Ask question

Ask question

All categories

3 years ago

14

The function y= 5^x passes through the point

1 answer:

wlad13 [49]3 years ago

4 0

The correct answer is c :)
the point is described by two variables: x and y. it is written like that: (x, y). so the first variable in brackets is the x variable. the other is y. so pick every point a-d, put these values into the equation, solve it and check if the values before and after = are the same. if so - you are correct!
look:
c) (1, 5) so x=1, y=5
equation: y=5^x
putting values: 5=5^1 and this is true!

You might be interested in

Use Euler's method with step size 0.2 to estimate y(1), where y(x) is the solution of the initial-value problem y' = x2y − 1 2 y

irina [24]

Answer:

Therefore the value of y(1)= 0.9152.

Step-by-step explanation:

According to the Euler's method

y(x+h)≈ y(x) + hy'(x) ....(1)

Given that y(0) =3 and step size (h) = 0.2.

$y'(x)= x^2y(x)-\frac12y^2(x)$

Putting the value of y'(x) in equation (1)

$y(x+h)\approx y(x) +h(x^2y(x)-\frac12y^2(x))$

Substituting x =0 and h= 0.2

$y(0+0.2)\approx y(0)+0.2[0\times y(0)-\frac12 (y(0))^2]$

$\Rightarrow y(0.2)\approx 3+0.2[-\frac12 \times3]$ [∵ y(0) =3 ]

$\Rightarrow y(0.2)\approx 2.7$

Substituting x =0.2 and h= 0.2

$y(0.2+0.2)\approx y(0.2)+0.2[(0.2)^2\times y(0.2)-\frac12 (y(0.2))^2]$

$\Rightarrow y(0.4)\approx 2.7+0.2[(0.2)^2\times 2.7- \frac12(2.7)^2]$

$\Rightarrow y(0.4)\approx 1.9926$

Substituting x =0.4 and h= 0.2

$y(0.4+0.2)\approx y(0.4)+0.2[(0.4)^2\times y(0.4)-\frac12 (y(0.4))^2]$

$\Rightarrow y(0.6)\approx 1.9926+0.2[(0.4)^2\times 1.9926- \frac12(1.9926)^2]$

$\Rightarrow y(0.6)\approx 1.6593$

Substituting x =0.6 and h= 0.2

$y(0.6+0.2)\approx y(0.6)+0.2[(0.6)^2\times y(0.6)-\frac12 (y(0.6))^2]$

$\Rightarrow y(0.8)\approx 1.6593+0.2[(0.6)^2\times 1.6593- \frac12(1.6593)^2]$

$\Rightarrow y(0.6)\approx 0.8800$

Substituting x =0.8 and h= 0.2

$y(0.8+0.2)\approx y(0.8)+0.2[(0.8)^2\times y(0.8)-\frac12 (y(0.8))^2]$

$\Rightarrow y(1.0)\approx 0.8800+0.2[(0.8)^2\times 0.8800- \frac12(0.8800)^2]$

$\Rightarrow y(1.0)\approx 0.9152$

Therefore the value of y(1)= 0.9152.

4 0

3 years ago

I need help on number 6 plz!

expeople1 [14]

Woah that looks hard

8 0

2 years ago

Evaluate the expression when y = 4. 5 y 8

Elden [556K]

I suppose you mean 5y^8.
When y = 4, we have,
5(4)^8 = 5 x 65,536 = 327,680

7 0

3 years ago

Im just going to post a picture

xz_007 [3.2K]

Answer:

the picture is hard

Step-by-step explanation:

5 0

3 years ago

Read 2 more answers

What are some rational numbers

Radda [10]

Answer: 1,2,3,4,5,6,7,8,9,

Step-by-step explanation: all numbers that can be expressed as the quotient or fraction

hope this helps please mark brainliest if it did

5 0

3 years ago