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3 years ago

Write the equation

Mathematics

2 answers:

Shkiper50 [21]3 years ago

3 0

$a)\ the\ slope-intercept\ form:\ y=mx+b\\--------------------------\\m=\frac{2}{3}\Rightarrow y=\frac{2}{3}x+b\\\\the\ point\ (-3;\ 6)\Rightarrow \frac{2}{3}\times(-3)+b=6\\\\-2+b=6\ \ \ \ \ |add\ 2\ to\ both\ sides\\\\b=8\\\\\boxed{y=\frac{2}{3}x+8}$

$b)\ the\ two-points\ formula:y-y_1=\frac{y_2-y_1}{x_2-x_1}(x-x_1)\\-----------------------------\\the\ points:\ (3;-1);\ (5;\ 7)\\\\y-(-1)=\frac{7-(-1)}{5-3}(x-3)\\\\y+1=\frac{7+1}{2}(x-3)\\\\y+1=\frac{8}{2}(x-3)\\\\y+1=4(x-3)\\\\y+1=4x-12\ \ \ \ \ |subtract\ 1\ from\ both\ sides\\\\\boxed{y=4x-13}$

musickatia [10]3 years ago

3 0

A. y=2/3x+c
6=((2/3)*-3)+c
6+2.33=c
8.33=c

y=2/3x+8 1/3

B. y=mx+c
m= y2-y1/x2-x1
m= (7--1)/(5-3)
m= 8/2 = 4

7=4x5+c
7-20=c
-13=c

y=4x-13

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What does the " commutative property" help us solve this equation: 6 + p = - 3<br><br> HELPPPP MEEEE

aliina [53]

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3 years ago

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Oxana [17]

Find the tangent of the angle, using the height (472) minus the elevated height of the camera (2) which would be 470 meters, and also the distance away from the shuttle, which would be 750. Opposite divided by adjacent would make the tangent 0.626...so using a triangle chart, the closest whole numbered angle 0.626 would be to is 32 degrees.

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3 years ago

2,17,82,257,626,1297 next one please ?

In-s [12.5K]

The easy thing to do is notice that 1^4 = 1, 2^4 = 16, 3^4 = 81, and so on, so the sequence follows the rule $n^4+1$ . The next number would then be fourth power of 7 plus 1, or 2402.

And the harder way: Denote the <em>n</em>-th term in this sequence by $a_n$ , and denote the given sequence by $\{a_n\}_{n\ge1}$ .

Let $b_n$ denote the <em>n</em>-th term in the sequence of forward differences of $\{a_n\}$ , defined by

$b_n=a_{n+1}-a_n$

for <em>n</em> ≥ 1. That is, $\{b_n\}$ is the sequence with

$b_1=a_2-a_1=17-2=15$

$b_2=a_3-a_2=82-17=65$

$b_3=a_4-a_3=175$

$b_4=a_5-a_4=369$

$b_5=a_6-a_5=671$

and so on.

Next, let $c_n$ denote the <em>n</em>-th term of the differences of $\{b_n\}$ , i.e. for <em>n</em> ≥ 1,

$c_n=b_{n+1}-b_n$

so that

$c_1=b_2-b_1=65-15=50$

$c_2=110$

$c_3=194$

$c_4=302$

etc.

Again: let $d_n$ denote the <em>n</em>-th difference of $\{c_n\}$ :

$d_n=c_{n+1}-c_n$

$d_1=c_2-c_1=60$

$d_2=84$

$d_3=108$

etc.

One more time: let $e_n$ denote the <em>n</em>-th difference of $\{d_n\}$ :

$e_n=d_{n+1}-d_n$

$e_1=d_2-d_1=24$

$e_2=24$

etc.

The fact that these last differences are constant is a good sign that $e_n=24$ for all <em>n</em> ≥ 1. Assuming this, we would see that $\{d_n\}$ is an arithmetic sequence given recursively by

$\begin{cases}d_1=60\\d_{n+1}=d_n+24&\text{for }n>1\end{cases}$

and we can easily find the explicit rule:

$d_2=d_1+24$

$d_3=d_2+24=d_1+24\cdot2$

$d_4=d_3+24=d_1+24\cdot3$

and so on, up to

$d_n=d_1+24(n-1)$

$d_n=24n+36$

Use the same strategy to find a closed form for $\{c_n\}$ , then for $\{b_n\}$ , and finally $\{a_n\}$ .

$\begin{cases}c_1=50\\c_{n+1}=c_n+24n+36&\text{for }n>1\end{cases}$

$c_2=c_1+24\cdot1+36$

$c_3=c_2+24\cdot2+36=c_1+24(1+2)+36\cdot2$

$c_4=c_3+24\cdot3+36=c_1+24(1+2+3)+36\cdot3$

and so on, up to

$c_n=c_1+24(1+2+3+\cdots+(n-1))+36(n-1)$

Recall the formula for the sum of consecutive integers:

$1+2+3+\cdots+n=\displaystyle\sum_{k=1}^nk=\frac{n(n+1)}2$

$\implies c_n=c_1+\dfrac{24(n-1)n}2+36(n-1)$

$\implies c_n=12n^2+24n+14$

$\begin{cases}b_1=15\\b_{n+1}=b_n+12n^2+24n+14&\text{for }n>1\end{cases}$

$b_2=b_1+12\cdot1^2+24\cdot1+14$

$b_3=b_2+12\cdot2^2+24\cdot2+14=b_1+12(1^2+2^2)+24(1+2)+14\cdot2$

$b_4=b_3+12\cdot3^2+24\cdot3+14=b_1+12(1^2+2^2+3^2)+24(1+2+3)+14\cdot3$

and so on, up to

$b_n=b_1+12(1^2+2^2+3^2+\cdots+(n-1)^2)+24(1+2+3+\cdots+(n-1))+14(n-1)$

Recall the formula for the sum of squares of consecutive integers:

$1^2+2^2+3^2+\cdots+n^2=\displaystyle\sum_{k=1}^nk^2=\frac{n(n+1)(2n+1)}6$

$\implies b_n=15+\dfrac{12(n-1)n(2(n-1)+1)}6+\dfrac{24(n-1)n}2+14(n-1)$

$\implies b_n=4n^3+6n^2+4n+1$

$\begin{cases}a_1=2\\a_{n+1}=a_n+4n^3+6n^2+4n+1&\text{for }n>1\end{cases}$

$a_2=a_1+4\cdot1^3+6\cdot1^2+4\cdot1+1$

$a_3=a_2+4(1^3+2^3)+6(1^2+2^2)+4(1+2)+1\cdot2$

$a_4=a_3+4(1^3+2^3+3^3)+6(1^2+2^2+3^2)+4(1+2+3)+1\cdot3$

$\implies a_n=a_1+4\displaystyle\sum_{k=1}^3k^3+6\sum_{k=1}^3k^2+4\sum_{k=1}^3k+\sum_{k=1}^{n-1}1$

$\displaystyle\sum_{k=1}^nk^3=\frac{n^2(n+1)^2}4$

$\implies a_n=2+\dfrac{4(n-1)^2n^2}4+\dfrac{6(n-1)n(2n)}6+\dfrac{4(n-1)n}2+(n-1)$

$\implies a_n=n^4+1$

4 0

3 years ago

Mariana received a bonus of $50 for working the day after thanksgiving ,plus his regular wage of 9.45 an hour .If his total wage

makkiz [27]

135.05 = 50 + 9.45x
85.05 = 9.45x
x = 85.05/9.45
x = 9
Mariana worked for 9 hours.

6 0

3 years ago

Read 2 more answers

Ammonia, nh3 (delta.hf = –46.2 kj), reacts with oxygen to produce water (delta.hf = –241.8 kj) and nitric oxide, no (delta.hf =

poizon [28]

The enthalpy change (ΔH) for the reaction given the data from the question is –900.8 KJ

<h3>Data obtained from the question</h3>

4NH₃ + 5O₂ —> 6H₂O + 4NO
Enthalpy of ammonia, NH₃ = –46.2 KJ
Enthalpy of Oxygen = 0 KJ
Enthalpy of water, H₂O = –241.8 KJ
Enthalpy of nitric oxide, NO = 91.3 KJ
Enthalpy change (ΔH) =?

<h3>How to determine the enthalpy change</h3>

ΔHrxn = ∑ΔH(products) - ∑ΔH(reactants)

ΔHrxn = ∑[H(H₂O) + H(NO)] - ∑[H(NH₃) + H(O₂)]

ΔHrxn = [(6 × –241.8) + (4 × 91.3)] – [(4 × –46.2) + (5×0)]

ΔHrxn = –1085.6 + 184.8

ΔHrxn = –900.8 KJ

Learn more about enthalpy change:

brainly.com/question/2364362

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2 years ago