about 2 × 10−6 in. A grain of sand measures about 1 × 10−4 in. About how many times longer i

taurus [48]

Answer : The grain of sand is 50 times greater than the phytoplankton.

Step-by-step explanation :

As we are given that:

Size of phytoplankton = 2 × 10⁻⁶ in

Size of sand grain = 1 × 10⁻⁴ in

Now we have to determine how many times longer the grain of sand is than the phytoplankton.

We are dividing the sand grain size by the phytoplankton size.

This can be done by equating it as follows:

$\frac{\text{Size of sand grain}}{\text{Size of phytoplankton}}=\frac{1\times 10^{-4}in}{2\times 10^{-6}in}$

$\frac{\text{Size of sand grain}}{\text{Size of phytoplankton}}=50$

Therefore, the grain of sand is 50 times greater than the phytoplankton.

5 0

3 years ago

Suppose that A and B are nonsingular matrices. Then AB is also nonsingular. Furthermore, a theorem from linear algebra then stat

kherson [118]

Answer:

A) Verified

B) Proved

Step-by-step explanation:

a) Let's verify it for 2 x 2 matrix,

$A=\left[\begin{array}{ccc}a&b\\c&d\end{array}\right]$ and $B=\left[\begin{array}{ccc}e&f\\g&h\end{array}\right]$

$AB=\left[\begin{array}{ccc}a&b\\c&d\end{array}\right]\left[\begin{array}{ccc}e&f\\g&h\end{array}\right]=\left[\begin{array}{ccc}a.e+b.g&a.f+b.h\\c.e+d.g&c.f+d.h\end{array}\right]$

$(AB)^{-1}=\frac{1}{(a.e+b.g)(c.f+d.h)-(a.f+b.h)(c.e+d.g)}\left[\begin{array}{ccc}c.f+d.h&-(a.f+b.h)\\-(c.e+d.g)&a.e+b.g\end{array}\right]$

$A^{-1}=\frac{1}{a.d-b.c} \left[\begin{array}{ccc}d&-b\\-c&a\end{array}\right]$

$B^{-1}=\frac{1}{e.h-f.g} \left[\begin{array}{ccc}h&-f\\-g&e\end{array}\right]$

$B^{-1}A^{-1}=\frac{1}{(a.e+b.g)(c.f+d.h)-(a.f+b.h)(c.e+d.g)}\left[\begin{array}{ccc}c.f+d.h&-(a.f+b.h)\\-(c.e+d.g)&a.e+b.g\end{array}\right]$

So it is proved that the results are same.

b) Now, let's prove it for any n x n matrix.

$(AB)(AB)^{-1}=I\\\\A^{-1}(AB)(AB)^{-1}=A^{-1}I\\\\IB(AB)^{=1}=A^{-1}I\\\\B(AB)^{=1}=A^{-1}\\\\B^{-1}B(AB)^{=1}=B^{-1}A^{-1}\\\\I(AB)^{=1}=B^{-1}A^{-1}\\\\(AB)^{=1}=B^{-1}A^{-1}$