0.6666 <-repeating 66% is both in a percent and a decimal
5:3
5units:3units
5+3=8
8 units
4qt=8units
divide by 8 both sides
1/2qt=1unit
red=5unit
red=5*1/2
red=5/2qt
d
lime green=3unit
lime green=3*1/2
lime green=3/2qt
not sure what the ratio of the blue paint to the yellow paint is
I do know that
when b=amount of blue paint in quarts
yo need 5/2qt red, b qt blue, and y qt yellow, where b+y=3/2qt
and the ratio is
5:b:y, where b+y=3
can't solve fully without ration of the the blue to yellow
There are 28 ways in which a couple can choose the name of the baby for its name.
<h3>What is defined as the combination?</h3>
- A combination is an algebraic technique for determining the number of possible arrangements in a set of items in which the order of the selection is irrelevant.
- You can choose the items in just about any order in combinations. Permutations and combinations are often confused.
If we need to choose objects from two groups of x and n objects so that one object from each group is chosen, we can do so by calculating the combinations possible by:
= ˣC₁ × ⁿC₁
Let 'x' be the set of first name = 7
Let 'n' be the set of second name = 4
Putting the values in formula;
= ⁷C₁ × ⁴C₁
= 7 × 4
= 28
Thus, there are 28 ways in which a couple can choose the name of the baby for its name.
To know more about the combination, here
brainly.com/question/12725706
#SPJ9
The complete question is-
A couple has narrowed down the choices of a name for their new baby to 7 first names and 4 second names.
How many different first- and second-name arrangements are possible?
keeping in mind that perpendicular lines have <u>negative reciprocal</u> slopes, let's find the slope of 3x + 4y = 9, by simply putting it in slope-intercept form.
![\bf 3x+4y=9\implies 4y=-3x+9\implies y=-\cfrac{3x+9}{4}\implies y=\stackrel{slope}{-\cfrac{3}{4}}x+\cfrac{9}{4} \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ \stackrel{\textit{perpendicular lines have \underline{negative reciprocal} slopes}} {\stackrel{slope}{-\cfrac{3}{4}}\qquad \qquad \qquad \stackrel{reciprocal}{-\cfrac{4}{3}}\qquad \stackrel{negative~reciprocal}{+\cfrac{4}{3}}\implies \cfrac{4}{3}}](https://tex.z-dn.net/?f=%20%5Cbf%203x%2B4y%3D9%5Cimplies%204y%3D-3x%2B9%5Cimplies%20y%3D-%5Ccfrac%7B3x%2B9%7D%7B4%7D%5Cimplies%20y%3D%5Cstackrel%7Bslope%7D%7B-%5Ccfrac%7B3%7D%7B4%7D%7Dx%2B%5Ccfrac%7B9%7D%7B4%7D%20%5C%5C%5C%5C%5B-0.35em%5D%20%5Crule%7B34em%7D%7B0.25pt%7D%5C%5C%5C%5C%20%5Cstackrel%7B%5Ctextit%7Bperpendicular%20lines%20have%20%5Cunderline%7Bnegative%20reciprocal%7D%20slopes%7D%7D%20%7B%5Cstackrel%7Bslope%7D%7B-%5Ccfrac%7B3%7D%7B4%7D%7D%5Cqquad%20%5Cqquad%20%5Cqquad%20%5Cstackrel%7Breciprocal%7D%7B-%5Ccfrac%7B4%7D%7B3%7D%7D%5Cqquad%20%5Cstackrel%7Bnegative~reciprocal%7D%7B%2B%5Ccfrac%7B4%7D%7B3%7D%7D%5Cimplies%20%5Ccfrac%7B4%7D%7B3%7D%7D%20)
so we're really looking for the equation of a line whose slope is 4/3 and runs through 8, -4.
1m =3.28 feet 1km= 1000m 1km=3 280 feet
120x3 280=393 600feet/h 1h=3 600sec
393 600/3 600=109.33feet/sec
