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2 years ago

15

In which number is the digit 8 ten times larger than it is in the number 18?

1 answer:

ollegr [7]2 years ago

8 0

Answer:

387 contains 80 and 8 x 10 = 80

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HELP!!!!!!!!!! PLEASE!!!!!!!WILL MARK AS BRAINLIEST

Goshia [24]

<t =33 since triangle RUT is isosceles (RU = TU)

<r + <u + <t = 180 triangle = 180

33 + <u + 33 = 180

66 + <u = 180

< u =114

<rus = < sut from the diagram

<rus + <sut = <u

x + x = 114

2x = 114

divide by 2

x = 57

6 0

3 years ago

Read 2 more answers

A triangular prism is 17 yards long and has a triangular face with a base of 12 yards and a height of 8 yards. The other two sid

Lisa [10]

Answer:

640 yard²

Step-by-step explanation:

From the question,

The surface are of a triangular prism is given as

A = bh+2ls+lb.................. Equation 1

Where A = Surface area of the triangular prism, b = base of of the triangular face, h = height of the triangular face, l = length of the triangular prism, s = length of one of the other sides of the triangle

Given: l = 17 yards, b = 12 yards, h = 8 yards, s = 10 yards.

Substitute these values into equation 1

A = [(12×8)+(2×17×10)+(17×12)]

A = 96+340+204

A = 640 Yard²

6 0

3 years ago

i f N(-7,-1) is a point on the terminal side of ∅ in standard form, find the exact values of the trigonometric functions of ∅.

Natali [406]

Answer:

$sin\ \varnothing = \frac{-1}{10}\sqrt 2$

$cos\ \varnothing = \frac{-7}{10}\sqrt 2$

$tan\ \varnothing = \frac{1}{7}$

$cot\ \varnothing = 7$

$sec\ \varnothing = \frac{-5}{7}\sqrt 2$

$csc\ \varnothing = -5\sqrt 2$

Step-by-step explanation:

Given

$N = (-7,-1)$ --- terminal side of $\varnothing$

Required

Determine the values of trigonometric functions of $\varnothing$ .

For $\varnothing$ , the trigonometry ratios are:

$sin\ \varnothing = \frac{y}{r}$ $cos\ \varnothing = \frac{x}{r}$ $tan\ \varnothing = \frac{y}{x}$

$cot\ \varnothing = \frac{x}{y}$ $sec\ \varnothing = \frac{r}{x}$ $csc\ \varnothing = \frac{r}{y}$

Where:

$r^2 = x^2 + y^2$

$r = \sqrt{x^2 + y^2$

In $N = (-7,-1)$

$x = -7$ and $y = -1$

So:

$r = \sqrt{(-7)^2 + (-1)^2$

$r = \sqrt{50$

$r = \sqrt{25 * 2$

$r = \sqrt{25} * \sqrt 2$

$r = 5 * \sqrt 2$

$r = 5 \sqrt 2$

<u>Solving the trigonometry functions</u>

$sin\ \varnothing = \frac{y}{r}$

$sin\ \varnothing = \frac{-1}{5\sqrt 2}$

Rationalize:

$sin\ \varnothing = \frac{-1}{5\sqrt 2} * \frac{\sqrt 2}{\sqrt 2}$

$sin\ \varnothing = \frac{-\sqrt 2}{5*2}$

$sin\ \varnothing = \frac{-\sqrt 2}{10}$

$sin\ \varnothing = \frac{-1}{10}\sqrt 2$

$cos\ \varnothing = \frac{x}{r}$

$cos\ \varnothing = \frac{-7}{5\sqrt 2}$

Rationalize

$cos\ \varnothing = \frac{-7}{5\sqrt 2} * \frac{\sqrt 2}{\sqrt 2}$

$cos\ \varnothing = \frac{-7*\sqrt 2}{5*2}$

$cos\ \varnothing = \frac{-7\sqrt 2}{10}$

$cos\ \varnothing = \frac{-7}{10}\sqrt 2$

$tan\ \varnothing = \frac{y}{x}$

$tan\ \varnothing = \frac{-1}{-7}$

$tan\ \varnothing = \frac{1}{7}$

$cot\ \varnothing = \frac{x}{y}$

$cot\ \varnothing = \frac{-7}{-1}$

$cot\ \varnothing = 7$

$sec\ \varnothing = \frac{r}{x}$

$sec\ \varnothing = \frac{5\sqrt 2}{-7}$

$sec\ \varnothing = \frac{-5}{7}\sqrt 2$

$csc\ \varnothing = \frac{r}{y}$

$csc\ \varnothing = \frac{5\sqrt 2}{-1}$

$csc\ \varnothing = -5\sqrt 2$

3 0

3 years ago

What is the value of this expression when d equals three

kotykmax [81]

Answer:

What is the expresion I am confused?

Step-by-step explanation:

8 0

3 years ago

How many 3/20 inch lengths are in 5 2/5 inches?

klasskru [66]

Answer:

0.81

Step-by-step explanation:

7 0

3 years ago