Ask question

Ask question

All categories

3 years ago

8

A squirrel found 48 acorns in a 3-day period. Each day the squirrel found 5 more acorns than the day before. How many acorns did

the squirrel find on the first day?

1 answer:

murzikaleks [220]3 years ago

7 0

Answer:

35

Step-by-step explanation:

3 days= 15 acorns so take 15 away from 48

You might be interested in

Whats your spotify user, i need some new playlists to listen to<br> mine is nllazzari

Firdavs [7]

Answer:

और किसने lil nas x रोबोक्स कॉन्सर्ट को हिट किया?

Step-by-step explanation:

5 0

3 years ago

Zolol [24]

Answer:

Exterior = 22.5 degrees

Interior = 157.5 degrees

Step-by-step explanation:

Exterior angles = 360/16 = 22.5

Interior Angles = 180 - 22.5 = 157.5

5 0

3 years ago

Help me with both questions pleaseeeee

12345 [234]

Answer:

Step-by-step explanation: not 100% sure about 5 but i think 5) 3p-2

6) a.2a+2a-2

3 0

3 years ago

What are the first five common multiples of five and six

Vladimir79 [104]

5 : 5, 10, 15, 20, 25
6 : 6, 12, 18, 24, 30

7 0

3 years ago

The point (1/3,1/4) lies on the terminal said of an angle. Find the exact value of the six trig functions and explain which func

katrin2010 [14]

Answer:

sine and cosec are inverse of each other.

cosine and sec are inverse of each other.

tan and cot are inverse of each other.

Step-by-step explanation:

Given point on terminal side of an angle ( $\frac{1}{3},\frac{1}4$ ).

Kindly refer to the attached image for the diagram of the given point.

Let it be point A( $\frac{1}{3},\frac{1}4$ )

Let O be the origin i.e. (0,0)

Point B will be ( $\frac{1}{3},0$ )

Now, let us consider the right angled triangle $\triangle OBA$ :

Sides:

$Base, OB = \frac{1}{3}\\Perpendicular, AB = \frac{1}{4}$

Using Pythagorean theorem:

$\text{Hypotenuse}^{2} = \text{Base}^{2} + \text{Perpendicular}^{2}\\\Rightarrow OA^{2} = OB^{2} + AB^{2}\\\Rightarrow OA^{2} = \frac{1}{3}^{2} + \frac{1}{4}^{2}\\\Rightarrow OA = \sqrt{\frac{1}{3}^{2} + \frac{1}{4}^{2}}\\\Rightarrow OA = \sqrt{\frac{4^2+3^2}{3^{2}.4^2 }}\\\Rightarrow OA = \frac{5}{12}$

$sin \angle AOB = \dfrac{Perpendicular}{Hypotenuse}$

$\Rightarrow sin \angle AOB = \dfrac{\frac{1}{4}}{\frac{5}{12}}\\\Rightarrow sin \angle AOB = \dfrac{3}{5}$

$cos\angle AOB = \dfrac{Base}{Hypotenuse}$

$\Rightarrow cos \angle AOB = \dfrac{\frac{1}{3}}{\frac{5}{12}}\\\Rightarrow cos\angle AOB = \dfrac{4}{5}$

$tan\angle AOB = \dfrac{Perpendicular}{Base}$

$\Rightarrow tan\angle AOB = \dfrac{3}{4}$

$cosec \angle AOB = \dfrac{Hypotenuse}{Perpendicular}$

$\Rightarrow cosec\angle AOB = \dfrac{5}{3}$

$sec\angle AOB = \dfrac{Hypotenuse}{Base}$

$\Rightarrow sec\angle AOB = \dfrac{5}{4}$

$cot\angle AOB = \dfrac{Base}{Perpendicular}$

$\Rightarrow cot\angle AOB = \dfrac{4}{3}$

3 0

3 years ago