I am so sorry! I am terrible at math!

Mathematics

2 answers:

eimsori [14]3 years ago

4 0

I have been learning this at school just add 6+6 and 10+10

Elis [28]3 years ago

3 0

Percent change=change/original times 100

perimiter
if length and width double then the perimiter doubles
because
P=2(L+W)
L=10
W=6
old one
P=2(10+6)
P=2(16)
P=32
new one
20 and 12
P=2(20+12)
P=2(32)
P=64

percent change=change/original times 100
change=64-32=32
original=32
percent change=32/32 times 100=100%

area
old are=legnth tiems wdith =6 times 10=60
nwe one
area=12 times 20=480
change is 240-60=180
percent change=180/60 times 100=3 times 100=300

perimiter is 100% change
area is 300% change

You might be interested in

URGENT <br> WILL GIVE BRAINLIEST <br> solve using matrices<br> 4x + 5y = 40 <br> x – y = 1

dalvyx [7]

Answer:

x = 5, y = 4

Step-by-step explanation:

$\left\{\begin{array}{ccc}4x+5y=40\\x-y=1\end{array}\right\\\\A=\left[\begin{array}{ccc}4&5\\1&-1\end{array}\right]\\\\detA=\left|\begin{array}{ccc}4&5\\1&-1\end{array}\right|=(4)(-1)-(1)(5)=-4-5=-9\\\\A^D=\left[\begin{array}{ccc}-1&-1\\-5&4\end{array}\right]\\\\\left(A^D\right)^T=\left[\begin{array}{ccc}-1&-5\\-1&4\end{array}\right]$

$A^{-1}=\dfrac{1}{detA}\left(A^D\right)^T\\\\A^{-1}=\dfrac{1}{-9}\left[\begin{array}{ccc}-1&-5\\-1&4\end{array}\right]=\left[\begin{array}{ccc}\dfrac{1}{9}&\dfrac{5}{9}\\\dfrac{1}{9}&-\dfrac{4}{9}\end{array}\right]$

$A\cdot\left[\begin{array}{ccc}x\\y\end{array}\right] =\left[\begin{array}{ccc}40\\1\end{array}\right] \\\\A^{-1}A\cdot\left[\begin{array}{ccc}x\\y\end{array}\right] =A^{-1}\cdot\left[\begin{array}{ccc}40\\1\end{array}\right] \\\\\left[\begin{array}{ccc}x\\y\end{array}\right] =A^{-1}\cdot\left[\begin{array}{ccc}40\\1\end{array}\right]$

$\left[\begin{array}{ccc}x\\y\end{array}\right] =\left[\begin{array}{ccc}\dfrac{1}{9}&\dfrac{5}{9}\\\dfrac{1}{9}&-\dfrac{4}{9}\end{array}\right]\cdot\left[\begin{array}{ccc}40\\1\end{array}\right] \\\\\left[\begin{array}{ccc}x\\y\end{array}\right] =\left[\begin{array}{ccc}\left(\frac{1}{9}\right)(40)+\left(\frac{5}{9}\right)(1)\\\\\left(\frac{1}{9}\right)(40)+\left(-\frac{4}{9}\right)(1)\end{array}\right]\\\\\left[\begin{array}{ccc}x\\y\end{array}\right] =\left[\begin{array}{ccc}5\\4\end{array}\right]\Rightarrow x=5,\ y=4$