1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
solmaris [256]
3 years ago
10

Write the equation of a line through points (5,5) and (1,0) in point-intercept form.

Mathematics
1 answer:
Sati [7]3 years ago
7 0

Answer:

The slope-intercept form is y = 5/4x - 5/4 and the point-slope form is y - 0 = 5/4(x - 1)

Step-by-step explanation:

There is no such thing as point-intercept form. There is point-slope form and slope-intercept form. To find either of those, you can follow the steps below.

First find the slope of the equation using the slope formula.

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

m = (5 - 0)/(5 - 1)

m = 5/4

Now we can use the slope along with a point to get the point-slope form.

y - y1 = m(x - x1)

y - 0 = 5/4(x - 1) ------->THIS IS POINT-SLOPE

Now to get slope-intercept, solve for y

y - 0 = 5/4(x - 1)

y = 5/4x - 5/4

You might be interested in
What is the relationship between exponentials and logarithms? How can you use these to solve equations? Provide an example in yo
Sveta_85 [38]

Answer:

Exponentials and logarithms are inverses of each other.

Step-by-step explanation:

Exponentials and logarithms are inverses of each other.

For logarithmic function:

Domain = \left ( 0,\infty  \right ), Range = \left ( -\infty ,\infty  \right )

Vertical asymptote is y - axis.

x - intercept is (1,0)

For exponential function:

Domain = \left ( -\infty ,\infty  \right ), Range = \left ( 0,\infty  \right )

Horizontal asymptote is x - axis.

y- intercept is (0,1)

Both exponential and logarithmic functions are increasing.

For example:

Solve: \log x=\frac{\log 5+\log 3}{\log 3^2}

\log x=\frac{\log 5+\log 3}{\log 3^2}\\\log x=\frac{\log (5\times 3)}{2\log 3}\,\,\left \{ \because \log (ab)=\log a+\log b\,,\,\log a^b=b\log a \right \}\\=\frac{\log 15}{2\log 3}

\Rightarrow \log x=\frac{\log 15}{2\log 3}\\\Rightarrow x=e^{\frac{\log 15}{2\log 3}}\,\,\left \{ \because \log x=y\Rightarrow x=e^y \right \}

4 0
3 years ago
Question 1 of 22
Masja [62]

Answer:

D

Step-by-step explanation:

Because it says Klay lunch cost is $2 less than Ella's.

==> Ella's - $2 = Klay's

==> x-2=6

plus both side by 2

so the answer is $8

8 0
2 years ago
Uhh help?
Leni [432]
Hope you can read that okay

5 0
3 years ago
Which is greater
olga55 [171]
They are even/equal
(please mark brainliest)
5 0
3 years ago
Read 2 more answers
HELP !!! | 20 Points.
RUDIKE [14]

Answer:

No. The ladder is too steep.

Step-by-step explanation:

Length of ladder = 17 feet.  This is the hypotenuse.

Side a of a right triangle = 16.6

To answer this, you need one of the trigonometric functions.  

Opposite = 16.5

hypotenuse = 17

The function you need is the sine function.

Sin(theta) = opposite / hypotenuse

sin(theta) = 16.5 / 17

Sin(theta) = 0.97059

theta = sin-1(0.07059)

theta = 76.07

No the ladder is too steep.

5 0
3 years ago
Read 2 more answers
Other questions:
  • No body helped me :'(
    5·1 answer
  • Find m/AEB<br><br> A.<br> 10<br> B.<br> 70<br> C.<br> 110<br> D.<br> 170
    9·2 answers
  • Find (f∘g)(x) if f(x) = 2x - 1 and g(x) = x + 2.<br><br>A.2x - 3<br>B.2x + 1<br>C.2x - 1<br>D.2x + 3
    14·1 answer
  • Right answer gets Brainlist
    8·1 answer
  • Find the equation of the quadratic function determined from the graph above.
    9·2 answers
  • Find the value of 7C7
    10·1 answer
  • 50,60,72 find the 8th term airthmetic and geometric sequence
    7·1 answer
  • Line PQ is represented as
    9·1 answer
  • Of the 28 students in class, about 65% are wearing sneakers. How many students are not wearing sneakers? Round to the nearest wh
    7·1 answer
  • Subtract. (4/5p+1/2)-(1/4p-3) <br><br> A: 21/20p-5/2 <br><br> B: 1/5p-7/2 <br><br> C: 11/20p+7/2
    13·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!