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2 years ago

Can anyone help me figure out the 2 square 3 is? and show step by step. Thank you

Mathematics

2 answers:

taurus [48]2 years ago

8 0

like 3 squared? or 2 cubed?

BlackZzzverrR [31]2 years ago

8 0

You mean something like 2^3 or 3^2?

If you mean 2^3, it's pretty much multiplying 2 three times. It would look like:

2 x 2 x 2, which would equal 8. You start off with 2 x 2 = 4, then you take the 4 and multiply it by 2 so it will equal 8.

If you mean 3^2, it's the same thing, except you're multiplying 3 two times. It would look like this:

3 x 3, which equals 9.

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2 Here are two equations:

MatroZZZ [7]

a. (3,4) is only the solution to Equation 1.

(4, 2.5) is the solution to both equations

(5,5) is the solution to Equation 2

(3,2) is not the solution to any equation.

b. No, it is not possible to have more than one (x,y) pair that is solution to both equations

Step-by-step explanation:

a. Decide whether each neither of the equations,

i (3,4)

ii. (4,2.5)

ill. (5,5)

iv. (3,2)

To decide whether each point is solution to equations or not we will put the point in the equations

Equations are:

Equation 1: 6x + 4y = 34

Equation 2: 5x – 2y = 15

i (3,4)

Putting in Equation 1:

$6(3) + 4(4) = 34\\18+16=34\\34=34\\$

Putting in Equation 2:

$5(3) - 2(4) = 15\\15-8 = 15\\7\neq 15$

ii. (4,2.5)

Putting in Equation 1:

$6(4) + 4(2.5) = 34\\24+10=34\\34=34\\$

Putting in Equation 2:

$5(4) - 2(2.5) = 15\\20-5 = 15\\15=15$

ill. (5,5)

$6(5) + 4(5) = 34\\30+20=34\\50\neq 34$

Putting in Equation 2:

$5(5) - 2(5) = 15\\25-10 = 15\\15=15$

iv. (3,2)

$6(3) + 4(2) = 34\\18+8=34\\26\neq 34$

Putting in Equation 2:

$5(3) - 2(2) = 15\\15-4 = 15\\11\neq 15$

Hence,

(3,4) is only the solution to Equation 1.

(4, 2.5) is the solution to both equations

(5,5) is the solution to Equation 2

(3,2) is not the solution to any equation.

b. Is it possible to have more than one (x, y) pair that is a solution to both

equations?

The simultaneous linear equations' solution is the point on which the lines intersect. Two lines can intersect only on one point. So a linear system cannot have more than one point as a solution

So,

a. (3,4) is only the solution to Equation 1.

(4, 2.5) is the solution to both equations

(5,5) is the solution to Equation 2

(3,2) is not the solution to any equation.

b. No, it is not possible to have more than one (x,y) pair that is solution to both equations

Keywords: Linear equations, Ordered pairs

Learn more about linear equations at:

#LearnwithBrainly

8 0

3 years ago

Solve for the equation below 4x^n + x^4n = ?

slavikrds [6]

4x + 6 = -10Simplifying 4x + 6 = -10 Reorder the terms: 6 + 4x = -10 Solving 6 + 4x = -10 Solving for variable 'x'. Move all terms containing x to the left, all other terms to the right. Add '-6' to each side of the equation. 6 + -6 + 4x = -10 + -6 Combine like terms: 6 + -6 = 0 0 + 4x = -10 + -6 4x = -10 + -6 Combine like terms: -10 + -6 = -16 4x = -16 Divide each side by '4'. x = -4 Simplifying x = -4

7 0

2 years ago

How do you solve this?

Katarina [22]

$\bf \textit{equation of a circle}\\\\ (x- h)^2+(y- k)^2= r^2 \qquad center~~(\stackrel{}{ h},\stackrel{}{ k})\qquad \qquad radius=\stackrel{}{ r} \\\\[-0.35em] ~\dotfill\\\\ (x-6)^2+(y+5)^2=16\implies [x-\stackrel{h}{6}]^2+[y-(\stackrel{k}{-5})]^2=\stackrel{r}{4^2}~\hfill \begin{cases} \stackrel{center}{(6,-5)}\\ \stackrel{radius}{4} \end{cases}$