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Montano1993 [528]
4 years ago
10

Which function family should be used to solve the following question?

Mathematics
1 answer:
Ulleksa [173]4 years ago
3 0

Answer:

I think d is the answer

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Dois triângulos semelhantes possuem razão entre suas áreas igual a 9. Se o perímetro de um deles é 10, o perímetro do outro deve
Digiron [165]

Answer:

O perímetro do outro triângulo deve ser ou de 30 unidades de comprimento, ou de 3.33 unidades de comprimento.

Step-by-step explanation:

Dois triângulos semelhantes possuem razão entre suas áreas igual a 9.

O perímetro tem grau um, enquanto a área tem grau 2. Isto implica que a razão entre os perímetros é a raiz quadrada da razão entre as áreas, então a razão entre os perímetros é de 3.

Se o perímetro de um deles é 10, o perímetro do outro deve ser:

Ou 10*3 = 30, ou \frac{10}{3} = 3.33

O perímetro do outro triângulo deve ser ou de 30 unidades de comprimento, ou de 3.33 unidades de comprimento.

5 0
3 years ago
2.5t=10 I need help lol
erik [133]

Answer:

t = 4

Step-by-step explanation:

10/2.5 = 4 how can you not get this ._.

I hope this was a joke

3 0
3 years ago
List the elements C={7n|n belongs to N, n<5}
Oksanka [162]

Answer:

C =  \{0,7,14,21,28\}

Step-by-step explanation:

Given

C \{7n: n < 5\}

Required

List the elements

Assume that n is not negative, the values of n are:

n = \{0,1,2,3,4\}

So, the values of C is:

C = 7 * n

C = 7 * \{0,1,2,3,4\}

Multiply each set element by 7

C =  \{7 *0,7 *1,7 *2,7 *3,7 *4\}

C =  \{0,7,14,21,28\}

5 0
3 years ago
A triangle has side lengths measuring 20 cm, 5 cm, and n cm. Which describes the possible values of n?
tekilochka [14]
Sum of two sides of the triangle should be more than 3d side.
n<20+5,
n<25

Third side should be more than difference of  two other sides.
n> 20-15
n>5

So, n should be
5<n<25.
6 0
4 years ago
Read 2 more answers
How many hours will Aaliyah finish the water? WILL MARK BRAINLIEST AND WILL REPORT ABSURD ANSWERS.
madam [21]
D 4 hours
4 times 7/8 equals 3.5 which is the same as 3 1/2
5 0
3 years ago
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