Ask question

Ask question

All categories

3 years ago

9

True or False: Only items that cannot be replaced or that would be expensive or hard to replace should be put in a safe deposit

box.

1 answer:

Masteriza [31]3 years ago

4 0

True. cuz if you don’t have the box they can’t really confirm it.

You might be interested in

In triangle JKL, angle K is a right angle. Write the sine of angle J as a ratio of side lengths

Evgen [1.6K]

Answer:

$Sin J =\frac{|KL|}{|JL|}$

Step-by-step explanation:

In Right Triangle JKL, with Right angle at K

$sin \theta =\frac{Opposite}{Hypotenuse} \\$

Opposite =|KL|

Hypotenuse=|JL|

Therefore, sin J as a ratio of side lengths is:

$Sin J =\frac{|KL|}{|JL|}$

5 0

3 years ago

7. The terminal side of theta, and angle in standard position, intersects the unit circle at

SIZIF [17.4K]

Complete question is;

The terminal side of angle θ in standard position, intersects the unit circle at P(-10/26, -24/26). What is the value of csc θ?

Answer:

csc θ = -13/12

Step-by-step explanation:

We know that in a unit circle;

(x, y) = (cos θ, sin θ)

Since the the terminal sides intersects P at the coordinates P(-10/26, -24/26), we can say that;

cos θ = -10/26

sin θ = -24/26

Now we want to find csc θ.

From trigonometric ratios, csc θ = 1/sin θ

Thus;

csc θ = 1/(-24/26)

csc θ = -26/24

csc θ = -13/12

7 0

3 years ago

For what value of x is line a parallel to line b

Oksanka [162]

Answer:

x = 30

Step-by-step explanation:

In order for the lines to be parallel,

2x + 20 = 80

2x = 60

x = 30

8 0

3 years ago

Read 2 more answers

Use synthetic division to determine f(-2) for the function f(x)=9x^3-18x^2-16x+32

Verizon [17]

Answer: -80

Step-by-step explanation:

7 0

3 years ago

What is the best result for h?

Kryger [21]

Answer: First option.

Step-by-step explanation:

To solve for "h" from the given the equation $S=2\pi rh+2\pi r^2$ , you need to:

Apply the Subtraction property of equality and subtract $2\pi r^2$ to both sides of the equation.

Then you need to apply the Division property of equality and divide both sides of the equation by $2\pi r$ .

Then:

$S-2\pi r^2=2\pi rh+2\pi r^2-2\pi r^2\\\\S-2\pi r^2=2\pi rh\\\\\frac{S}{2\pi r}-\frac{2\pi r^2}{2\pi r}=\frac{2\pi rh}{2\pi r}\\\\\frac{S}{2\pi r}-r=h$

6 0

3 years ago

Read 2 more answers