Ask question

Ask question

All categories

3 years ago

7

Two consecutive odd integers have a sum of 36

1 answer:

harkovskaia [24]3 years ago

3 0

Two consecutive odd numbers are n and n+2.

n+n+2=36

2n+2=36

2n=34

n=17

So the two integers are 17 and 19

You might be interested in

Write the rational number as a decimal -8/5 -9 3/8 -9/11 6 7/27

marissa [1.9K]

$-\frac{8}{5}$ = -1.6

$-9\frac{3}{8}$ = -11.625

$-\frac{9}{11}$ = -0.81818...

$6\frac{7}{27}$ = 2.4814

7 0

2 years ago

Tyrianne next solved a quadratic equation her work shown below in which step did Tyrianne make an error 1/2(x+4)^2-3=29

Alisiya [41]

Answer:

step 2

Step-by-step explanation:

3 0

3 years ago

How do you get 85 tens in 23 ones in standard form ?

Andrew [12]

The answer is 873 hope this will help you.

7 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

Simplify this expression.<br> 3^-3

Sphinxa [80]

Answer:

1/27 is the answer

Step-by-step explanation:

It is 1/(3^3) which is 1/27

4 0

2 years ago