Ask question

Ask question

All categories

2 years ago

11

write the equation of the line that passes through the point (-2,4) and (5,8). Put your answer in fully reduced point slope form

, unless it is a vertical or horizontal line.

1 answer:

sergij07 [2.7K]2 years ago

8 0

Answer:

y-4=4(x+2)/7

Step-by-step explanation:

slope(m) = (8-4)/(5-(-2)) =4/7

b = 4-(4/7)×(-2) = 36/7

y = mx+b

y = 4x/7+36/7

in point-slope form,

y-4=4(x+2)/7

You might be interested in

WILL GIVE BEST RESPONSE

tino4ka555 [31]

We will use demonstration of recurrences<span>1) for n=1, 10= 5*1(1+1)=5*2=10, it is just
2) assume that the equation </span>10 + 30 + 60 + ... + 10n = 5n(n + 1) is true, <span>for all positive integers n>=1
</span>3) let's show that the equation<span> is also true for n+1, n>=1
</span><span>10 + 30 + 60 + ... + 10(n+1) = 5(n+1)(n + 2)
</span>let be N=n+1, N is integer because of n+1, so we have
<span>10 + 30 + 60 + ... + 10N = 5N(N+1), it is true according 2)
</span>so the equation<span> is also true for n+1,
</span>finally, 10 + 30 + 60 + ... + 10n = 5n(n + 1) is always true for all positive integers n.
<span>
</span>

4 0

2 years ago

Elena surveyed her classmates to see how many like to make jewelry. When she asked 20 of her classmates, 50 percent said they li

nordsb [41]

10 students of Elena's classmates like to make jewellery.

<u>Step-by-step explanation:</u>

Let total no.of classmates who liked the jewellery = x

Here, in the given problem, given she asked her classmates '20'. So, it clearly shows that total number of students/classmates = 20. Also, in that, 50% said like to make jewellery. Now find 'x' as below,

$x = 50 \% \text { of } 20$

$x = \frac{50}{100} \times 20$

$x = 10$

Hence, the total number of classmates who liked to make Jewellery = 10

5 0

3 years ago

12x+y=−4 y=2x+16 im really confused on how to do this, i need to find the solution

ololo11 [35]

Answer:

here

Step-by-step explanation:

5 0

2 years ago

Read 2 more answers

Consider the problem of finding the line of symmetry and vertex of the quadratic equation f(x) =x^2-8x+15 What is the error in t

Elenna [48]

$\bf \textit{ vertex of a vertical parabola, using coefficients}\\\\
\begin{array}{llccll} f(x) = &{{ 1}}x^2&{{ -8}}x&{{ +15}}\\
&\uparrow &\uparrow &\uparrow \\
&a&b&c
\end{array}\qquad 
\left(-\cfrac{{{ b}}}{2{{ a}}}\quad ,\quad {{ c}}-\cfrac{{{ b}}^2}{4{{ a}}}\right)
\\\\\\
\textit{so the line of symmetry will be at }x=-\cfrac{{{ b}}}{2{{ a}}}$

3 0

3 years ago

Read 2 more answers

Can I have some help with these? Thanks.

9966 [12]

The x^2 over the x is like subtracting x from x^2, so the new equation would be 9-x/-3. You divide 9 by -3 to get the final answer of -3-x, or -x-3, which is E.

5 0

3 years ago