Evaluate the expression.<br> 9!/3!<br> A. 3<br> B. 6<br> C. 60,480<br> D. 362,874

pogonyaev

Start by writing the ratio of:

0.125/1 (since a number divided by 1 is itself)

Then multiply the numerator and denominator by 10, 100, or 1000, etc. until the numerator becomes an integer: (in this case multiplying by 1000 does the trick)

(0.125*1000)/(1*1000)
125/1000

Now simplify by dividing by common factors. For this problem, 5 is a common factor of both the numerator and denominator:

(125/5)/(1000/5)
25/200

Do this again with another common factor if possible. For this problem, 25 is another common factor:

(25/25)/(200/25)
1/8

Now there are no more common factors between the numerator and denominator. When this is the case, the fraction can be considered to have been simplified:

Answer is: 1/8

8 0

3 years ago

The diagram shows a right-angled triangle.

Agata [3.3K]

Answer:

$\sin( \alpha ) = \frac{8}{14} = \frac{4}{7} \\ \\ \alpha = 34.84 \: degee$

6 0

3 years ago

Read 2 more answers

Y varies directly as x and k = 5 <br> Y=kx<br><br>Find y when x = 5

SpyIntel [72]

Answer:

y = 25

Step-by-step explanation:

Given y = kx and k = 5 then

y = 5x ← equation of variation

When x = 5 , then

y = 5 × 5 = 25

5 0

3 years ago

NO LINKS!!! Find the arc measure and arc length of AB. Then find the area of the sector ABQ.

Norma-Jean [14]

Answer:

<u>Arc Measure</u>: equal to the measure of its corresponding central angle.

<u>Formulas</u>

$\textsf{Arc length}=2 \pi r\left(\dfrac{\theta}{360^{\circ}}\right)$

$\textsf{Area of a sector of a circle}=\left(\dfrac{\theta}{360^{\circ}}\right) \pi r^2$

$\textsf{(where r is the radius and the angle }\theta \textsf{ is measured in degrees)}$

<h3><u>Question 39</u></h3>

Given:

r = 7 in
$\theta$ = 90°

Substitute the given values into the formulas:

Arc AB = 90°

$\textsf{Arc length of AB}=2 \pi (7) \left(\dfrac{90^{\circ}}{360^{\circ}}\right)=3.5 \pi=11.00\:\sf in\:(2\:d.p.)$

$\textsf{Area of the sector AQB}=\left(\dfrac{90^{\circ}}{360^{\circ}}\right) \pi (7)^2=\dfrac{49}{4} \pi=38.48\:\sf in^2\:(2\:d.p.)$

<h3><u>Question 40</u></h3>

Given:

r = 6 ft
$\theta$ = 120°

Substitute the given values into the formulas:

Arc AB = 120°

$\textsf{Arc length of AB}=2 \pi (6) \left(\dfrac{120^{\circ}}{360^{\circ}}\right)=4\pi=12.57\:\sf ft\:(2\:d.p.)$

$\textsf{Area of the sector AQB}=\left(\dfrac{120^{\circ}}{360^{\circ}}\right) \pi (6)^2=12 \pi=37.70\:\sf ft^2\:(2\:d.p.)$

<h3><u>Question 41</u></h3>

Given:

r = 12 cm
$\theta$ = 45°

Substitute the given values into the formulas:

Arc AB = 45°

$\textsf{Arc length of AB}=2 \pi (12) \left(\dfrac{45^{\circ}}{360^{\circ}}\right)=3 \pi=9.42\:\sf cm\:(2\:d.p.)$

$\textsf{Area of the sector AQB}=\left(\dfrac{45^{\circ}}{360^{\circ}}\right) \pi (12)^2=18 \pi=56.55\:\sf cm^2\:(2\:d.p.)$

8 0

2 years ago

Drag the tiles to the correct boxes to complete the pairs. Add each pair of monomials and match it to the correct sum. -15u3 + 5

Grace [21]

Consider the given monomials. Let us determine the sum of these monomials,

1. $-15u^3+5u^3$

= $-10u^3$

Therefore, the sum of the given monomials is $-10u^3$ .

2. $10u^3+(-5u^3)$

= $10u^3-5u^3$

= $5u^3$

Therefore, the sum of the given monomials is $5u^3$ .

3. $10u^3+5u^3$

= $15u^3$

Therefore, the sum of the given monomials is $15u^3$ .

4. $5u^3+(-10u^3)$

= $5u^3-10u^3$

= $-5u^3$

Therefore, the sum of the given monomials is $-5u^3$ .

7 0

4 years ago

Read 2 more answers

Evaluate the expression. 9!/3! A. 3 B. 6 C. 60,480 D. 362,874