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Artemon [7]
3 years ago
13

Write the standard form of the equation of the line through the given point with the given slope.

Mathematics
1 answer:
lisov135 [29]3 years ago
6 0

Answer:

The equation for this slope will be y=7x-5

Step-by-step explanation:

Hope this helped

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4x + y = 7 find the slope and write the equation <br><br> PLEASE HELP!!!!!
Alexus [3.1K]

Answer:

3

Step-by-step explanation:

4x + 3 = 7

6 0
3 years ago
What the recursive formula for the sequence
klemol [59]

\bf n^{th}\textit{ term of an arithmetic sequence} \\\\ a_n=a_1+(n-1)d\qquad \begin{cases} n=n^{th}\ term\\ a_1=\textit{first term's value}\\ d=\textit{common difference} \end{cases} \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ a_n=2-5(n-1)\implies a_n=\stackrel{\stackrel{a_1}{\downarrow }}{2}+(n-1)(\stackrel{\stackrel{d}{\downarrow }}{-5})


so, we know the first term is 2, whilst the common difference is -5, therefore, that means, to get the next term, we subtract 5, or we "add -5" to the current term.

\bf \begin{cases} a_1=2\\ a_n=a_{n-1}-5 \end{cases}


just a quick note on notation:

\bf \stackrel{\stackrel{\textit{current term}}{\downarrow }}{a_n}\qquad \qquad \stackrel{\stackrel{\textit{the term before it}}{\downarrow }}{a_{n-1}} \\\\[-0.35em] ~\dotfill\\\\ \stackrel{\textit{current term}}{a_5}\qquad \quad \stackrel{\textit{term before it}}{a_{5-1}\implies a_4}~\hspace{5em}\stackrel{\textit{current term}}{a_{12}}\qquad \quad \stackrel{\textit{term before it}}{a_{12-1}\implies a_{11}}

8 0
3 years ago
What is the answer of K &lt; -5 ?
alexira [117]
To your question, "K<-5", that is the answer. So the answer is K<-5
8 0
3 years ago
What is the value of X?<br><br> Enter your answer in the box.
morpeh [17]
I got 2.17 but I'm not sure that's right.

180 - (71 + 90) = (6x + 4)
4 0
3 years ago
Read 2 more answers
How do you tell if four points are coplanar
wariber [46]
You know that three points A,B,CA,B,C (two vectors A⃗ BA→B, A⃗ CA→C) form a plane. If you want to show the fourth one DD is on the same plane, you have to show that it forms, with any of the other point already belonging to the plane, a vector belonging to the plane (for instance A⃗ DA→D).

Since the cross product of two vectors is normal to the plane formed by the two vectors (A⃗ B×A⃗ CA→B×A→C is normal to the plane ABCABC), you just have to prove your last vector A⃗ DA→D is normal to this cross product, hence the triple product that should be equal to 00:

A⃗ D⋅(A⃗ B×AC)=0
3 0
3 years ago
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