How to find a perfect square

Mathematics

1 answer:

Rzqust [24]3 years ago

8 0

Answer:

Step-by-step explanation:

The square of a number n is denoted by n^2.

Examples: The square of 2 is 2^2, or 4; the square of 5 is 5^2, or 25. To find a perfect square of a number n, multiply the number by itself (obtaining the square of n).

You might be interested in

5 and below is having a discount sale for 8% off on all items. She wanted to get a shirt that was originally $25.00. How much wi

Lena [83]

AnswerNaruto fan?

Step-by-step explanation:

3 0

4 years ago

Read 2 more answers

HELP ME NOW I SPENT 99 POINTS TRYING LAST TIEM

marshall27 [118]

2.30 per gallon yghehhf

5 0

3 years ago

In this question, i is a unit vector due east and j is a unit vector due north. A cyclist rides at a speed of 4 m/s on a bearing

d1i1m1o1n [39]

Answer:

a) 1.04i + 3.86j

b) magnitude = 8; bearing = 302.7°

Step-by-step explanation:

The first diagram represents the velocity vector of the cyclist.

To express this vector in the form xi + yj, we have to find the components of the vector in the horizontal (i) and vertical (j) directions.

If we consider the horizontal component of the vector to be $x$ , and the vertical component to be $y$ , then:

• horizontal component ⇒ $sin (15^{\circ}) = \frac{x}{4}$

⇒ $x = 4\space\ sin(15^{\circ})$

⇒ $x \approx \bf 1.04$

• vertical component ⇒ $cos(15^{\circ}) = \frac{y}{4}$

⇒ $y = 4 \space\ cos(15^{\circ})$

⇒ $y \approx \bf 3.86$

Now that we have the values of both the horizontal and vertical component, we can write the vector in the form of xi + yj:

vector ⇒ 1.04i + 3.86j

The second diagram shows the first vector (red), the second vector (blue), and the resultant vector v (black). The dashed lines represent the components of the respective vectors.

To add two vectors given their magnitudes and direction, we have to add their components.

In order to find the horizontal and vertical components of the given vectors, we can use a method similar to that used above, so that:

○ For the first vector (magnitude 6):

• horizontal component ⇒ $x = 6 \space\ sin (60^{\circ})$