All categories

3 years ago

The of choice are ways epress 0.4 which is not

Mathematics

1 answer:

Ann [662]3 years ago

6 0

Answer:

NO IT IS 4

Step-by-step explanation:

You might be interested in

Simplify completely"""

Wewaii [24]

Answer:

$\displaystyle \frac{2(x + 2)}{x - 10}$

General Formulas and Concepts:

Pre-Algebra

Order of Operations: BPEMDAS

Brackets
Parenthesis
Exponents
Multiplication
Division
Addition
Subtraction

Left to Right

Algebra I

Terms/Coefficients
Factoring

Step-by-step explanation:

Step 1: Define

$\displaystyle \frac{2x^2 + 20x + 32}{x^2 - 2x - 80}$

Step 2: Simplify

[Fraction] Factor numerator: $\displaystyle \frac{2(x + 2)(x + 8)}{x^2 - 2x - 80}$
[Fraction] Factor denominator: $\displaystyle \frac{2(x + 2)(x + 8)}{(x - 10)(x + 8)}$
[Fraction] Divide: $\displaystyle \frac{2(x + 2)}{x - 10}$

4 0

2 years ago

What is the solution of -27 = m/9 ? A. -243 B. -3 C. 3 D. 243

asambeis [7]

Answer: m = -243

Explanation:

• Solve

-27 = m/9

•Multiply 9 by 27

m/9 * -27

• Result

m = -243

8 0

3 years ago

Read 2 more answers

Convert into 24 hours clock.4.17pm

yulyashka [42]

Answer:

16:17

Step-by-step explanation:

when it's pm, add 12 to the 12 hr time

8 0

2 years ago

The solution system to 3y-2x=-9 and y=-2x+5

Alecsey [184]

Answer:

$\boxed{(3,-1)}$

Step-by-step explanation:

Hey there!

Well to find the solution the the given system,

3y - 2x = -9

y = -2x + 5

So to find x lets plug in -2x + 5 for y in 3y - 2x = -9.

3(-2x + 5) - 2x = -9

Distribute

-6x + 15 - 2x = -9

-8x + 15 = -9

-15 to both sides

-8x = -24

Divide -8 to both sides

x = 3

Now that we have x which is 3, we can plug in 3 for x in y = -2x + 5.

y = -2(3) + 5

y = -6 + 5

y = -1

So the solution is (3,-1).

Hope this helps :)

4 0

2 years ago

let a = (a1, a2) and b = (b1, b2) and c = (c1,c2) be three non zero vectors. if a1b2 - a2b1 is not equal to 0. then show three a

Ksenya-84 [330]

Consider the contrapositive of the statement you want to prove.

The contrapositive of the logical statement

p ⇒ q

¬q ⇒ ¬p

In this case, the contrapositive claims that

"If there are no scalars α and β such that c = αa + βb, then a₁b₂ - a₂b₁ = 0."

The first equation is captured by a system of linear equations,

$\begin{cases}c_1 = \alpha a_1 + \beta b_1\\ c_2 = \alpha a_2 + \beta b_2\end{cases}$

or in matrix form,

$\begin{pmatrix}c_1\\c_2\end{pmatrix} = \begin{pmatrix}a_1&b_1\\a_2&b_2\end{pmatrix}\begin{pmatrix}\alpha\\\beta\end{pmatrix}$

If this system has no solution, then the coefficient matrix on the right side must be singular and its determinant would be

$\begin{vmatrix}a_1&b_1\\a_2&b_2\end{vmatrix} = a_1b_2-a_2b_1 = 0$

and this is what we wanted to prove. QED

3 0

2 years ago