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3 years ago

How do you find x and y in a word problem

1 answer:

3 0

You replace one letter with 0 so you can cancel it out. So for example, when you are solving for y, you put x as 0 so then you can find the y. If the problem is x+y=24. You put x as 0 and you solve for y.

You might be interested in

Determine if each equation is a linear equation.

Lady_Fox [76]

Answer:

Linear-1,2,3,6

Nin-Linear-4,5

Step-by-step explanation:

A linear equation is an equation in which the highest power of the variable is 1.

Out of the given equations, the following are linear.

1) $5x_1 + x_2 - 7x_3 = 4$

2) $5x_1 + 4.6x_2 + 12x_3 = -5$

3) $3x_1 - x_3 = -5x_2 + 7$

6) $6x_1 - 4x_2 + x_3 = 0$

These equations on the other hand are nonlinear.

4) $3x_1x_2 + x_3 = 5$

5) $4x_1^3 + x_2 = 5$

No 4 contains a term which is the product of two variables therefore it is not linear.

No 5 contains a term whose highest power is 3. Such an equation is a Cubic Equation.

4 0

3 years ago

Sec squared 55 - tan squared 55

sergejj [24]

Answer:

sec squared 55 – tan squared 55 = 1

Explanation:

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 =

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