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rewona [7]
3 years ago
13

Solve system of equations 14x + y = -4 and y = 3x^2 - 11x - 4 

Mathematics
1 answer:
Keith_Richards [23]3 years ago
7 0
Hello here is a solution : 

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Simplify (x − 4)(3x2 − 6x + 2). 3x3 + 6x2 − 22x + 8 3x3 − 18x2 + 26x − 8 3x3 − 18x2 − 22x − 8 3x3 + 6x2 + 22x + 8
Eduardwww [97]

Step-by-step explanation:

you know the answerv noow

7 0
4 years ago
I need help it is due tonight
ddd [48]

Answer:

2) x= 1/2

3) x= 1/2

4) x= 1/2

5) x= 1/10

6) x=2/3

7) x= 3/4

8) x= 1/3

9) x= 5/9

Also use Math_way ^_^



5 0
3 years ago
G(x) = -10x – 8<br> a function
wlad13 [49]

Answer:

yes that is a function and not an equation

7 0
3 years ago
A train travels a distance of 60 km at uniform speed. If the speed of the train was reduced by 10 kmh-1, the time taken to trave
tester [92]

Answer:

<u>Initial</u><u> </u><u>speed</u><u> </u><u>is</u><u> </u><u>3</u><u>2</u><u> </u><u>m</u><u>/</u><u>s</u>

At uniform speed, acceleration is 0, (a = 0).

When speed reduced, (v - u) = 2.78 ms-¹, t = 1800 sec, s = 60 ,000 metres.

From first equation of motion:

{ \boxed{ \bf{v = u + at}}} \\  { \tt{(v - u) = at}}

substitute:

{ \tt{2.78 = (a \times 1800)}} \\ { \tt{acceleration = 0.0015 \:  {ms}^{ - 2} }}

from second equation of motion:

{ \boxed{ \bf{s = ut +  \frac{1}{2} a {t}^{2} }}}

substitute:

{ \tt{60000 = 1800u + ( \frac{1}{2} \times 0.0015 \times  {1800}^{2})  }} \\ { \tt{1800u = 57570}} \\ { \tt{u = 32 \: m {s}^{ - 1} }}

5 0
3 years ago
Enlarge the triangle by scale factor 0.5 using 3,1 as the centre of enlargement
NISA [10]

Answer:

the points are :

(0,-1)

(0,0)

(-1,3)

(If you are using MathsWatch) For this question the triangle gets smaller so i'm not sure why it says Enlarge the triangle.

6 0
3 years ago
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