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3 years ago

Graph the system of equations.

1 answer:

8 0

You can soustracte the first or the second one

and y cancel

x = 6

now you put 6 in x
you can take the first or the second one

6 - y = -2
6 - y = -2
y = -8

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Find the points on the lemniscate where the tangent is horizontal. 8(x2 + y2)2 = 81(x2 − y2)

slamgirl [31]

The answers to this problem are:(±5√3/8,±5/8)Here is the solution:
Step 1: x2+y2=2516[2]x2+y2=2516[2]

Step 2: Substitute:
8(25/16)^2=25(x^2−y^2)
8(25/16)^2=25(x^2−y^2)
x^2−y^2=25/32.

Add [2] and [3]:
2x^2=75/32
x^2=75/74
x=±5√3/8
Substitute into [2]:
75/64+y^2=50/32
y^2=25/64
y=±5/8

3 0

4 years ago

Read 2 more answers

In this figure m<3=130 What is m<6 ?

Georgia [21]

Answer: $m\angle6=50\°$

Step-by-step explanation:

You can observe in the figure two parallel lines that are intersected by a transversal and several angles are formed.

The angles m∠3 and m∠6 are located inside the parallel lines and on one side of the transversal, this angles are known as "Consecutive interior angles" and they are supplementary (which means that they add up 180°).

Therefore, you know that:

$m\angle3+m\angle6=180\°$

So you can substitute m∠3=130° and solve for m∠6. Then you get:

$130\°+m\angle6=180\°\\m\angle6=180\°-130\°\\m\angle6=50\°$

4 0

3 years ago

Read 2 more answers

√44 is between what 2 numbers? A.4 and 5 B.5 and 6 C.6 and 7 D.7 and 8

Gemiola [76]

I believe the answer is C.

6 0

3 years ago

Please help me with these, oh sweet jesus

Lelechka [254]

Answer:

77. $\cot^{6} x = \cot^{4} x \csc^{2}x - \cot^{4} x$ Proved

78. $\sec^{4}x \tan^{2} x = \sec^{2}x [\tan^{2}x + \tan^{4}x ]$ Proved

79. $\cos^{3} x\sin^{2} x = [\sin^{2}x - \sin^{4}x] \cos x$ Proved.

80. $\sin^{4}x - \cos^{4}x = 1 - 2\cos^{2}x + 2 \cos^{4} x$ Proved.

Step-by-step explanation:

77. Left hand side

= $\cot^{6} x$

= $\cot^{4} x \times \cot^{2} x$

= $\cot^{4}x [\csc^{2}x - 1]$

{Since we know, $\csc^{2} x - \cot^{2}x = 1$ }

= $\cot^{4} x \csc^{2}x - \cot^{4} x$

= Right hand side (Proved)

78. Left hand side

= $\sec^{4}x \tan^{2} x$

= $\sec^{2} x [1 + \tan^{2}x] \tan^{2} x$

{Since $\sec^{2}x - \tan^{2}x = 1$ }

= $\sec^{2}x [\tan^{2}x + \tan^{4}x ]$

= Right hand side (Proved)

79. Left hand side

= $\cos^{3} x\sin^{2} x$

= $\cos x[1 - \sin^{2} x] \sin^{2} x$

{Since $\sin^{2}x + \cos^{2} x = 1$ }

= $[\sin^{2}x - \sin^{4}x] \cos x$

= Right hand side

80. Left hand side

= $\sin^{4}x - \cos^{4}x$

= $[\sin^{2}x + \cos^{2}x]^{2} - 2\sin^{2} x \cos^{2}x$

{Since $\sin^{2}x + \cos^{2} x = 1$ }

= $1 - 2\cos^{2} x[1 - \cos^{2}x ]$

= $1 - 2\cos^{2}x + 2 \cos^{4} x$

= Right hand side. (Proved)

7 0

3 years ago

Verify identity: (sec(x)-csc(x))/(sec(x)+csc(x))=(tan(x)-1)/(tan(x)+1)

Nikitich [7]

So hmmm let's do the left-hand-side first

$\bf \cfrac{sec(x)-csc(x)}{sec(x)+csc(x)}\implies \cfrac{\frac{1}{cos(x)}-\frac{1}{sin(x)}}{\frac{1}{cos(x)}+\frac{1}{sin(x)}}\implies 
\cfrac{\frac{sin(x)-cos(x)}{cos(x)sin(x)}}{\frac{sin(x)+cos(x)}{cos(x)sin(x)}}
\\\\\\
\cfrac{sin(x)-cos(x)}{cos(x)sin(x)}\cdot \cfrac{cos(x)sin(x)}{sin(x)+cos(x)}\implies \boxed{\cfrac{sin(x)-cos(x)}{sin(x)+cos(x)}}$

now, let's do the right-hand-side then

$\bf \cfrac{tan(x)-1}{tan(x)+1}\implies \cfrac{\frac{sin(x)}{cos(x)}-1}{\frac{sin(x)}{cos(x)}+1}\implies \cfrac{\frac{sin(x)-cos(x)}{cos(x)}}{\frac{sin(x)+cos(x)}{cos(x)}}
\\\\\\
\cfrac{sin(x)-cos(x)}{cos(x)}\cdot \cfrac{cos(x)}{sin(x)+cos(x)}\implies \boxed{\cfrac{sin(x)-cos(x)}{sin(x)+cos(x)}}$

7 0

3 years ago