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3 years ago

Help QUICKLY please ill mark brainliest

Mathematics

2 answers:

mash [69]3 years ago

8 0

Answer:

the height of NBA basketball players, in meters

Step-by-step explanation:

I believe we can be pretty sure that here's no NBA players of heights 1.1m or 3.9m - the second height would be way over the height of the tallest man on earth

natta225 [31]3 years ago

7 0

Answer:

D but I'm not 100 percent sure

Step-by-step explanation:

so sorry if I'm wrong

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I need to change the subject to x

ira [324]

Answer:

h) (x+4t²)/4= 3V

x+4t²= 4*3V

x+4t²=12V

x=12V-4t²

MN=√(3x-2)

(MN)²=3x-2

3x-2=(MN)²

3x=(MN)²+2

x= [(MN)²+2]/3

7 0

3 years ago

Help..................

Sindrei [870]

Its h the answer is h

4 0

3 years ago

Read 2 more answers

3x- 5=1 what does x represent

V125BC [204]

Answer:

x=2

Step-by-step explanation:

3x=5+1

3x=6

x=6/3

x=2

8 0

3 years ago

Read 2 more answers

If a circle has a diameter of 30 meters, which expression gives its area in

sattari [20]

Answer:

C. 15²π

General Formulas and Concepts:

Pre-Algebra

Order of Operations: BPEMDAS

Brackets
Parenthesis
Exponents
Multiplication
Division
Addition
Subtraction

Left to Right

Geometry

Diameter: d = 2r
Area of a Circle: A = πr²

Step-by-step explanation:

Step 1: Define

d = 30 m

Step 2: Find Area

Substitute [D]: 30 m = 2r
Isolate r: 15 m = r
Rewrite: r = 15 m
Substitute [AC]: A = π(15 m)²
Rearrange: A = 15²π

3 0

3 years ago

URGENTE, POR FAVOR

Firlakuza [10]

$\mathbf A=\begin{bmatrix}a_{1,1}&a_{1,2}\\a_{2,1}&a_{2,2}\end{bmatrix}$

$a_{i,j}=2i-j\implies\mathbf A=\begin{bmatrix}2\cdot1-1&2\cdot1-2\\2\cdot2-1&2\cdot2-2\end{bmatrix}=\begin{bmatrix}1&0\\3&2\end{bmatrix}$

Por la misma razón,

$\mathbf B=\begin{bmatrix}1-1&2-1\\1-2&2-2\end{bmatrix}=\begin{bmatrix}0&1\\-1&0\end{bmatrix}$

Entonces,

$4\mathbf A=\begin{bmatrix}4\cdot1&4\cdot0\\4\cdot3&4\cdot2\end{bmatrix}=\begin{bmatrix}4&0\\12&8\end{bmatrix}$

$\mathbf A-\mathbf B=\begin{bmatrix}1&0\\3&2\end{bmatrix}-\begin{bmatrix}0&1\\-1&0\end{bmatrix}=\begin{bmatrix}1-0&0-1\\3-(-1)&2-0\end{bmatrix}=\begin{bmatrix}1&-1\\4&2\end{bmatrix}$

$\mathbf B^\top=(b_{j,i})_{2\times2}=\begin{bmatrix}b_{1,1}&b_{2,1}\\b_{1,2}&b_{2,2}\end{bmatrix}=\begin{bmatrix}0&-1\\1&0\end{bmatrix}$

Yo no reconozco la notación "Ae"... Puedes explicarla?

$a_{1,1}\cdot b_{1,1}=1\cdot0=0$

$a_{2,2}\cdot b_{1,2}=2\cdot1=2$

6 0

3 years ago