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3 years ago

Математика срочно пожалуйста

Mathematics

1 answer:

azamat3 years ago

7 0

Step-by-step explanation:

2,3÷2,5=23÷25=0,92

0,3÷0,8=3÷8=0,375

1,7÷025=170÷25=6,8

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Use tens and ones to the 62 two different ways

sineoko [7]

Use 6 tens to get 60 and then 2 ones
use 5 tens to get 50 and 12 ones

4 0

3 years ago

PLZ HELP!

Tanya [424]

Answer:

B. 60cm^3

Step-by-step explanation:

The volume of a right triangular prism is

V = BH where B is the area of the triangle and H is height

B = 1/2 bh where b =5 and h =8

B = 1/2 *40 = 20

V = (20) * 3

V = 60 cm^3

We need to be careful to do the area of the triangle first

8 0

4 years ago

What are the solutions to 2 * (x - 7) ^ 2 = 32 ?

amm1812

Answer:

11 is answer

2 * (x - 7) ^ 2 = 32

(x-7)²=32/2

(x-7)²=16

x-7=√16

x-7=4

x=7+4=11

4 0

3 years ago

Read 2 more answers

Solve the initial value problems.

slavikrds [6]

Both equations are linear, so I'll use the integrating factor method.

The first ODE

$xy' + (x+1)y = 0 \implies y' + \dfrac{x+1}x y = 0$

has integrating factor

$\exp\left(\displaystyle \int\frac{x+1}x \, dx\right) =\exp\left(x+\ln(x)\right) = xe^x$

In the original equation, multiply both sides by eˣ :

$xe^x y' + (x+1) e^x y = 0$

Observe that

d/dx [xeˣ] = eˣ + xeˣ = (x + 1) eˣ

so that the left side is the derivative of a product, namely

$\left(xe^xy\right)' = 0$

Integrate both sides with respect to x :

$\displaystyle \int \left(xe^xy\right)' \, dx = \int 0 \, dx$

$xe^xy = C$

Solve for y :

$y = \dfrac{C}{xe^x}$

Use the given initial condition to solve for C. When x = 1, y = 2, so

$2 = \dfrac{C}{1\cdot e^1} \implies C = 2e$

Then the particular solution is

$\boxed{y = \dfrac{2e}{xe^x} = \dfrac{2e^{1-x}}x}$

The second ODE

$(1+x^2)y' - 2xy = 0 \implies y' - \dfrac{2x}{1+x^2} y = 0$

has integrating factor

$\exp\left(\displaystyle \int -\frac{2x}{1+x^2} \, dx\right) = \exp\left(-\ln(1+x^2)\right) = \dfrac1{1+x^2}$

Multiply both sides of the equation by 1/(1 + x²) :

$\dfrac1{1+x^2} y' - \dfrac{2x}{(1+x^2)^2} y = 0$

and observe that

d/dx[1/(1 + x²)] = -2x/(1 + x²)²

Then

$\left(\dfrac1{1+x^2}y\right)' = 0$

$\dfrac1{1+x^2}y = C$

$y = C(1 + x^2)$

When x = 0, y = 3, so

$3 = C(1+0^2) \implies C=3$

$\implies \boxed{y = 3(1 + x^2) = 3 + 3x^2}$

7 0

2 years ago

What is a “ridgid movement”

solniwko [45]

Definition 1. A rigid motion of the plane ( or an isometry ) is a motion whichpreserves distance.There are four basic rigid motions:(1) Reflection(2) Glide Reflection(3) Rotation(4) Translation

3 0

3 years ago

Математика срочно пожалуйста​ ​

Математика срочно пожалуйста