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Ganezh [65]
3 years ago
15

What is the equation for the graph in slope-intercept?

Mathematics
1 answer:
Julli [10]3 years ago
6 0

Answer:

y=7/4x+7

Step-by-step explanation:

You might be interested in
-12 41/24 in simplest form.
Temka [501]

Answer:

-13 17/24

Step-by-step explanation:

-12 41/24 is equivalent to -(-12 41/24), or - (12 + 24/24 + 17/24)

Combining 12 and 24/24, we get 13, and thus have:  -(13 17/24), or

-13 17/24

7 0
3 years ago
Read 2 more answers
Is it possible for two different numbers, when squared, to give the same result?
lawyer [7]

Actually, yes, it is possible for two different numbers to give the same result when squared.

In my last answer, I wrote that it wasn't, but I realize now where my mistake was made.

When a number like positive 4 is squared, the answer is 16. When a number like negative 4 is squared, the answer is also 16. I think that the only time when two different squared numbers have the same result is when they are the same number but have a different positive/negative sign.

I hope this helps.

7 0
3 years ago
Solve and graph the solution set<br> 4x + 5+x&gt; 2 + 2x-3
qaws [65]

Answer:

I think it's true...

Step-by-step explanation:

3x>-6

8 0
3 years ago
1. if csc β = 7/3 and cot β = - 2√10 / 3, Find sec β
slava [35]

Step-by-step explanation:

1.

\tan \beta  =  \frac{1}{ \cot \beta }  =  -  \frac{3}{2 \sqrt{10} }  =  -  \frac{3 \sqrt{10} }{20}

\csc \beta  \tan \beta  =  \frac{1}{ \cos \beta  }  =  \sec \beta

Therefore,

\sec \beta  = ( \frac{7}{3} )( -  \frac{3 \sqrt{10} }{20} ) =  -  \frac{7 \sqrt{10} }{20}

2.

\csc y =  \frac{1}{ \sin y}  =  -  \frac{ \sqrt{6} }{2}

=  >  \sin y =  -  \frac{ \sqrt{6} }{3}

Use the identity

\cos y =   \sqrt{1 -  \sin ^{2} y}    \:  \:  \: \:  \:  \:  \:  \:  \:  \:  \:  \\ =  \sqrt{1 -  {( -  \frac{ \sqrt{6} }{3}) }^{2} }  =  -  \frac{ \sqrt{3} }{3}

We chose the negative value of the cosine because of the condition where cot y > 0. Otherwise, choosing the positive root will yield a negative cotangent value. Now that we know the sine and cosine of y, we can now solve for the tangent:

\tan \beta  =  \frac{ \sin y}{ \cos y} =( -  \frac{ \sqrt{6} }{3} )( -  \frac{3}{ \sqrt{3} } ) =  \sqrt{2}

3. Recall that sec x = 1/cos x, therefore cos x = 5/6. Solving for sin x,

\sin x =   \sqrt{1 -  \cos ^{2} x} =  \sqrt{ \frac{11}{6} }

Solving for tan x:

\tan x =  \frac{ \sin x}{ \cos x}  =  (\frac{ \sqrt{11} }{ \sqrt{6} } )( \frac{6}{5} ) =  \frac{ \sqrt{66} }{5}

5 0
3 years ago
Need help solving this differentiation problem please. Thank you.
DIA [1.3K]
First convert the terms to fractional exponents

u  =  t^2/3  - 3t^3/2

differentiating

u'  =  2/3 t^ (2/3 - 1) - 3* 3/2  t^(3/2 - 1)

     =  2/3 t ^(-1/3) - 9/2 t ^(1/2)

     =  2 / (3∛t) -  9 √ t / 2   in radical form


7 0
3 years ago
Read 2 more answers
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