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3 years ago

Find an equation of the horizontal line through (9,1) The equation is _____

Mathematics

1 answer:

Elena L [17]3 years ago

4 0

Answer:

Step-by-step explanation:

horizontal eq . of a line is y=k,

here y=1

slope of horizontal line =0

eq. of line through (9,1) is

y-1=0(x-9)

or y-1=0

or y=1

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Point U is on line segment TV . given TU=4 UV=5x and TV=3x+6 determine the numerical lenth of TV

adelina 88 [10]

TU=13; UV= 13; TV=26

Step-by-step explanation:

Hope this helped!

6 0

3 years ago

Read 2 more answers

Log(x-2)+log2=2logy<br>log(x-3y+3)=0

In-s [12.5K]

Answer:

Step-by-step explanation:

$\left\{\begin{array}{ccc}\log(x-2)+\log2=2\log y&(1)\\\log(x-3y+3)=0&(2)\end{array}\right$

$===========================\\\text{DOMAIN:}\\\\x-2>0\to x>2\\y>0\\x-3y+3>0\to x-3y>-3\\=====================\\\\\text{We know:}\ \log_ab=c\iff a^c=b,\\\\\text{therefore}\ \log_ab=0\iff a^0=b\to b=1\\\\\text{From this we have}\\\\\log(x-3y+3)=0\iff x-3y+3=1\qquad\text{add}\ 3y\ \text{to both sides}\\\\x+3=3y+1\qquad\text{subtract 3 from both isdes}\\\\\boxed{x=3y-2}\qquad(*)\\\\\text{Substitute it to (1)}$

$\log(3y-2-2)+\log2=2\log(y)\qquad\text{use}\ \log_ab^n=n\log_ab\\\\\log(3y-4)+\log2=\log(y^2)\qquad\text{subtract}\ \log(y^2)\ \text{from both sides}\\\\\log(3y-4)+\log2-\log(y^2)=0\\\\\text{Use}\ \log_ab+\log_ac=\log_a(bc)\ \text{and}\ \log_ab-\log_ac=\log_a\dfrac{b}{c}\\\\\log\dfrac{(3y-4)(2)}{y^2}=0\qquad\text{use}\ \log_ab=0\Rightarrow b=1\\\\\dfrac{(3y-4)(2)}{y^2}=1\iff y^2=(3y-4)(2)\\\\\text{Use the distributive property}\ a(b+c)=ab+ac$

$y^2=(3y)(2)+(-4)(2)\\\\y^2=6y-8\qquad\text{subtract}\ 6y\ \text{from both sides}\\\\y^2-6y=-8\qquad\text{add 8 to both sides}\\\\y^2-6y+8=0\\\\y^2-2y-4y+8=0\\\\y(y-2)-4(y-2)=0\\\\(y-2)(y-4)=0\iff y-2=0\ \vee\ y-4=0\\\\\boxed{y=2\ \vee\ y=4}\in D$

$\text{Put the values of}\ y\ \text{to }\ (*):\\\\\text{for}\ y=2:\\\\x=3(2)-2\\\\x=6-2\\\\\boxed{x=4}\in D\\\\\text{for}\ y=4:\\\\x=3(4)-2\\\\x=12-2\\\\\boxed{x=10}\in D$

4 0

3 years ago

In segment AC, the midpoint is B. If segments AC = 5x-9 and AB = 2x, what is the measure of segment AC.

IgorLugansk [536]

Answer:

The measure of segment AC is 36 units

Step-by-step explanation:

- The mid-point divides the segment into two equal parts in length

- B is the mid point of segment AC

- That means B divides segment AC into two equal parts in length

∴ AB = BC

∵ AC = 5x - 9

∵ AB = 2x

- The two parts AB and BC are equal in length

∴ BC = 2x

∵ AC = AB + BC

- Substitute the values of AB and BC in the expression of AC

∴ AC = 2x + 2x

∴ AC = 4x

∵ AC = 5x - 9

- Equate the two values of AC

∴ 5x - 9 = 4x

- Add 9 to both sides

∴ 5x = 4x + 9

- Subtract 4x from both sides

∴ x = 9

- Substitute the value of x in any expression of AC

∵ AC = 4x

∵ x = 9

∴ AC = 4(9) = 36

* The measure of segment AC is 36 units

8 0

3 years ago

Kelsey has already spent 42 minutes on the phone, and she expects to spend 2 more minutes

grandymaker [24]

Answer:

11 calls

Step-by-step explanation:

She has already spent 42 min. Also she wants to talk for 2 min per call. We can change calls on x. So, only 2x min for her calls left.

42 + 2x = 64

2x = 64 - 42

2x= 22

x=11

Answer: 11 calls

8 0

3 years ago

Micah was asked to add the following expressions: 3x2−x−9x2+3x+2+−2x2+2x+5x2+3x+2 First, he combined like terms in the numerator

shtirl [24]

Micah was asked to add the following expressions:

$\frac{3x^2-x-9}{x^2+3x+2} + \frac{-2x^2+2x+5}{x^2+3x+2}$

First, he combined like terms in the numerator and kept the common denominator

First step is correct. He added the like terms in the numerator, because the denominators are same.

$3x^2 -x -9 -2x^2+2x+5becomes x^2 +x -4$

So he got , $\frac{x^2+x-4}{x^2+3x+2}$

In the next step, he cannot cancel out x^2 from the top and bottom . Because x-4 and 3x+2 are added with x^2

If we have x^2 is multiplied with other terms at the top and bottom , then we can cancel out x^2.

So Micah added the expression incorrectly. Final answer is not correct.

5 0

3 years ago