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3 years ago

Is this equation a linear relation, and please solve the equation? -3x + 2y = 1

Mathematics

1 answer:

Liono4ka [1.6K]3 years ago

8 0

Answer:

Step-by-step explanation:

It is a linear equation and the gradient equation is given as:

y=mx+c

Where m is the gradient and c is the y intercept

-3x+2y=1

2y=1+3x

Dividing all through by 2

y=1/2 + 3x/2

Therefore the gradient of this linear equation is 3/2 and the y intercept is 1/2

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3 years ago

Violet made cookies for the bake sale. 7/10 of the cookies are chocolate chip. 3/4 of the remaining cookies are sugar. What frac

FromTheMoon [43]

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3 years ago

HELP WILL GIVE BRAINLIEST!!!!

iris [78.8K]

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Step-by-step explanation:

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3 years ago

Read 2 more answers

Sec squared 55 - tan squared 55

sergejj [24]

<u>Answer: </u>

sec squared 55 – tan squared 55 = 1

<u>Explanation:</u>

Given, sec square 55 – tan squared 55

We know that,

$\sec \Theta=\frac{\text {hypotenuse}}{\text {base}}$

And,

$\tan \theta=\frac{\text { perpendicular }}{\text { base }}$

where Ө is the angle

Substituting the values

$\left(\frac{\text {hypotenuse}}{\text {base}}\right)^{2}-\left(\frac{\text { perpendicular }}{\text {base}}\right)^{2}$

Solving,

$\frac{(\text {hypotenuse})^{2}-(\text {perpendicular})^{2}}{(\text {base}) *(\text {base})}$

According to Pythagoras theorem,

$\text { (hypotenuse) }^{2}-\text { (perpendicular) }^{2}=(\text { base })^{2}$

Putting this in the equation;

squared 55 - tan squared 55 =

$\frac{(\text {hypotenuse})^{2}-(\text {perpendicular})^{2}}{(\text {base}) *(\text {base})}=\frac{(\text {base})^{2}}{(\text {base}) *(\text {base})}=1$

Therefore, sec squared 55 – tan squared 55 = 1

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3 years ago

Please hurry for quiz

fenix001 [56]

The third answer is correct

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3 years ago