Find the sum without using a number line -6 1/3 + 3=

Draw a polygon with the given conditions in a coordinate plane of rectangle with a perimeter of 20 units

Anarel [89]

Question says to draw a polygon with the given conditions in a coordinate plane. It is talking about Rectangle with perimeter of 20 units.

So we can use formula of perimeter of rectangel to find it's possible dimensions.

We knwo that perimeter of the rectangle is given by 2(L+W)

where L is lenght and W is width.

Given perimeter is 20 so we get:

2(L+W)=20

L+W=10

L=10-W

So basically we can use any number for W which can be from 0 to 10 as side length can't be negative.

Let W=4

Then L=10-W=10-4=6

Hence we just have to draw a rectangle having length 6 and width 4.

So the final answer will be picture in the attached graph.

3 0

2 years ago

(x+3)(3x² - 5x - 10) multiply polynomial

brilliants [131]

Answer:

$\boxed{\sf{3x^3+4x^2-25x-30}}$

Step-by-step explanation:

$\underline{\text{SOLUTION:}}$

To isolate the term of x from one side of the equation, you must multiply by a polynomial.

$\underline{\text{GIVEN:}}$

$:\Longrightarrow: \sf{(x+3)(3x^2 - 5x - 10)}$

You have to solve with parentheses first.

$:\Longrightarrow \sf{x\cdot \:3x^2+x\left(-5x\right)+x\left(-10\right)+3\cdot \:3x^2+3\left(-5x\right)+3\left(-10\right)}$

Solve.

$\sf{x*3x=3x^3}$

x(-5x)=-5x²

$\sf{x(-10)=-10x}$

3*3x²=9x²

3(-5x)=-15x

3(-10)=-30

Then, rewrite the problem down.

$\sf{3x^3-5x^2-10x+9x^2-15x-30}$

Combine like terms.

$\Longrightarrow: \sf{3x^3-5x^2+9x^2-10x-15x-30}$

Add/subtract the numbers from left to right.

-5x²+9x²=4x²

$\Longrightarrow: \sf{3x^3+4x^2-10x-15x-30}$

Solve.

$\sf{-10x-15x=-25x}$

Then rewrite the problem.

$\Longrightarrow: \boxed{\sf{3x^3+4x^2-25x-30}}$

Therefore, the correct answer is 3x³+4x²-25x-30.

I hope this helps! Let me know if you have any questions.

6 0

2 years ago

Read 2 more answers

Helpppppp pleaseeee anyoneeeee!!!!!

kakasveta [241]

Answer:

Create a function representing number of published articles per month:

$a=12m+60$

Where:

a = total number of articles published
m = number of months
12 = slope = number of articles published per month
60 = y-intercept = number of published articles at the start (month 0)

Now substitute in the values of m to get the a-value:

m = 1 → a = 12(1) + 60 = 72
m = 3 → a = 12(3) + 60 = 36 + 60 = 96
m = 4 → a = 12(4) + 60 = 48 + 60 = 108
m = 9 → a = 12(9) + 60 = 108 + 60 = 168

6 0

2 years ago

Find the point that is 1/5 the way from A to B where A(-7,4) and B(3, 10).

mestny [16]

Answer:

The coordinates of the point that is 1/5 the way from A to B is $(x,y) = (-5,\frac{26}{5})$

Step-by-step explanation:

Here, the given points are: A (-7,4) and B (3,10)

Let us assume the point M(x,y) on AB is such that

AM : AB = 1 : 5

⇒ AM : (AB - AM) = 1 : (5-1) = 1: 4

⇒ AM : MB = 1 : 4

Now, The Section Formula states the coordinates of point (x,y) on any line dividing the line in the ratio m1 : m2

$(x,y) = (\frac{m_2x_1+m_1x_2}{m_1+m_2} ,\frac{m_2y_1+m_1y_2}{m_1+m_2} )$

Here, in the given equation, m1: m2 = 1:4

So, the coordinates M(x,y) is given as:

$(x,y) = (\frac{(-7)(4) + 1 (3)}{1+ 4} ,\frac{4(4) + 1(10)}{1+4} )\\\implies (x,y) = (\frac{-28+3}{5} ,\frac{16+10}{5} ) = (\frac{-25}{5} ,\frac{26}{5} )\\\implies (x,y) = (-5,\frac{26}{5} )$

Hence, the coordinates of the point that is 1/5 the way

from A to B is $(x,y) = (-5,\frac{26}{5})$

5 0

3 years ago

The graph shown is a translation of the graph of f(x)=x^2. Write the function in vertex form.

Gnesinka [82]

$\sf \longrightarrow \: f(x) = {x}^{2}$

Write the parent function in the standard form I.e.

$\boxed{ \tt \:y =a(x - h) \times 2 + k }$

Where,

a = 1
h = 0
k = 0

$\sf \longrightarrow \:y= 1(x - 0)^{2} \times 2 + 0$

$\sf \longrightarrow \: y = (x - 0) ^{2} \times 2 + 0$

$\sf \longrightarrow \: y = {x}^{2} \times 2$

$\sf \longrightarrow \: y = 2{x}^{2}$

4 0

2 years ago