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3 years ago

3 + 1.2 x > 13.50 can someone please help me

Mathematics

1 answer:

saveliy_v [14]3 years ago

8 0

Answer:

The answer is x > 8.75.

Step-by-step explanation:

1) Subtract 3 from both sides.

$1.2x > 13.50 - 3$

2) Simplify 13.50 - 3 to 10.5.

$1.2x > 10.5$

3) Divide both sides by 1.2.

$x > \frac{10.5}{1.2}$

4) Simplify 10.5/1.2 to 8.75.

$x > 8.75$

Thus, the answer is x > 8.75.

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Which of the following expressions are equivalent to (9 7/8 + 2 4/5)-1/2

nikklg [1K]

Answer:A and B. Step-by-step explanation: C) 9 7/8 - (- 2 4/5) - (- 1/2) = = 9 7/8 + 2 4/5 + 1/2 (F). D) - (- 9 7/8 + 2 4/5) - 1/2) =.

I hope this helped^.^

May i have brainliest? ^.^

Have a great day buddy! ~hailey

7 0

2 years ago

Read 2 more answers

If f(3) = 24, what is f-1 (24)?

True [87]

Answer:

f=25

Step-by-step explanation:

24+1=25

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2 years ago

Find the maximum and minimum values of the function f(x,y)=2x2+3y2−4x−5 on the domain x2+y2≤100. The maximum value of f(x,y) is:

ryzh [129]

First find the critical points of f :

$f(x,y)=2x^2+3y^2-4x-5=2(x-1)^2+3y^2-7$

$\dfrac{\partial f}{\partial x}=2(x-1)=0\implies x=1$

$\dfrac{\partial f}{\partial y}=6y=0\implies y=0$

so the point (1, 0) is the only critical point, at which we have

$f(1,0)=-7$

Next check for critical points along the boundary, which can be found by converting to polar coordinates:

$f(x,y)=f(10\cos t,10\sin t)=g(t)=295-40\cos t-100\cos^2t$

Find the critical points of g :

$\dfrac{\mathrm dg}{\mathrm dt}=40\sin t+200\sin t\cos t=40\sin t(1+5\cos t)=0$

$\implies\sin t=0\text{ OR }1+5\cos t=0$

$\implies t=n\pi\text{ OR }t=\cos^{-1}\left(-\dfrac15\right)+2n\pi\text{ OR }t=-\cos^{-1}\left(-\dfrac15\right)+2n\pi$

where n is any integer. We get 4 critical points in the interval [0, 2π) at

$t=0\implies f(10,0)=155$

$t=\cos^{-1}\left(-\dfrac15\right)\implies f(-2,4\sqrt6)=299$

$t=\pi\implies f(-10,0)=235$

$t=2\pi-\cos^{-1}\left(-\dfrac15\right)\implies f(-2,-4\sqrt6)=299$

So f has a minimum of -7 and a maximum of 299.

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3 years ago

Find the derivative of g(y)=(y-4)*(2y+y^2)

sasho [114]

Answer:

$g'(y)=3y^2-4y-8$

Step-by-step explanation:

start by foiling out the given function

$g(y)=(y-4)(2y+y^2)\\=2y^2+y^3-8y-4y^2\\=y^3-2y^2-8y$

next, use the power rule to find the derivative

power rule: To use the power rule, multiply the variable's exponent n, by its coefficient a, then subtract 1 from the exponent. If there's no coefficient (the coefficient is 1), then the exponent will become the new coefficient.

$g'(y)=3y^2-4y-8$

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3 years ago