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3 years ago

6 ⅔ × (-5 ¹/¹⁶) ___answer

Mathematics

1 answer:

olya-2409 [2.1K]3 years ago

6 0

I got 24 as the answer to ur question

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System of equations 9x+8y=-19 ; 7x+9y=-12

klasskru [66]

Answer:

(-3,1)

Step-by-step explanation:

9x+8y=-19

7x+9y=-12

I'll assume the question is to find the solution to these equations. The solution will be the point (x,y) where the two lines intersect. The intersection is the one point that satisfies both equations (the smae value of (x,y) works in both.

We can either solve matematically of graph to find the intersection. I'll do both, and hope the answers are identical.

Matematically

Rearrange either equation to isolate one of the variables (either x or y). I'll take the second and isolate x:

7x+9y=-12

7x = -9y - 12

x = (-9y - 12)/7

Now use this definition of x in the other equation:

9x+8y=-19

9((-9y - 12)/7) + 8y = -19

(-81y - 108)/7 + 8y = -19

-81y - 108 + 56y = - 133

-25y = -25

y = 1

If y = 1, then:

9x+8y=-19

9x+8(1)=-19

9x = -27

x = -3

The solution is (-3,1)

Graphing

Graph both lines and look for the intersection. The attached graph shows the lines cross at (-3,1).

The solution, bu both approachjes, is (-3,1)

6 0

2 years ago

You are given the polar curve r = e^θ

galina1969 [7]

The tangent to $r(\theta)=e^\theta$ has slope $\frac{\mathrm dy}{\mathrm dx}$ , where

$\begin{cases}x(\theta)=r(\theta)\cos\theta\\y(\theta)=r(\theta)=\sin\theta\end{cases}$

By the chain rule, we have

$\dfrac{\mathrm dy}{\mathrm dx}=\dfrac{\frac{\mathrm dy}{\mathrm d\theta}}{\frac{\mathrm dx}{\mathrm d\theta}}$

and by the product rule,

$\dfrac{\mathrm dx}{\mathrm d\theta}=\dfrac{\mathrm dr}{\mathrm d\theta}\cos\theta-r(\theta)\sin\theta$

$\dfrac{\mathrm dy}{\mathrm d\theta}=\dfrac{\mathrm dr}{\mathrm d\theta}\sin\theta+r(\theta)\cos\theta$

so that with $\frac{\mathrm dr}{\mathrm d\theta}=e^\theta$ , we get

$\dfrac{\mathrm dy}{\mathrm dx}=\dfrac{e^\theta\sin\theta+e^\theta\cos\theta}{e^\theta\cos\theta-e^\theta\sin\theta}=\dfrac{\sin\theta+\cos\theta}{\cos\theta-\sin\theta}=-\dfrac{1+\sin(2\theta)}{\cos(2\theta)}$

The tangent line is horizontal when the slope is 0; this happens for

$-\dfrac{1+\sin(2\theta)}{\cos(2\theta)}=0\implies\sin(2\theta)=-1\implies2\theta=-\dfrac\pi2+2n\pi\implies\theta=-\dfrac\pi4+n\pi$

where $n$ is any integer. In the interval $0\le\theta\le2\pi$ , this happens for $n=1,2$ , or

$\theta=\dfrac{3\pi}4\text{ and }\theta=\dfrac{7\pi}4$

i.e at the points

$(r,\theta)=\left(e^{3\pi/4},\dfrac{3\pi}4\right)$

and

$(r,\theta)=\left(e^{7\pi/4},\dfrac{7\pi}4\right)$

8 0

3 years ago

Please help me guysthank you this question is really hard

disa [49]

The answer is 18
I hope I helped! ♥

7 0

3 years ago

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Helppp asapppp!!!!right Answers please

evablogger [386]

Translation is 4 units left and 2 units down

6 0

3 years ago

How do I put the slope and y-intercept on the graph

mestny [16]

So as you can see, -3 is the y-intercept. You would make a point on (-3,0) first. Then from the point, you would go up 1 and go right 2 because the slope is a positive 1/2 slope.

7 0

3 years ago

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6 ⅔ × (-5 ¹/¹⁶) ___answer​

6 ⅔ × (-5 ¹/¹⁶) ___answer