All categories

2 years ago

Given the function f defined by the formula f(x)=2x+1 find the following

Mathematics

2 answers:

alexgriva [62]2 years ago

6 0

$f(4)=2\cdot4+1=9\\
f(-5)=2\cdot(-5)+1=-9\\
f(0)=2\cdot0+1=1\\
f\left(\dfrac{1}{2}\right)=2\cdot\dfrac{1}{2}+1=2$

Goryan [66]2 years ago

6 0

This is my answer to help you, go to the picture I uploaded.

You might be interested in

A bottle of water is 2/3 full. You drink the given portion of

Citrus2011 [14]

Answer:

1/2

Step-by-step explanation:

Step 1: You would multiply 2/3 by 3/4.

(Hint: if it is ____ of ____, you would most likely multiply)

Step 2: 2×3/3×4

Step 3: 2×3=6

Step 4: 3×4=12

Step 5: 6/12

Step 6: Simplify 6/12 = 3/6 = 1/2

Hope this helps :)

3 0

1 year ago

Read 2 more answers

Last year, the average monthly revenue at stamford times newspaper was 35,700. So far this year, the average has been 29% more t

monitta

Answer:

$46,053

Step-by-step explanation:

Convert 29% into a decimal, which is 0.29.
Multiply $35,700 by 29% as a decimal (0.29), which is 10,353.
Since the keyword is more, we want to add $35,700 and 10,353 (29% of $35,700). The answer will be $46,053.

6 0

2 years ago

What is an estimate for 17 divided by 1569

Y_Kistochka [10]

It will be 93 because it 92 but you have to round up with deimal number

5 0

3 years ago

Read 2 more answers

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

2 years ago

You are the new manager at Taco bell. A customer told you that they had to wait 13 minutes for their food.You approach your staf

Nady [450]

3900 ㅤ
Yes, it is unusual

ㅤ

6 0

1 year ago