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3 years ago

Find the coordinates of the point that is located 8 units below the x-axis and 3 units to the left of the y-axis

Mathematics

1 answer:

4vir4ik [10]3 years ago

4 0

-8,-3 if i read that correctly, that should be the answer.

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Write an equation of the line in slope-intercept form.

DerKrebs [107]

Answer:

Maybe D?

Step-by-step explanation:

7 0

3 years ago

1. Solve by factoring and using the zero product property. (Diamond &<br> x2 – 2x – 15 = 0

Harlamova29_29 [7]

The factors are: x-5 and x+3

Step-by-step explanation:

Given

$x^2-2x-15 = 0$

Simplifying

$x^2-5x+3x-15 = 0\\x(x-5) + 3(x-5) = 0\\(x+3)(x-5) = 0$

Let us define the zero product property first

The zero product property states that

if ab=0 then either a=0 or b=0

So,

$x+3 = 0\\x+3-3 = 0-3\\x = -3\\AND\\x-5 = 0\\x-5+5 = 0+5\\x = 5$

Hence,

the factors are: x-5 and x+3

Keywords: Factorization, Zero product property

Learn more about factorization at:

#LearnwithBrainly

3 0

3 years ago

A car travelling along a horizontal road accelerates from rest to a speed of 120km/h in T seconds, covering a distance of 88 8/9

scoray [572]

Answer:

a = 6.25 m/s²

T = 5⅓ seconds

Step-by-step explanation:

a = (v - u)/t

v = 120×1000/3600 = 100/3

a = 100/3t

v² = u² + 2as

(100/3)² = 2(100/3t)(88 8/9)

10000/9 = (200/3t)(800/9)

200/3t = 12.5

t = 5⅓ seconds

a = 6.25

8 0

3 years ago

What times 5 equals 62?

Anni [7]

12.4 because 62 divided by 5 = 12.4

5 0

3 years ago

Read 2 more answers

Start with the geometric power series LaTeX: 1+x+x^2+x^3+x^4+\cdots Substitute LaTeX: -x^3in place of LaTeX: x, and then take a

valina [46]

Answer:

$-3x^2+6x^5-9x^8+12x^{11}+...=\sum_{n=1}^{\infty }\left ( -1 \right )^n\,\,3n\,\,x^{3n-1}$

Step-by-step explanation:

Given : $1+x+x^2+x^3+x^4+\cdots$

On putting $-x^3$ in place of x , we get $1+\left ( -x^3 \right )+\left ( -x^3 \right )^2+\left ( -x^3 \right )^3+\left ( -x^3 \right )^4+...$

On simplifying , we get $1-x^3+x^6-x^9+x^{12}+...$

On differentiating , we get $\left ( 1-x^3+x^6-x^9+x^{12}+... \right )'=-3x^2+6x^5-9x^8+12x^{11}+...$

Here ,

$-3x^2=\left ( -1 \right )^1\,3(1)x^{3(1)-1}\\6x^5=\left ( -1 \right )^2\,\,3(2)\,\,x^{3(2)-1}\\-9x^8=\left ( -1 \right )^3\,\,3(3)\,\,x^{3(3)-1}\\12x^{11}=\left ( -1 \right )^4\,\,3(4)\,\,x^{3(4)-1}$

Now , we need to express it using summation notation.

$\left ( 1-x^3+x^6-x^9+x^{12}+... \right )'=-3x^2+6x^5-9x^8+12x^{11}+...=\sum_{n=1}^{\infty }\left ( -1 \right )^n\,\,3n\,\,x^{3n-1}$

4 0

3 years ago