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borishaifa [10]
3 years ago
11

Please help ASAP 1. 21:24 = 28:x

Mathematics
1 answer:
dexar [7]3 years ago
3 0

Answer:

Step-by-step explanation:

I think it is negative 2

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Answer the question for 20 points!!!!!
Ira Lisetskai [31]

Answer:

4/6

Step-by-step explanation:

ok so you have 2/3 for 1 group. so for 2 groups youll just have to add 2/3 + 2/3 and that gives u 4/6

8 0
2 years ago
Hola alguien puede ayudarme a resolver estos ejercicios por favor Ejercicios de intervalos 1.- Escribe en todas las formas posib
yawa3891 [41]

Answer:

Tenemos 3 formas de escribir cada intervalo, una es con solo texto, la otra es usando los símbolos { <, >, ≤, ≥ }, y la última es con intervalos, donde:

( se usa para " < x "

) se usa para " > x "

[ se usa para  " ≤ x "

] se usa para   " ≥ x "

a) -2 ≤ x ≤3

Esto se escribe como:

"números mayores o iguales que -2 y menores o iguales que 3"

y

[-2, 3]

b) Números mayores que –1

Las otras formas de escribir esto son:

-1 < x

y no tenemos limite mayor, entonces:

(-1, ∞)

con -∞ y ∞ siempre se usan los símbolos ( o )  

c) (–[infinity], –5]  

Las otras notaciones son:

"Un numero menor o igual que -5"

x ≤ -5

d)  Números mayores o iguales que –7 y menores que 19

Las otras notaciones son:

-7 ≤ x < 19

[-7, 19)

e) Números mayores que 9 y menores que 5.

Las otras formas de escribir esto son:

5 < x < 9

(5, 9)

8 0
3 years ago
What are 2 points on the graph for 6x-5y=25
zepelin [54]
To find any two random point on the graph of 6x-5y=25, we can just plug in any two values for x and solve for y, and find the 2 points in the form of (x,y).

So lets plug in x=0.

\sf 6x-5y=25
\sf 6(0)-5y=25
\sf -5y=25
\sf y=-5
So one point is (0,-5)

And to find another point, we can plug in x=1

\sf 6x-5y=25
\sf 6(1)-5y=25
\sf 6-5y=25
\sf -5y=25-6
\sf -5y=19
\sf y=19/-5
\sf y=-3.8
So another point is (1,-3.8)
8 0
3 years ago
Find the interval(s) where the function is increasing and the interval(s) where it is decreasing. (Enter your answers using inte
Mashcka [7]

Answer:

f'(x) > 0 on (-\infty ,\frac{1}{e}) and f'(x)<0 on(\frac{1}{e},\infty)

Step-by-step explanation:

1) To find and interval where any given function is increasing, the first derivative of its function must be greater than zero:

f'(x)>0

To find its decreasing interval :

f'(x)

2) Then let's find the critical point of this function:

f'(x)=\frac{\mathrm{d} }{\mathrm{d} x}[6-2^{2x}]=\frac{\mathrm{d} }{\mathrm{d}x}[6]-\frac{\mathrm{d}}{\mathrm{d}x}[2^{2x}]=0-[ln(2)*2^{2x}*\frac{\mathrm{d}}{\mathrm{d}x}[2x]=-ln(2)*2^{2x}*2=-ln2*2^{2x+1\Rightarrow }f'(x)=-ln(2)*2^{2x}*2\\-ln(2)*2^{2x+1}=-2x^{2x}(ln(x)+1)=0

2.2 Solving for x this equation, this will lead us to one critical point since x'  is not defined for Real set, and x''=\frac{1}{e}≈0.37 for e≈2.72

x^{2x}=0 \ni \mathbb{R}\\(ln(x)+1)=0\Rightarrow ln(x)=-1\Rightarrow x=e^{-1}\Rightarrow x\approx 2.72^{-1}x\approx 0.37

3) Finally, check it out the critical point, i.e. f'(x) >0 and below f'(x)<0.

8 0
3 years ago
What Is the sum 6x^2 + 2x - 4 and 2x^2 - 4x - 6 ​
Anni [7]

Answer:

8x^2-2x-10 is the answer

5 0
3 years ago
Read 2 more answers
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