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Mariana [72]
3 years ago
6

HELP I NEED TO DO THIS NOW

Mathematics
1 answer:
Elan Coil [88]3 years ago
3 0

Answer: The answer is 1:2, i hope it helps.

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Given the sequence 2, 5, 8, 11, 14, ..., which term is 59? (Hint: find n.)
____ [38]
N = 3  ( the next number is 3 plus the previous number
 sequence formula = an = 3(n-1) +2

so for term the term that =59

59 = 3(n-1) +2

distribute:
59 = 3n -3+2

59 = 3n -1

60 = 3n
n = 60/3
n = 20

59 is the 20th term



8 0
3 years ago
There are 7 cows and 5 chicken in the farmer's filed
Vanyuwa [196]

Answer:

7:12

Step-by-step explanation:

Total no of animals in the field = 7 + 5 = 12

Ratio of cows 2 all d animals in the field =

\frac{no \: of \: cows}{total \: no \: of \: animals \: on \: the \: field}

Ratio = 7/12 = 7:12

8 0
3 years ago
En un triángulo rectángulo A es un ángulo agudo y Sen A = 4/5 ¿Cuál será el valor de Tan A?
Nonamiya [84]

Answer:

\displaystyle \tan A=\frac{4}{3}

Step-by-step explanation:

<u>Funciones Trigonométricas</u>

La identidad principal en trigonometría es:

sen^2A+cos^2A=1

Si sabemos que A es un ángulo agudo (que mide menos de 90°), su seno y coseno son positivos.

Dado que Sen A = 4/5, calculamos el coseno:

cos^2A=1-sen^2A

Sustituyendo:

\displaystyle cos^2A=1-\left(\frac{4}{5}\right)^2

\displaystyle cos^2A=1-\frac{16}{25}

\displaystyle cos^2A=\frac{25-16}{25}

\displaystyle cos^2A=\frac{9}{25}

Tomando raíz cuadrada:

\displaystyle cos\ A=\sqrt{\frac{9}{25}}=\frac{3}{5}

La tangente se define como:

\displaystyle \tan A=\frac{sen\ A}{cos\ A}

Substituyendo:

\displaystyle \tan A=\frac{\frac{4}{5}}{\frac{3}{5}}

\displaystyle \tan A=\frac{4}{3}

6 0
3 years ago
1-2m-5m=15 multistep homework
Montano1993 [528]

1-2m-5m=15

combine like terms

1 -7m = 15

subtract 1 from each side

1-1-7m = 15-1

simplify

-7m = 14

divide by -7

-7m/-7 = 14/-7

simplify

m = -2


6 0
3 years ago
Help please with this question
choli [55]

Answer:

They lose about 2.79% in purchasing power.

Step-by-step explanation:

Whenever you're dealing with purchasing power and inflation, you need to carefully define what the reference is for any changes you might be talking about. Here, we take <em>purchasing power at the beginning of the year</em> as the reference. Since we don't know when the 6% year occurred relative to the year in which the saving balance was $200,000, we choose to deal primarily with percentages, rather than dollar amounts.

Each day, the account value is multiplied by (1 + 0.03/365), so at the end of the year the value is multiplied by about

... (1 +0.03/365)^365 ≈ 1.03045326

Something that had a cost of 1 at the beginning of the year will have a cost of 1.06 at the end of the year. A savings account value of 1 at the beginning of the year would purchase one whole item. At the end of the year, the value of the savings account will purchase ...

... 1.03045326 / 1.06 ≈ 0.9721 . . . items

That is, the loss of purchasing power is about ...

... 1 - 0.9721 = 2.79%

_____

If the account value is $200,000 at the beginning of the year in question, then the purchasing power <em>normalized to what it was at the beginning of the year</em> is now $194,425.14, about $5,574.85 less.

7 0
3 years ago
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