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Mathematics

2 answers:

Alex Ar [27]2 years ago

6 0

A^2 + 9 = 36 now solve subtract 9 from both sides to get a^2 = 27 then find the square root of 27 and that should be your answer so like 5.196 something

AysviL [449]2 years ago

5 0

You have to use the Pythagorean theorem to solve for the missing side. Here’s the work shown! Hope this helps!

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Matthew is making a sequence in which -3 is added to each successive term. The 6th term in his sequence is -1. What is the 1st t

Hatshy [7]

Answer:

The first term is 14.

Step-by-step explanation:

The general formula for this kind of sequence (arithmetic) is a(n) = a(1) + (n-1)c, where c is the common difference. Here, we have a(6) = -1 = a(1) + (6-1)(-3), or -1 +15 = a(1). The first term is 14.

3 0

2 years ago

Use chain rule to find dw/dt w=ln(x^2 + y^2 + z^2)^(1/2) , x=sint, y=cost, z=tant

son4ous [18]

The chain rule states that

$\dfrac{\mathrm dw}{\mathrm dt}=\dfrac{\partial w}{\partial x}\dfrac{\mathrm dx}{\mathrm dt}+\dfrac{\partial w}{\partial y}\dfrac{\mathrm dy}{\mathrm dt}+\dfrac{\partial w}{\partial z}\dfrac{\mathrm dz}{\mathrm dt}$

Since $\ln(x^2+y^2+z^2)^{1/2}=\dfrac12\ln(x^2+y^2+z^2)$ , you have

$\dfrac{\partial w}{\partial x}=\dfrac x{x^2+y^2+z^2}$
$\dfrac{\partial w}{\partial x}=\dfrac y{x^2+y^2+z^2}$
$\dfrac{\partial w}{\partial z}=\dfrac z{x^2+y^2+z^2}$

and

$\dfrac{\mathrm dx}{\mathrm dt}=\cos t$
$\dfrac{\mathrm dy}{\mathrm dt}=-\sin t$
$\dfrac{\mathrm dz}{\mathrm dt}=\sec^2t$

Also, since $x^2+y^2+z^2=\sin^2t+\cos^2t+\tan^2t=1+\tan^2t=\sec^2t$ , the derivative is

$\dfrac{\mathrm dw}{\mathrm dt}=\dfrac{\sin t\cos t}{\sec^2t}-\dfrac{\sin t\cos t}{\sec^2t}+\dfrac{\tan t\sec^2t}{\sec^2t}$
$\dfrac{\mathrm dw}{\mathrm dt}=\tan t$

4 0

2 years ago

PLS DO THIS 20 PTS AND BRAINLY

deff fn [24]

The table would be y = 6x

Divide the Y values by the X values and they all equal 6, so you multiply the X value by 6 to get y.

Look at the dots on the graph ( 1,5) (2,10) (3,15) (4,20)

Divide the Y value by the X and they all equal 5, soy = 5x

3 0

3 years ago

Read 2 more answers

Find the equation of all tangent lines having slope of -1 that are tangent to the curve y=(9)/(x+1)

fredd [130]

Answer:

$f(x)=\frac9{x+1}\\ f'(x)=-\frac9{(x+1)^2}\\ f'(x)=-1\ \iff\ -\frac9{(x+1)^2}=-1\ \to \ \frac9{(x+1)^2}=1\ \to \ (x+1)^2=9\\ |x+1|=3\ \to \ x+1=3\ \vee\ x+1=-3\\ x_1=2\ \vee\ x_2=-4\\ f(x_1)=f(2)=\frac9{2+1}=3\\ f(x_2)=f(-4)=\frac9{-4+1}=-3$