A number y increased by 5 is at most 28

Mathematics

2 answers:

White raven [17]3 years ago

5 0

The answer to your question is this:

Iteru [2.4K]3 years ago

4 0

$y + 5\leq 28\\y\leq 23$

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If 3x^2 + y^2 = 7 then evaluate d^2y/dx^2 when x = 1 and y = 2. Round your answer to 2 decimal places. Use the hyphen symbol, -,

S_A_V [24]

Taking $y=y(x)$ and differentiating both sides with respect to $x$ yields

$\dfrac{\mathrm d}{\mathrm dx}\bigg[3x^2+y^2\bigg]=\dfrac{\mathrm d}{\mathrm dx}\bigg[7\bigg]\implies 6x+2y\dfrac{\mathrm dy}{\mathrm dx}=0$

Solving for the first derivative, we have

$\dfrac{\mathrm dy}{\mathrm dx}=-\dfrac{3x}y$

Differentiating again gives

$\dfrac{\mathrm d}{\mathrm dx}\bigg[6x+2y\dfrac{\mathrm dy}{\mathrm dx}\bigg]=\dfrac{\mathrm d}{\mathrm dx}\bigg[0\bigg]\implies 6+2\left(\dfrac{\mathrm dy}{\mathrm dx}\right)^2+2y\dfrac{\mathrm d^2y}{\mathrm dx^2}=0$

Solving for the second derivative, we have

$\dfrac{\mathrm d^2y}{\mathrm dx^2}=-\dfrac{3+\left(\frac{\mathrm dy}{\mathrm dx}\right)^2}y=-\dfrac{3+\frac{9x^2}{y^2}}y=-\dfrac{3y^2+9x^2}{y^3}$

Now, when $x=1$ and $y=2$ , we have

$\dfrac{\mathrm d^2y}{\mathrm dx^2}\bigg|_{x=1,y=2}=-\dfrac{3\cdot2^2+9\cdot1^2}{2^3}=\dfrac{21}8\approx2.63$

3 0

3 years ago

Please help this is timed

elena-14-01-66 [18.8K]

Answer:

option d is correct

de=9 and ef=15

6 0

3 years ago

Read 2 more answers

Carlos has 38 apples.He puts 8 into each bag.If he fills many bags as possible ,how many apples are left?

leva [86]