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3 years ago

Need urgent help, will give brainliest.

Mathematics

2 answers:

prisoha [69]3 years ago

8 0

Answer:

I would say D

Step-by-step explanation:

With the given information I can make 2 expressions:

g>b

b≥1

They do not tell us anything about 2b or what the sum of g+b would be equal to

pashok25 [27]3 years ago

7 0

Answer:

Option D can be the answer.

Step-by-step explanation:

A bag has more green balls than blue balls, and there is at least one blue ball. We can say that -

b is greater than or equal to 1

and g is always greater than b.

With respect too g, we can say b is:

g-1 or g-2, or generally g-n.

The equation thus becomes: 2(g-n) = g-n + g

which is equal to: 2g - 2n = 2g - n.

From this equation we can conclude that the left side is always smaller than the right side so:

2g - 2n < 2g - n.

Which becomes: 2b < g + b.

Hence, option D can be the answer.

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3 years ago

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Alenkinab [10]

Answer:

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Step-by-step explanation:

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Double negative = Positive:

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3 years ago

HELPPP!!!!!!! The graph of F(x) shown below resembles the graph of G(x) = x^2 but it has been stretched somewhat and shifted. Wh

Fittoniya [83]

Answer:

Option C.

Step-by-step explanation:

Equation of the parent function graphed in the figure is,

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By reflecting the parent function over the x-axis,

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By shifting G'(x) by 2 units down over the y-axis,

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Therefore, Option C will be the answer.

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3 years ago

Find the tangent plane to the given surface of f(x,y)=6- 6/5 x-y at the point (5, -1, 1). Make sure tat your final answer for th

HACTEHA [7]

Answer:

Required equation of tangent plane is $z=\frac{6}{5}(x-5y-11)$ .

Step-by-step explanation:

Given surface function is,

$f(x,y)=6-\frac{6}{5}(x-y)$

To find tangent plane at the point (5,-1,1).

We know equation of tangent plane at the point $(x_0,y_0,z_0)[/tex] is,

$z=f(x_0,y_0)+f_x(x_0,y_0)(x-x_0)+f_y(x_0,y_0)(y-y_0)\hfill (1)$

So that,

$f(x_0,y_0)=6-\frac{6}{5}(5+1)=-\frac{6}{5}$

$f_x=-\frac{6}{5}y\implies f_x(5,-1,1)=\frac{6}{5}$

$f_y=-\frac{6}{5}x\implies f_y(5,-1,1)=-6$

Substitute all these values in (1) we get,

$z=\frac{6}{5}(x-5)-6(y+1)-\frac{6}{5}$

$\therefore z=\frac{6}{5}(x-5y-11)$

Which is the required euation of tangent plane.

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3 years ago

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Mazyrski [523]

Answer:

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