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3 years ago

8

Write a description of the three different possibilities that may occur when graphing a system of two linear equations.

1 answer:

Gelneren [198K]3 years ago

3 0

There could be an infinite number of solutions, one solution, or no solution at all.

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Answer question 4 pls for math

brilliants [131]

Answer:

Rule: y = 2x

m = 2

b = 0

Step-by-step explanation:

(17, 34) and (22, 44)

<u>Slope:</u>

m=(y2-y1)/(x2-x1)

m=(44-34)/(22-17)

m= (10)/5

m = 2

<u>Slope-intercept:</u>

y - y1 = m(x - x1)

y - 34 = 2(x - 17)

y - 34 = 2x - 34

y = 2x

y = mx + b

y = 2x

b = 0

3 0

2 years ago

4. Find the value of x.<br> A) 6 √ 2<br> B) 12 √ 3<br> C) 24 √ 3<br> D) 12

Schach [20]

A^2+b^2=c^2
24^2=12^2+x^2
sqrt(576-144)=x
sqrt(432)=x
12sqrt(3)=x
the answer is c

7 0

2 years ago

Read 2 more answers

ln(y+1)-lny=1+3lnx . Express y in terms of x, in a form not involving logarithms. (4 marks, Maths CIE A level)

Delicious77 [7]

$remember$
$ln(x)=log_e(x)$
$and$
$log_x(a)-log_x(b)= log_x( \frac{a}{b} )$
$and$
$log_a(b)=c$ means $a^c=b$
$and$
$blog(a)=log(a^b)$

$so$

$ln(y+1)-ln(y)=1+3ln(x)$
$ln( \frac{y+1}{y} )=1+ln(x^3)$

$minus$ $ln(x^3)$ $from$ $both$ $sides$

$ln( \frac{y+1}{y}-ln(x^3) )=1$
$ln( \frac{y+1}{yx^3})=1$
$log_e( \frac{y+1}{yx^3})=1$

$translate$

$e^1=\frac{y+1}{yx^3}$
$e=\frac{y+1}{yx^3}$

$times$ $both$ $sides$ $by$ $yx^3$

$eyx^3=y+1$

$minus$ $y$ $from$ $both$ $sides$

$eyx^3-y=1$
$y(ex^3-1)=1$

$divide$ $both$ $sides$ $by$ $(ex³-1)$

$y= \frac{1}{ex^3-1}$

6 0

2 years ago

Read 2 more answers

Write the fraction as a percent. Round to the nearest tenth of a percent of necessary. 1/7

balu736 [363]

It's 14.285714%

So, simplified it would be 14.3% because you round the 2 up since theres an 8 behind it.

8 0

3 years ago

Math Story problem. I need help

Olin [163]

Answer:

i believe it would be 12

hope this helps !!

Step-by-step explanation:

96 divided by 8 would equal 12 so therefore the answer is 12

6 0

2 years ago