Which list shows the sides in order from shortest to longest

Mathematics

2 answers:

avanturin [10]3 years ago

6 0

Hey it would actually be C

GREYUIT [131]3 years ago

5 0

D is the right answer

hope it helps you ❣❣

Mark me as brainliest

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Help plz !!! thx have a good day!

Aleksandr-060686 [28]

Answer:

Substitution

Step-by-step explanation:

tbh i havent checked so sorry if its wrong

3 0

3 years ago

Rewrite the quadratic function in vertex form.<br> Y=3x^2-12x+4

chubhunter [2.5K]

Answer:

$\large\boxed{y=3(x-2)^2-8}$

Step-by-step explanation:

The vertex form of an equation of a parabola y = ax² + bx + c:

$f(x)=a(x-h)^2+k$

(h, k) - vertex

$h=\dfrac{-b}{2a},\ k=f(h)$

We have the equation:

$y=3x^2-12x+4\\\\a=3,\ b=-12,\ c=4$

Substitute:

$h=\dfrac{-(-12)}{2(3)}=\dfrac{12}{6}=2\\\\k=f(2)=3(2^2)-12(2)+4=3(4)-24+4=12-24+4=-8$

Finally:

$y=3(x-2)^2+(-8)=3(x-2)^2-8$

4 0

3 years ago

How many ways are there to select a 5-card hand from a regular deck such that the hand contains at least one card from each suit

Snezhnost [94]

Answer:

There are 685464 ways of selecting the 5-card hand

Step-by-step explanation:

Since the hand has 5 cards and there should be at least 1 card for each suit, then there should be 3 suits that appear once in the hand, and one suit that apperas twice.

In order to create a possible hand, first we select the suit that will appear twice. There are 4 possibilities for this. For that suit, we select the 2 cards that appear with the respective suit. Since there are 13 cards for each suit, then we have ${13 \choose 2} = 78$ possibilities. Then we pick one card of all remaining 3 suits. We have 13 ways to pick a card in each case.