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Vika [28.1K]
3 years ago
6

(05.03 LC) 15 POINTS + BRANLIEST PLEASE HELP!!!!!

Mathematics
2 answers:
Vanyuwa [196]3 years ago
8 0
Initial value is the value of y where x = 0 (y-intercept).

You are already given that line contains point (0, 5). The y-coordinate is 5 therefore answer is C.
Please mark Brainliest
meriva3 years ago
4 0

Answer: 5

Step-by-step explanation:

We know that when a liner function is represented by a graph i.e. by a line, then its initial value is represented by the y intercept of the line.

In the given graph , we can see that the y-intercept of the function is point (0,5).

The y -intercept of the function : y=5

Then , the initial value of the function will also be 5.

Hence, the initial value of the function represented by this graph=5

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If the general solution of a differential equation is ​y(t)equals=Upper C e Superscript negative 4 t Baseline plus 7Ce−4t+7​, wh
Maru [420]

Answer:

y(t) = 3e^{-4t}+7

Step-by-step explanation:

y(t) = Ce^{-4t}+7

At <em>y</em>(0) = 10, when <em>t</em> = 0, <em>y</em> = 10.

y(0) = 10 = Ce^{-4\times0}+7

10 = C+7

C = 3

Hence, the solution is

y(t) = 3e^{-4t}+7

8 0
3 years ago
Read 2 more answers
Expand (2x+2)^6<br> How would you find the answer using the binomial theorem?
Yanka [14]

Answer:

Step-by-step explanation:

\displaystyle\\\sum\limits _{k=0}^n\frac{n!}{k!*(n-k)!}a^{n-k}b^k .\\\\k=0\\\frac{n!}{0!*(n-0)!}a^{n-0}b^0=C_n^0a^n*1=C_n^0a^n.\\\\ k=1\\\frac{n!}{1!*(n-1)!} a^{n-1}b^1=C_n^1a^{n-1}b^1.\\\\k=2\\\frac{n!}{2!*(n-2)!} a^{n-2}b^2=C_n^2a^{n-2}b^2.\\\\k=n\\\frac{n!}{n!*(n-n)!} a^{n-n}b^n=C_n^na^0b^n=C_n^nb^n.\\\\C_n^0a^n+C_n^1a^{n-1}b^1+C_n^2a^{n-2}b^2+...+C_n^nb^n=(a+b)^n.

\displaystyle\\(2x+2)^6=\frac{6!}{(6-0)!*0!} (2x)^62^0+\frac{6!}{(6-1)!*1!} (2x)^{6-1}2^1+\frac{6!}{(6-2)!*2!}(2x)^{6-2}2^2+\\\\ +\frac{6!}{(6-3)!*3!} (2a)^{6-3}2^3+\frac{6!}{(6-4)*4!} (2x)^{6-4}b^4+\frac{6!}{(6-5)!*5!}(2x)^{6-5} b^5+\frac{6!}{(6-6)!*6!}(2x)^{6-6}b^6. \\\\

(2x+2)^6=\frac{6!}{6!*1} 2^6*x^6*1+\frac{5!*6}{5!*1}2^5*x^5*2+\\\\+\frac{4!*5*6}{4!*1*2}2^4*x^4*2^2+  \frac{3!*4*5*6}{3!*1*2*3} 2^3*x^3*2^3+\frac{4!*5*6}{2!*4!}2^2*x^2*2^4+\\\\+\frac{5!*6}{1!*5!} 2^1*x^1*2^5+\frac{6!}{0!*6!} x^02^6\\\\(2x+2)^6=64x^6+384x^5+960x^4+1280x^3+960x^2+384x+64.

8 0
1 year ago
An equation is shown below:
Ad libitum [116K]

Answer:

solution given:

4(2x − 5) = 4

2x-5=4/4

2x=1+5

x=6

Step-by-step explanation:

Part A:One

Part B:x=6

3 0
2 years ago
What is the reverse of add 1
Aleksandr-060686 [28]

Subtract 1

Hope this helps! ;)

4 0
2 years ago
Read 2 more answers
HELP ME PLEASE ASAP!!
timama [110]

Answer:

8

Step-by-step explanation:


8 0
3 years ago
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