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3 years ago

If tan 0 = 6.25, then cot 0 =

Mathematics

1 answer:

Aliun [14]3 years ago

4 0

Answer:

cotΘ = 0.16

Step-by-step explanation:

Using the trigonometric identity

cot x = $\frac{1}{tanx}$ , then

cotΘ = $\frac{1}{6.25}$ = 0.16

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You live on a triangular piece of land with a base of 121 yards and

morpeh [17]

The area of triangular piece of land in acres is 1 acre.

Further Explanation:

In order to find the area of the triangular piece of land in acres we have to find the area in yards

<u>Given:</u>

Base = 121 yards

Height = 80 yards

$Area = \frac{1}{2} * base * height\\ = 0.5 * 121 *80\\=4840\ square\ yards\\Square\ Yards\ in\ one\ acre=4840\ square\ yards\\So,\\Area\ in\ acres = \frac{4840}{4840} \\= 1\ acre$

So, the area of triangular piece of land in acres is 1 acre.

Keywords: Area of triangle, conversion from yards to acre

Learn More at:

#LearnwithBrainly

6 0

3 years ago

A concession stand sells hamburgers. The revenue from the hamburgers is (30x + 45) dollars. The price of a hamburger is $5. Writ

kiruha [24]

Answer:

5(6x+9)

Step-by-step explanation:

(30x+45)

30=(5)6x

45=(5)9

5(6x+9)

5 0

3 years ago

Which expression is equivalent to 3 (x-9) + 6?

vlada-n [284]

Answer: 3x-27

Step-by-step explanation:

6 0

3 years ago

Read 2 more answers

Tina was asked to determine the possible dimensions of a given rectangle whose area is 12a4b3-18a6b3-72a7b3. Tina stated that th

Rufina [12.5K]

Answer:

Tina is correct

Step-by-step explanation:

Given

$Area = 12a^4b^3 - 18a^6b^3 - 72a^7b^3$

Required

State if $3a^4b^3(4-6a^2-24a^3)$ is a possible dimension

To do this, we simply expand $3a^4b^3(4-6a^2-24a^3)$

$3a^4b^3(4-6a^2-24a^3)$

$3a^4b^3 * 4-3a^4b^3 * 6a^2-3a^4b^3 * 24a^3$

$12a^4b^3 - 18a^{4+2}b^3 * -72a^{4+3}b^3$

$12a^4b^3 - 18a^{6}b^3 * -72a^{7}b^3$

By comparison, the result of the expansion

$12a^4b^3 - 18a^{6}b^3 * -72a^{7}b^3$

and the given expression

$Area = 12a^4b^3 - 18a^6b^3 - 72a^7b^3$

are the same.

<em>Hence, Tina is correct</em>

7 0

3 years ago

50 points, question is down below.

maks197457 [2]

$\dfrac{\frac{3}{x-1}-4}{2-\frac{2}{x-1}}=\left(\dfrac{3}{x-1}-\dfrac{4(x-1)}{x-1}\right):\left(\dfrac{2(x-1)}{x-1}-\dfrac{2}{x-1}\right)\\\\=\dfrac{3-4x+4}{x-1}\cdot\dfrac{2x-2-2}{x-1}=\dfrac{-4x+7}{x-1}\cdot\dfrac{x-1}{2x-4}\\\\=\dfrac{-4x+7}{1}\cdot\dfrac{1}{2(x-2)}=\dfrac{-4x+7}{2(x-2)}$

6 0

3 years ago

Read 2 more answers