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Bezzdna [24]
3 years ago
8

Find the length indicated.​

Mathematics
1 answer:
svetlana [45]3 years ago
7 0
The answer is 6 because if the total length is 10 and part of the total length is 4, 10-4 is 6
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Is an estimate for the qoutient of a division problem involving decimals always sometimes or never less than the actual qoutient
SVETLANKA909090 [29]
Well it actually is less than quotient because the number you're dividing by is always smaller than the quotient.
3 0
3 years ago
When solving negative one over three (x − 15) = −4, what is the correct sequence of operations?
iogann1982 [59]
-1/3(x - 15) = -4

step 1 : multiply both sides by -3, step 2 : add 15 to both sides

because when u multiply both sides by -3, this cancels out the -1/3 on the left side leaving u with : x - 15 = -4 * -3.....x - 15 = 12.....and then u would add 15 to both sides....giving u : x = 12 + 15.....x = 27
7 0
3 years ago
2. Un auto viaja a 60 km/h y se detiene a los 2 segundos. qué distancia recorrió en el
posledela

Se calcula que la aceleración es - 8,35 m/s.

<h3 /><h3>¿Qué es la aceleración?</h3>

Ahora sabemos lo siguiente de la pregunta;

Velocidad inicial (u) = 60 km/h o 16,7 m/s

Tiempo empleado (t) = 2 s

Dado que;

v = tu + en

v = 0 m/s porque el carro se detenía

tu = -en

a = -(u/t)

 a= -1(6,7 m/s/2 s)

a =- 8,35 m/s

La aceleración es negativa porque la velocidad disminuye con el tiempo.

Obtenga más información sobre la aceleración: brainly.com/question/12550364

#SPJ1

5 0
1 year ago
Which of the following can be used to create a regular tessellation?
svlad2 [7]
I think it’s d because it’s like similar
5 0
3 years ago
Read 2 more answers
What is the nonpermissible value of x in 5x/6x+3
cestrela7 [59]

Answer: -\frac{1}{2}

<u>Step-by-step explanation:</u>

\frac{5x}{6x+3}

Restriction: 6x + 3 ≠ 0     <em>denominator cannot be zero</em>

                   <u>     -3  </u>  <u> -3 </u>

                    6x      ≠ -3

                 <u> ÷6       </u>   <u>÷6 </u>

                      x       ≠ -\frac{1}{2}

4 0
3 years ago
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