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2 years ago

Need help with math question ??

Mathematics

1 answer:

AVprozaik [17]2 years ago

4 0

Answer:

g(x)=(x+7)^2-2

Step-by-step explanation:

The number inside the parenthesis moves the parabola left and right. It didn't move in any of those directions, so 7 stays the same. Since it moved 2 units, 2 goes outside the parenthesis. Since it moved down, you subtract the 2.

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Factorise a.x3+4.a.x

Contact [7]

Answer:

${ax}^{3} + 4ax \\$

factorise out a and x :

${ \boxed{answer{= ax( {x}^{2} + 4) }}}\\ but \: farther \: more : \\ = ax( {x}^{2} + {2}^{2} )$

but from general factorization:

${ \boxed{( {a}^{2} + {b}^{2}) = {(a + b)}^{2} - 2ab }}$

a » x

b » 2

therefore:

$= ax \{ {(x + 2)}^{2} - 2(x)(2) \} \\ \\ = { \boxed{ \boxed{ax( {x + 2)}^{2} - 4x }}}$

4 0

2 years ago

32 lb 12 ounces equals in pounds

Radda [10]

Answer:

32 3/4 lb

Step-by-step explanation:

32 lb 12 oz = ? lb

Rewrite 12 oz as (12/16) lb.

Then:

32 lb 12 oz = 32 lb + (3/4) lb, or 32 3/4 lb

4 0

2 years ago

Elena and Noah used different methods to compute 0.23 x 1.5. Both calculations were correct. Analyze the 2 methods.

mafiozo [28]

Answer:

i have this same question. im sorry but i don't get it either. sorry..

Step-by-step explanation:

5 0

2 years ago

The gradiant y=2x³-5x+1 at point (2,7)

olga2289 [7]

Given:

The equation of the curve is:

$y=2x^3-5x+1$

To find:

The gradient (slope) of the given curve at point (2,7).

Solution:

We have,

$y=2x^3-5x+1$

Differentiate the given equation with respect to x.

$y=2(3x^2)-5(1)+(0)$

$y'=6x^2-5$

Now we need to find the value of this derivative at (2,7).

$y'_{(2,7)}=6(2)^2-5$

$y'_{(2,7)}=6(4)-5$

$y'_{(2,7)}=24-5$

$y'_{(2,7)}=19$

Therefore, the gradient (slope) of the given curve at point (2,7) is 19.

8 0

2 years ago

PLEASE HELP ASAP!!! CORRECT ANSWER ONLY PLEASE!!!

Mrrafil [7]

The answer is D, (x + 1) (x + 2) (x - 3)

7 0

2 years ago

Read 2 more answers