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Romashka-Z-Leto [24]
3 years ago
8

3(x-4)=-5 solve for x

Mathematics
2 answers:
tiny-mole [99]3 years ago
5 0

Answer:

x = 7/3

Step-by-step explanation:

3(x-4)=-5

Distribute

3x - 12 = -5

Add 12 to each side

3x -12+12 = -5+12

3x = 7

Divide by 3

3x/3 = 7/3

x = 7/3

Harlamova29_29 [7]3 years ago
3 0

Answer:

x = 7/3

Step-by-step explanation:

3(x-4)=-5

3x -12= -5

3x = 7

x =7/3

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Step-by-step explanation:

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lisabon 2012 [21]

Answer:

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Sveta_85 [38]

I'll do Problem 8 to get you started

a = 4 and c = 7 are the two given sides

Use these values in the pythagorean theorem to find side b

a^2 + b^2 = c^2\\\\4^2 + b^2 = 7^2\\\\16 + b^2 = 49\\\\b^2 = 49 - 16\\\\b^2 = 33\\\\b = \sqrt{33}\\\\

With respect to reference angle A, we have:

  • opposite side = a = 4
  • adjacent side = b = \sqrt{33}
  • hypotenuse = c = 7

Now let's compute the 6 trig ratios for the angle A.

We'll start with the sine ratio which is opposite over hypotenuse.

\sin(\text{angle}) = \frac{\text{opposite}}{\text{hypotenuse}}\\\\\sin(A) = \frac{a}{c}\\\\\sin(A) = \frac{4}{7}\\\\

Then cosine which is adjacent over hypotenuse

\cos(\text{angle}) = \frac{\text{adjacent}}{\text{hypotenuse}}\\\\\cos(A) = \frac{b}{c}\\\\\cos(A) = \frac{\sqrt{33}}{7}\\\\

Tangent is the ratio of opposite over adjacent

\tan(\text{angle}) = \frac{\text{opposite}}{\text{adjacent}}\\\\\tan(A) = \frac{a}{b}\\\\\tan(A) = \frac{4}{\sqrt{33}}\\\\\tan(A) = \frac{4\sqrt{33}}{\sqrt{33}*\sqrt{33}}\\\\\tan(A) = \frac{4\sqrt{33}}{(\sqrt{33})^2}\\\\\tan(A) = \frac{4\sqrt{33}}{33}\\\\

Rationalizing the denominator may be optional, so I would ask your teacher for clarification.

So far we've taken care of 3 trig functions. The remaining 3 are reciprocals of the ones mentioned so far.

  • cosecant, abbreviated as csc, is the reciprocal of sine
  • secant, abbreviated as sec, is the reciprocal of cosine
  • cotangent, abbreviated as cot, is the reciprocal of tangent

So we'll flip the fraction of each like so:

\csc(\text{angle}) = \frac{\text{hypotenuse}}{\text{opposite}} \ \text{ ... reciprocal of sine}\\\\\csc(A) = \frac{c}{a}\\\\\csc(A) = \frac{7}{4}\\\\\sec(\text{angle}) = \frac{\text{hypotenuse}}{\text{adjacent}} \ \text{ ... reciprocal of cosine}\\\\\sec(A) = \frac{c}{b}\\\\\sec(A) = \frac{7}{\sqrt{33}} = \frac{7\sqrt{33}}{33}\\\\\cot(\text{angle}) = \frac{\text{adjacent}}{\text{opposite}} \ \text{  ... reciprocal of tangent}\\\\\cot(A) = \frac{b}{a}\\\\\cot(A) = \frac{\sqrt{33}}{4}\\\\

------------------------------------------------------

Summary:

The missing side is b = \sqrt{33}

The 6 trig functions have these results

\sin(A) = \frac{4}{7}\\\\\cos(A) = \frac{\sqrt{33}}{7}\\\\\tan(A) = \frac{4}{\sqrt{33}} = \frac{4\sqrt{33}}{33}\\\\\csc(A) = \frac{7}{4}\\\\\sec(A) = \frac{7}{\sqrt{33}} = \frac{7\sqrt{33}}{33}\\\\\cot(A) = \frac{\sqrt{33}}{4}\\\\

Rationalizing the denominator may be optional, but I would ask your teacher to be sure.

7 0
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two verticies of paralellogram ABCD are A(-6,-1) and B(-5,3). The intersection of the diagonals, E, are (-2, 1/2). Determine the
masya89 [10]

Answer:

D(2,2)

Step-by-step explanation:

The diagonals of a parallelogram gram bisects each other.

Therefore E is the midpoint of AD.

Let the coordinates of D be (a,b).

By the midpoint rule:

( \frac{x_1+x_2}{2} ,\frac{y_1+y_2}{2}) = ( - 2, \frac{1}{2} )

This implies that:

( \frac{ - 6+a}{2} ,\frac{ - 1+b}{2}) = ( - 2, \frac{1}{2} )

This implies that:

( \frac{ - 6+a}{2} =  - 2 ,\frac{ - 1+b}{2} =  \frac{1}{2} )

( - 6+a =- 4 , - 1+b = 1 )  \\ ( a =- 4 + 6 , b = 1 + 1 )  \\ ( a =2 , b =2 )

7 0
3 years ago
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