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creativ13 [48]
3 years ago
7

3 12 27 48. what is the rule?​

Mathematics
2 answers:
Shkiper50 [21]3 years ago
8 0

Answer:

Step-by-step explanation:

3(1)^2=3, 3(2)^2=12, 3(3)^2=27, 3(4)^2=48

f(n)=3n^2, n≥1

pshichka [43]3 years ago
3 0

Answer:

this rule follow the airthmatics progression.

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A scientist conducted an experiment. Due to chemical reaction, the temperature of a compound rose 1/10 degree every 1/4 hour for
klio [65]

Answer:

The rate is \frac{2}{5} degree per hour.

Step-by-step explanation:

Given : A scientist conducted an experiment. Due to chemical reaction, the temperature of a compound rose \frac{1}{10} degree every \frac{1}{4} hour for a limited time.

To find : What was the rate, in degrees per hour that the temperature of the compound rose ?

Solution :

In every \frac{1}{4} hour the temperature of a compound rose \frac{1}{10} degree

To get  the rate, in degrees per hour,

In every 1 hour the temperature of a compound rose is

r=\dfrac{\frac{1}{10}}{\frac{1}{4}} degree

r=\dfrac{4}{10} degree/hr

r=\dfrac{2}{5} degree/hr

Therefore, the rate is \frac{2}{5} degree per hour.

5 0
3 years ago
Identify each shape as translation, rotation and reflection​
Wewaii [24]
1 reflection, 2 rotation,
7 0
3 years ago
X³ by a³ + a³by x³=p x³by a³- a³by ³=q eliminate x
sasho [114]

Answer:

Step-by-step explanation:

\frac{x^3}{a^3} +\frac{a^3}{x^3} =p\\\frac{x^3}{a^3} -\frac{a^3}{x^3} =q\\adding\\2\frac{x^3}{a^3} =p+q\\\frac{x^3}{a^3} =\frac{p+q}{2} \\substitute~in~first\\\frac{p+q}{2} +\frac{2}{p+q} =p\\\frac{(p+q)^2+4}{2(p+q)} =p\\(p+q)^2+4=2p(p+q)\\p^2+q^2+2pq+4-2p^2-2pq=0\\q^2-p^2+4=0\\q^2=p^2-4

6 0
2 years ago
A bullet travels at 1128 feet per second. How many miles per hour does that bullet go?
Katena32 [7]
1128 feet  = 1128/5280   miles
it travels at 1128/5280 miles per second

to convert to miles / hour  we multiply by 3600

so speed of the bullet  = (1128 * 3600) / 5280   =  769.1 mph to nearest tenth.

7 0
3 years ago
Prove the identity secxcscx(tanx+cotx)=2+tan^2x+cot^2x
svetlana [45]
Hello,

sec(x)= \dfrac{1}{cos(x)} \\

cosec(x)= \dfrac{1}{sin(x)} \\

sec(x)*cosec(x)*(tg(x)+cotg(x))=\dfrac{1}{cos(x)}* \dfrac{1}{sin(x)}*( \frac{sin(x)}{cos(x)} +\frac{cos(x)}{sin(x)})\\

= \dfrac{sin^2(x)+cos^2(x)}{sin^2x*cos^2x} \\

= \dfrac{1}{sin^2x*cos^2x} \\


==============================================================
2+tg^2(x)+cotg^2(x)=2+ \dfrac{sin^2x}{cos^2x} + \dfrac{cos^2x}{sin^2x} \\

=2+ \dfrac{sin^4x+cos^4x}{sin^2x*cos^2x} \\

=\dfrac{2*sin^2x*cos^2x+sin^4x+cos^4x}{sin^2x*cos^2x} \\

= \dfrac{(sin^2x+cos^2x)^2}{sin^2x*cos^2x}} \\

= \dfrac{1}{sin^2x*cos^2x}} 

8 0
3 years ago
Read 2 more answers
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