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bazaltina [42]
3 years ago
11

Pls do 4 and 5 for me <3

Mathematics
1 answer:
WINSTONCH [101]3 years ago
8 0

yuhhdhisjehe yup yup yup yup yup yup
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9
Gwar [14]

Answer:

It is A and I am positive its A

Step-by-step explanation:

6 0
3 years ago
The question is there
Katarina [22]

Answer:

3 cm

Step-by-step explanation:

Set up an equation that solves for the surface area, "112cm^{2}" :

10(2)+10(2)+2x+2x+10x+10x = 112

20+20+4x+20x = 112

24x+40=112

24x=72

x=3

So, 3 cm.

7 0
3 years ago
What is the standard form of the equation y=-5/2x-3 ?
butalik [34]

The standard form of an equation is the form where that equation has no fractions and is written in the form ax + by = c.


To get rid of the fractions in y = -5/2 x - 3, we need to multiply both sides by 2. That makes the equation become:


2y = -5x - 6


Now we need to get it into the form ax + by = c. We can do this by adding 5x to each side.


5x + 2y = -6 is our answer

6 0
3 years ago
Easy 6th grade math
Grace [21]

Answer:

18 in

Step-by-step explanation:

3 0
3 years ago
Read 2 more answers
Find a and b if the point p(6,0) and Q(3,2) lie on the graph of ax+ by=12
tiny-mole [99]

to get the equation of any straight line we only need two points off of it, hmmm let's use P and Q here and then let's set the equation in standard form, that is

standard form for a linear equation means

• all coefficients must be integers, no fractions

• only the constant on the right-hand-side

• all variables on the left-hand-side, sorted

• "x" must not have a negative coefficient

(\stackrel{x_1}{6}~,~\stackrel{y_1}{0})\qquad (\stackrel{x_2}{3}~,~\stackrel{y_2}{2}) ~\hfill \stackrel{slope}{m}\implies \cfrac{\stackrel{rise} {\stackrel{y_2}{2}-\stackrel{y1}{0}}}{\underset{run} {\underset{x_2}{3}-\underset{x_1}{6}}}\implies \cfrac{2}{-3}\implies -\cfrac{2}{3} \\\\\\ \begin{array}{|c|ll} \cline{1-1} \textit{point-slope form}\\ \cline{1-1} \\ y-y_1=m(x-x_1) \\\\ \cline{1-1} \end{array}\implies y-\stackrel{y_1}{0}=\stackrel{m}{-\cfrac{2}{3}}(x-\stackrel{x_1}{6})

\stackrel{\textit{multiplying both sides by }\stackrel{LCD}{3}}{3(y-0)~~ = ~~3\left( -\cfrac{2}{3}(x-6) \right)}\implies 3y=-2(x-6) \\\\\\ 3y=-2x+12 \implies \stackrel{a}{2} x+\stackrel{b}{3} y=12

5 0
2 years ago
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