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3 years ago

My other account just got banned for 40 something hour's how lame and I did nun so idk why I got banned but it's whatever

Mathematics

2 answers:

Anestetic [448]3 years ago

6 0

Answer:

move around it is what it is and no I'm not cooping that other girl

Alexxx [7]3 years ago

4 0

Answer:

it is what is iss

Step-by-step explanation:

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PLEASE HELP ME! GIVING AWAY POINTS! Solve. 2+y−8=19

navik [9.2K]

Y=25 because 19-2+8 equals 25 so y = 25

5 0

3 years ago

Read 2 more answers

What is the distance between (2,-6) and (-4,3)

ExtremeBDS [4]

Oh okie so cute ☺️ cute lol

8 0

2 years ago

Letters A and B represent a question]MATH

LiRa [457]

Answer:

@Genius102 has answered this already, but here it is

Step-by-step explanation:

a) the scale should go by 1,000 because that would show that the slope is growing by time.

b) the intervals should go by 500 because 18,500 is the last number, so it should be 500

7 0

3 years ago

Classify the real number. <img src="https://tex.z-dn.net/?f=%20%5Csqrt%7B15%7D%20" id="TexFormula1" title=" \sqrt{15} " alt="

lilavasa [31]

Answer:

<h2>Irrational </h2>

Step-by-step explanation:

Firstly, 

According to rational and irrational,

$\sqrt{15} \: is \: irrational$

Since,

Natural numbers, Whole Numbers and Integers all come under Rational number.

Hence,

 $\sqrt{15}$ 

Is an irrational number. 

5 0

3 years ago

Find the two intersection points

bogdanovich [222]

Answer:

Our two intersection points are:

$\displaystyle (3, -2) \text{ and } \left(-\frac{53}{25}, \frac{46}{25}\right)$

Step-by-step explanation:

We want to find where the two graphs given by the equations:

$\displaystyle (x+1)^2+(y+2)^2 = 16\text{ and } 3x+4y=1$

Intersect.

When they intersect, their x- and y-values are equivalent. So, we can solve one equation for y and substitute it into the other and solve for x.

Since the linear equation is easier to solve, solve it for y:

$\displaystyle y = -\frac{3}{4} x + \frac{1}{4}$

Substitute this into the first equation:

$\displaystyle (x+1)^2 + \left(\left(-\frac{3}{4}x + \frac{1}{4}\right) +2\right)^2 = 16$

Simplify:

$\displaystyle (x+1)^2 + \left(-\frac{3}{4} x + \frac{9}{4}\right)^2 = 16$

Square. We can use the perfect square trinomial pattern:

$\displaystyle \underbrace{(x^2 + 2x+1)}_{(a+b)^2=a^2+2ab+b^2} + \underbrace{\left(\frac{9}{16}x^2-\frac{27}{8}x+\frac{81}{16}\right)}_{(a+b)^2=a^2+2ab+b^2} = 16$

Multiply both sides by 16:

$(16x^2+32x+16)+(9x^2-54x+81) = 256$

Combine like terms:

$25x^2+-22x+97=256$

Isolate the equation:

$\displaystyle 25x^2 - 22x -159=0$

We can use the quadratic formula:

$\displaystyle x = \frac{-b\pm\sqrt{b^2-4ac}}{2a}$

In this case, a = 25, b = -22, and c = -159. Substitute:

$\displaystyle x = \frac{-(-22)\pm\sqrt{(-22)^2-4(25)(-159)}}{2(25)}$

Evaluate:

$\displaystyle \begin{aligned} x &= \frac{22\pm\sqrt{16384}}{50} \\ \\ &= \frac{22\pm 128}{50}\\ \\ &=\frac{11\pm 64}{25}\end{aligned}$

Hence, our two solutions are:

$\displaystyle x_1 = \frac{11+64}{25} = 3\text{ and } x_2 = \frac{11-64}{25} =-\frac{53}{25}$

We have our two x-coordinates.

To find the y-coordinates, we can simply substitute it into the linear equation and evaluate. Thus:

$\displaystyle y_1 = -\frac{3}{4}(3)+\frac{1}{4} = -2$

And:

$\displaystyle y _2 = -\frac{3}{4}\left(-\frac{53}{25}\right) +\frac{1}{4} = \frac{46}{25}$

Thus, our two intersection points are:

$\displaystyle (3, -2) \text{ and } \left(-\frac{53}{25}, \frac{46}{25}\right)$

6 0

3 years ago