All categories

3 years ago

I neeeeeeeeeddddd helpppppppp

Mathematics

1 answer:

aksik [14]3 years ago

7 0

5/8 paper chains used more than 2 containers but less than 2 3/4 containers of glitter.

You might be interested in

what is 2+200 plz help btw who ever helps gets 21 pts and it is so easy i know the anwser i am just a bit bored so trying to giv

nlexa [21]

Its 202 becasue you just add two more so yeah

5 0

3 years ago

Read 2 more answers

1/5,0,-1/5,what two answers comes next?

jenyasd209 [6]

The next 2 answers are: -2/5 and -3/5

4 0

3 years ago

X/7+x-2/14=1/7<br> Please answer x as a fraction!

V125BC [204]

Answer:

x = $\frac{1}{4}$

Step-by-step explanation:

Given

$\frac{x}{7}$ + x - $\frac{2}{14}$ = $\frac{1}{7}$

Multiply through by 14 to clear the fractions

2x + 14x - 2 = 2 ( add 2 to both sides )

16x = 4 ( divide both sides by 16 )

x = $\frac{4}{16}$ = $\frac{1}{4}$

4 0

3 years ago

A box contains 144 marbles, which are either red or blue. if there are 3 times as many red marbles as there are blue marbles in

kicyunya [14]

Given:
X: The blue marbles.
3X: The red marbles.
A box contains 144 marbles.
SOLVING:
X + 3X = 144
4X = 144
X = $\frac{144}{4}$
X = 36 ===> The blue marbles.

If there are 3 times as many red marbles as there are blue marbles:
3(36) = 108 ===> The red marbles. SOLUTION

TESTING:
$36+108=144$
$\boxed{144}=\boxed{144}\quad\checkmark\ The\ value\ is\ correct.$
GOOD LUCK...!!!

3 0

3 years ago

What situations would solve by graphing be your preferred choice? Give an example.

Damm [24]

Answer:

1) The solve by graphing will the preferred choice when the equation is complex to be easily solved by the other means

Example;

y = x⁵ + 4·x⁴ + 3·x³ + 2·x² + x + 3

2) Solving by substitution is suitable where we have two or more variables in two or more (equal number) of equations

2x + 6y = 16

x + y = 6

We can substitute the value of x = 6 - y, into the first equation and solve from there

3) Solving an equation be Elimination, is suitable when there are two or more equations with coefficients of the form, 2·x + 6·y = 23 and x + y = 16

Multiplying the second equation by 2 and subtracting it from the first equation as follows

2·x + 6·y - 2×(x + y) = 23 - 2 × 16

2·x - 2·x + 6·y - 2·y = 23 - 32

0 + 4·y = -9

4) An example of a linear system that can be solved by all three methods is given as follows;

2·x + 6·y = 23

x + y = 16

Step-by-step explanation:

4 0

3 years ago