59:52

Mathematics

2 answers:

Vera_Pavlovna [14]3 years ago

7 0

A line is an undefined geometric term that is described as a two dimensional set of points that has no beginning or end

Natali5045456 [20]3 years ago

6 0

Answer: A line is the undefined geometric term is described as a two-dimensional set of points that has no beginning or end.

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Describe in words how you would solve the linear system y = 3x + 1 and y = - 2x + 3.

Sergio [31]

We are using substitution to solve this problem.

y=3x+1
y=-2x+3
Make these equal each other: 3x+1=-2x+3
Add 2x to both sides: 3x+2x+1=-2x+2x+3
= 5x+1=3
Subtract 1 from both sides; 5x+1-1=3-1
= 5x=2
Divide 5 from both sides: 5x/5=2/5
X= 2/5
Plug X into one of the equations: y=3(2/5) +1
y=11/5(2.2)

(2/5,11/5)

5 0

3 years ago

Select the correct answer. During the summer, Jody earns $10 per hour babysitting and $15 per hour doing yardwork. This week she

omeli [17]

Answer:

A. x + y = 34; 10x + 15y = 410

Step-by-step explanation:

Jody's total hours will be the sum of the hours spent on each job:

x + y = 34 . . . . . total hours

Jody's total earnings will be the sum of the earnings from each job:

10x +15y = 410 . . . total earnings

So, the system of equations is ...

x + y = 34
10x + 15y = 410

3 0

2 years ago

Find the following:

Butoxors [25]

Answer:

Step-by-step explanation:

Limit refers to the value that the function approaches as the input approaches some value.

We say $\displaystyle \lim_{x\rightarrow a}f(x)=L$ , if f(x) approaches L as x approaches 'a'.

(a)

$\displaystyle \lim_{x\rightarrow 5}\left ( \frac{f(x)-8}{x-5} \right )=4\\\frac{\displaystyle \lim_{x\rightarrow 5}f(x)-\displaystyle \lim_{x\rightarrow 5}8}{\displaystyle \lim_{x\rightarrow 5}x-\displaystyle \lim_{x\rightarrow 5}5}=4\\$

$\frac{\displaystyle \lim_{x\rightarrow 5}f(x)-8}{\displaystyle \lim_{x\rightarrow 5}x-5}=4\\\displaystyle \lim_{x\rightarrow 5}f(x)-8=4\left ( \displaystyle \lim_{x\rightarrow 5}x-5 \right )\\\displaystyle \lim_{x\rightarrow 5}f(x)-8=4\displaystyle \lim_{x\rightarrow 5}x-4(5)\\\displaystyle \lim_{x\rightarrow 5}f(x)-8=4(5)-4(5)\\$

$\displaystyle \lim_{x\rightarrow 5}f(x)-8=20-20=0\\\displaystyle \lim_{x\rightarrow 5}f(x)=8$

(b)

$\displaystyle \lim_{x\rightarrow 5}\left ( \frac{f(x)-8}{x-5} \right )=7\\\frac{\displaystyle \lim_{x\rightarrow 5}f(x)-\displaystyle \lim_{x\rightarrow 5}8}{\displaystyle \lim_{x\rightarrow 5}x-\displaystyle \lim_{x\rightarrow 5}5}=7\\\frac{\displaystyle \lim_{x\rightarrow 5}f(x)-8}{\displaystyle \lim_{x\rightarrow 5}x-5}=7\\$

$\displaystyle \lim_{x\rightarrow 5}f(x)-8=7\left ( \displaystyle \lim_{x\rightarrow 5}x-5 \right )\\\displaystyle \lim_{x\rightarrow 5}f(x)-8=7\displaystyle \lim_{x\rightarrow 5}x-7(5)\\\displaystyle \lim_{x\rightarrow 5}f(x)-8=7(5)-7(5)\\\displaystyle \lim_{x\rightarrow 5}f(x)-8=35-35=0\\\displaystyle \lim_{x\rightarrow 5}f(x)=8$