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3 years ago

9x+3x-10=3(3x+x) identity or solution

Mathematics

2 answers:

k0ka [10]3 years ago

8 0

There is no solution. The output of this equation is equal to -10 = 0 which is obviously not true.

romanna [79]3 years ago

3 0

There is no solution

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COS Find the quotient when the polynomial expression is divided by a binomial Find the quotient ANSWER SOLUTION CHOICES X2+11x+

ryzh [129]

Answer:

x=−1 or x=−9

Step-by-step explanation:

x2+11x+10=x+1

Step 1: Subtract x+1 from both sides.

x2+11x+10−(x+1)=x+1−(x+1)

x2+10x+9=0

Step 2: Factor left side of equation.

(x+1)(x+9)=0

Step 3: Set factors equal to 0.

x+1=0 or x+9=0

x=−1 or x=−9

8 0

3 years ago

What is the distance between the points (-4,2) and (3,-5)

34kurt

Answer:

7√2 units

Step-by-step explanation:

The distance between two points having coordinates (x1, y1) and (x2, y2) is given by,

$\sqrt{ {(x2 - x1)}^{2} + {(y2 - y1)}^{2} }$

Here, x1=-4 and y1=2

x2=3 and y2=-5

$\sqrt{ {(3 - ( - 4))}^{2} + {( - 5 - 2)}^{2} }$

$\sqrt{ {(3 + 4)}^{2} + {( - 7)}^{2} }$

$\sqrt{ {(7)}^{2} + {( - 7)}^{2} }$

$\sqrt{49 + 49} = 7 \sqrt{2}$

6 0

3 years ago

Read 2 more answers

What's is the converse of the statement?If x is odd, then 2x is even.

dsp73

If you have two statements p and q and $p\Rightarrow q$ is true, then $\neg q\Rightarrow \neg p$ is also true.

In your case, statements are p - "x is odd" and q - "2x is even".

Then $\neg p$ - "x is not odd" and $\neg q$ - "2x is not even."

Hence $\neg q\Rightarrow \neg p$ sounds as: "If 2x is not even, then x is not odd".
Answer: correct choice is D.

7 0

3 years ago

Indicate the equation of the given line in standard form. Show all your work for full credit. the line containing the median of

alukav5142 [94]

Answer:

* The equation of the median of the trapezoid is 10x + 6y = 39

Step-by-step explanation:

* Lets explain how to solve the problem

- The slope of the line whose end points are (x1 , y1) , (x2 , y2) is

$m=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}$

- The mid point of the line whose end point are (x1 , y1) , (x2 , y2) is

$(\frac{x_{1}+x_{2}}{2},\frac{y_{1}+y_{2}}{2})$

- The standard form of the linear equation is Ax + BC = C, where

A , B , C are integers and A , B ≠ 0

- The median of a trapezoid is a segment that joins the midpoints of

the nonparallel sides

- It has two properties:

# It is parallel to both bases

# Its length equals half the sum of the base lengths

* Lets solve the problem

- The trapezoid has vertices R (-1 , 5) , S (! , 8) , T (7 , -2) , U (2 , 0)

- Lets find the slope of the 4 sides two find which of them are the

parallel bases and which of them are the non-parallel bases

# The side RS

∵ $m_{RS}=\frac{8-5}{1 - (-1)}=\frac{3}{2}$

# The side ST

∵ $m_{ST}=\frac{-2-8}{7-1}=\frac{-10}{6}=\frac{-5}{3}$

# The side TU

∵ $m_{TU}=\frac{0-(-2)}{2-7}=\frac{2}{-5}=\frac{-2}{5}$

# The side UR

∵ $m_{UR}=\frac{5-0}{-1-2}=\frac{5}{-3}=\frac{-5}{3}$

∵ The slope of ST = the slop UR

∴ ST// UR

∴ The parallel bases are ST and UR

∴ The nonparallel sides are RS and TU

- Lets find the midpoint of RS and TU to find the equation of the

median of the trapezoid

∵ The median of a trapezoid is a segment that joins the midpoints of

the nonparallel sides

∵ The midpoint of RS = $(\frac{-1+1}{2},\frac{5+8}{2})=(0,\frac{13}{2})$

∵ The median is parallel to both bases

∴ The slope of the median equal the slopes of the parallel bases = -5/3

∵ The form of the equation of a line is y = mx + c

∴ The equation of the median is y = -5/3 x + c

- To find c substitute x , y in the equation by the coordinates of the

midpoint of RS

∵ The mid point of Rs is (0 , 13/2)

∴ 13/2 = -5/3 (0) + c

∴ 13/2 = c

∴ The equation of the median is y = -5/3 x + 13/2

- Multiply the two sides by 6 to cancel the denominator

∴ The equation of the median is 6y = -10x + 39

- Add 10x to both sides

∴ The equation of the median is 10x + 6y = 39

* The equation of the median of the trapezoid is 10x + 6y = 39

7 0

3 years ago

Can u please help me please. Please

jonny [76]

The answer is
C = 4 ÷ 1/5

6 0

3 years ago