A small submarine travels at a speed of 2.4 km per hour. How far will the submarine travel in 0.25 hours

A textbook has 428 pages numbered in order starting with 1. You flip

BartSMP [9]

Here's link to the answer:

linkcutter.ga/gyko

8 0

2 years ago

Read 2 more answers

9 inches

Semmy [17]

What is 9 inches? Is 9 inches the diameter? Or the radius? What?

The equation for circumference is pi x diameter.

So assuming the 9 inches is the diameter:

Pi x 9 = 28.2743339

Circumference = 28.274 (3 d.p.)

7 0

3 years ago

I’m sure this is easy but i'm gonna ask anyway

Arada [10]

No, it's always okay to ask if you don't understand a question. Anyway, for this problem, we just have to plug in the value for V to solve for P :

$P = 4V^{ \frac{1}{3}} \\ P = 4(125)^{ \frac{1}{3}}$

A number to the power of 1/3 is the same as taking a cube root.

$P = 4(125)^{ \frac{1}{3}} \\ P = 4(5) \\ P=20$

The perimeter of one face of the cube is 20m.

5 0

3 years ago

Find sin(a)&cos(B), tan(a)&cot(B), and sec(a)&csc(B).

Reil [10]

Answer:

Part A) $sin(\alpha)=\frac{4}{7},\ cos(\beta)=\frac{4}{7}$

Part B) $tan(\alpha)=\frac{4}{\sqrt{33}},\ tan(\beta)=\frac{4}{\sqrt{33}}$

Part C) $sec(\alpha)=\frac{7}{\sqrt{33}},\ csc(\beta)=\frac{7}{\sqrt{33}}$

Step-by-step explanation:

Part A) Find $sin(\alpha)\ and\ cos(\beta)$

we know that

If two angles are complementary, then the value of sine of one angle is equal to the cosine of the other angle

In this problem

$\alpha+\beta=90^o$ ---> by complementary angles

so

$sin(\alpha)=cos(\beta)$

Find the value of $sin(\alpha)$ in the right triangle of the figure

$sin(\alpha)=\frac{8}{14}$ ---> opposite side divided by the hypotenuse

simplify

$sin(\alpha)=\frac{4}{7}$

therefore

$sin(\alpha)=\frac{4}{7}$

$cos(\beta)=\frac{4}{7}$

Part B) Find $tan(\alpha)\ and\ cot(\beta)$

we know that

If two angles are complementary, then the value of tangent of one angle is equal to the cotangent of the other angle

In this problem

$\alpha+\beta=90^o$ ---> by complementary angles

so

$tan(\alpha)=cot(\beta)$

Find the value of the length side adjacent to the angle alpha

Applying the Pythagorean Theorem

Let

x ----> length side adjacent to angle alpha

$14^2=x^2+8^2\\x^2=14^2-8^2\\x^2=132$

$x=\sqrt{132}\ units$

simplify

$x=2\sqrt{33}\ units$

Find the value of $tan(\alpha)$ in the right triangle of the figure

$tan(\alpha)=\frac{8}{2\sqrt{33}}$ ---> opposite side divided by the adjacent side angle alpha

simplify

$tan(\alpha)=\frac{4}{\sqrt{33}}$

therefore

$tan(\alpha)=\frac{4}{\sqrt{33}}$

$tan(\beta)=\frac{4}{\sqrt{33}}$

Part C) Find $sec(\alpha)\ and\ csc(\beta)$

we know that

If two angles are complementary, then the value of secant of one angle is equal to the cosecant of the other angle

In this problem

$\alpha+\beta=90^o$ ---> by complementary angles

so

$sec(\alpha)=csc(\beta)$

Find the value of $sec(\alpha)$ in the right triangle of the figure

$sec(\alpha)=\frac{1}{cos(\alpha)}$

Find the value of $cos(\alpha)$

$cos(\alpha)=\frac{2\sqrt{33}}{14}$ ---> adjacent side divided by the hypotenuse

simplify

$cos(\alpha)=\frac{\sqrt{33}}{7}$

therefore

$sec(\alpha)=\frac{7}{\sqrt{33}}$

$csc(\beta)=\frac{7}{\sqrt{33}}$

6 0

3 years ago

Correct answer will get brainliest

sineoko [7]

Answer:

B 2/3

Step-by-step explanation:

it's positive 2 over 3

up two and over three

because it's rise over run

so your answer would be the 2nd option or B

I hope this helps! have a nice day/night, blessings, xx, nm <3 :)

8 0

2 years ago