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2 years ago

-6n+4= -5 solve for n

Mathematics

1 answer:

ElenaW [278]2 years ago

5 0

Answer:

Answer is below

Step-by-step explanation:

Subtract 4 from both sides.

$-6n+4(-4)=-5(-4)\\-6n=-9\\$

Divide both sides by -6.

$\frac{-6n}{-6}=\frac{-9}{-6} \\n=\frac{3}{2} Decimal 1.5$

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Y= -(x-1)^2+1 please helppppp

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Answer:x:

y=-5/4

Step-by-step explanation:

Vertex (1,-1)

focus(1,-3/4)

Axis of symmetry(X=1)

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2 years ago

Read 2 more answers

Find the global maximum and global minimum values (if they exist) of x 2 + y 2 in the region x + y = 1. If there is no global ma

Grace [21]

Given that $x+y=1$ , we have $y=1-x$ , so that

$f(x,y)=x^2+y^2\iff g(x)=x^2+(1-x)^2$

Take the derivative and find the critical points of $g$ :

$g'(x)=2x-2(1-x)=4x-2=0\implies x=\dfrac12$

Take the second derivative and evaluate it at the critical point:

$g''(x)=4>0$

Since $g''$ is positive for all $x$ , the critical point is a minimum.

At the critical point, we get the minimum value $g\left(\frac12\right)=f\left(\frac12,\frac12\right)=\frac12$ .

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2 years ago

Evaluate the following integral using trigonometric substitution

serg [7]

Answer:

The result of the integral is:

$\arcsin{(\frac{x}{3})} + C$

Step-by-step explanation:

We are given the following integral:

$\int \frac{dx}{\sqrt{9-x^2}}$

Trigonometric substitution:

We have the term in the following format: $a^2 - x^2$ , in which a = 3.

In this case, the substitution is given by:

$x = a\sin{\theta}$

$dx = a\cos{\theta}d\theta$

In this question:

$a = 3$

$x = 3\sin{\theta}$

$dx = 3\cos{\theta}d\theta$

$\int \frac{3\cos{\theta}d\theta}{\sqrt{9-(3\sin{\theta})^2}} = \int \frac{3\cos{\theta}d\theta}{\sqrt{9 - 9\sin^{2}{\theta}}} = \int \frac{3\cos{\theta}d\theta}{\sqrt{9(1 - \sin^{\theta})}}$

We have the following trigonometric identity:

$\sin^{2}{\theta} + \cos^{2}{\theta} = 1$

$1 - \sin^{2}{\theta} = \cos^{2}{\theta}$

Replacing into the integral:

$\int \frac{3\cos{\theta}d\theta}{\sqrt{9(1 - \sin^{2}{\theta})}} = \int{\frac{3\cos{\theta}d\theta}{\sqrt{9\cos^{2}{\theta}}} = \int \frac{3\cos{\theta}d\theta}{3\cos{\theta}} = \int d\theta = \theta + C$

Coming back to x:

We have that:

$x = 3\sin{\theta}$

$\sin{\theta} = \frac{x}{3}$

Applying the arcsine(inverse sine) function to both sides, we get that:

$\theta = \arcsin{(\frac{x}{3})}$

The result of the integral is:

$\arcsin{(\frac{x}{3})} + C$

8 0

2 years ago

6x squared plus 12x plus 13 equals 2x squared plus 4

Andrews [41]

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2 years ago

How many hundreths equal 6 tenths

Tresset [83]

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3 years ago