Ask question

Ask question

All categories

3 years ago

15

The amount of snow on the ground increased by 4 inches between 1 p.m. and 6 p.m. What integer represents the amount of snow that

fell during those five hours?

1 answer:

scZoUnD [109]3 years ago

3 0

Answer:

The snow from each event melts off before the next accumulation and no snow is on the ground at your scheduled time of observation. The total snowfall for that reporting 24-hour day is the sum of the three separate snow squalls, 6.7 inches, even though the snow depth on your board at observation time was zero. Snow often melts as it lands. your welcome!

You might be interested in

I'm stuck on a question I'm helping my brother with. We need to find out if y = -x squared is linear or not.

Anna71 [15]

Y=-x^2 is non linear due to the x being squared

4 0

3 years ago

I need this to be specific and quick! 35 points!

g100num [7]

D) these two figures are similar they don't look similar at first because the X'Z'Y' is more upwards. But they are congruent meaning they are similar or exactly the same almost. This basically tells me that I need to use translation.

E) I can use reflection to get XYZ into X'Y'Z' by reflecting it across the x-axis. I could also use translation and move it up 6 and then move it to the right.

C) The last one I could use is rotation and move it around around 45* or something close to the the other one.

7 0

4 years ago

Enrico is filling his pool. The pool has 3,000 gallons of water now. The water hose used by Enrico puts 500 gallons per hour in

Feliz [49]

Answer:

3,000=500x

Step-by-step explanation:

6 0

3 years ago

Read 2 more answers

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

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

<em>Find the value of the length side adjacent to the angle alpha</em>

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

How do I find x on the attached right triangle<br> ?

ioda

The value of x from the given triangle is 32√3/3

<h3>SOH CAH TOA identity</h3>

In order to determine the value of x, first we need to determine the hypotenuse side of the smaller triangle.

sin 60 = 8√2/H

H = 8√2/sin60

H = 8√2 * 2/√3

H = 16√2/√3
H = 16√6/3

For the value of x, we will use the expression

sin45 = (16√6/3)/x

1/√2 = 16√6/3x
3x = 16√12
3x = 32√3

x = 32√3/3

Hence the value of x from the given triangle is 32√3/3

Learn more on SOH CAH TOA here: brainly.com/question/20734777

#SPJ1

4 0

2 years ago