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2 years ago

69. The height of a triangle is 1 in more than the base. If the height is increased by 2 in while the base remains

Mathematics

1 answer:

Rainbow [258]2 years ago

6 0

Answer:

Base and height

b= 5

h=6

B) area

15square in

Step-by-step explanation:

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Solve for x. x^3=8/125 Enter your answer in the box as a fraction in simplest form.

stepan [7]

x^3=8/125
x^3 = 2^3 / 5^3
x = 2/5

hope it helps

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4 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 ⇒ $x = 6 \space\ sin (60^{\circ})$

⇒ $\bf 5.2$

• vertical component ⇒ $y = 6 \space\ cos(60^{\circ})$

⇒ $y = \bf 3$

○ For the second vector (magnitude 2):

• horizontal component ⇒ $x = 2 \space\ cos (40^{\circ})$

⇒ $\bf 1.5$

• vertical component ⇒ $y = 2 \space\ sin(40^{\circ})$

⇒ $\bf 1.3$

Now we can add the respective components together:

v = 5.2i + 3j + 1.5i + 1.3j

⇒ (5.2 + 1.5)i + (3 + 1.3)j

⇒ 6.7i + 4.3j

∴ Magnitude of v ⇒ $|v| = \sqrt{(6.7)^2 + (4.3)^2}$

⇒ $|v| \approx \bf 8$

To find the bearing of v, we have to first calculate the angle marked $\alpha$ :

$tan \alpha = \frac{4.3}{6.7}$

⇒ $\alpha = tan^{-1}(\frac{4.3}{6.7})$

⇒ $\alpha = \bf 32.7^{\circ}$

∴ Bearing = 270° + 32.7°

= 302.7°

8 0

2 years ago

What is the slope of the line?

CaHeK987 [17]

Answer:

#1 Slope= -3 #2 Slope=

Step-by-step explanation:

4-1/1-2=

3/-1=

= -3

-4-1/4-3=?

= -5/1

= -5

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3 years ago

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