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2 years ago

Use the number 751,486

Mathematics

1 answer:

vodomira [7]2 years ago

6 0

Billionare Jeff Bezos spent $751,486 on new robot technology last March.

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Please help me answer this!!

Alex Ar [27]

Answer:

SUPER SIMPLE =

if r//s then the two angles which in this case are 110 and 80 should be equal but clearly in this case are not so the the lines r and s arent parallel to each other .

5 0

2 years ago

Solve for f.<br> –9 + 2f > – 3

kozerog [31]

F>3
That’s the answer

6 0

2 years ago

A source of laser light sends rays AB and AC toward two opposite walls of a hall. The light rays strike the walls at points B an

Nataly [62]

Answer:

The distance between the walls is 70 m.

Step-by-step explanation:

Given: A source of laser light is at point A on the ground between two parallel walls BE and CD . The walls are perpendicular to the ground that is

BE ⊥ ED and CD ⊥ ED

AB is a ray of light which strikes the wall on the left at point B which is 30 meters above the ground. that is BE = 30 m

AC is a ray of light which strikes the wall on the right at point C. The length of AC = 80 meters.

The ray AB makes an angle of 45 degrees with the ground that is m∠BAE = 45°

The ray AC makes an angle of 60 degrees with the ground that is m∠CAD = 60°

As shown is figure attached below.

WE have to find the distance between the walls that is Length of ED

Length of ED = EA + AD

Consider the Δ AEB,

Using trigonometric ratio,

$\tan\theta=\frac{perpendicular}{base}$

Here $\theta=45^{\circ}$ , perpendicular = 30 m and base we can find.

thus,

$\tan 45^{\circ}=\frac{30}{EA}$

We know $\tan 45^{\circ}=1$

thus, EA = 30 m

Consider the Δ AEB,

Using trigonometric ratio,

$\cos\theta=\frac{base}{hypotenuse}$

Here $\theta=60^{\circ}$ , hypotenuse = 80 m and base we can find.

thus, $\cos 60^{\circ}=\frac{base}{80}$

We know, $\cos 60^{\circ}=\frac{1}{2}$

thus, Base = 40 m

AD = 40 m

Thus, the distance between the walls that is the length of ED = 30 + 40 = 70 m

4 0

2 years ago

DATA ANALYSIS AND STATISTICSProbabilities involving two rolls of a dieAn ordinary (falr) die is a cube with the numbers 1 throug

Zepler [3.9K]

Solution

The table below is the required sample space of the to fair die

From the above table

The sample space contain 36 outcomes

Event A: The sum is greater than 9

we will look at the table and count all the elements that are greater than 9

There are 6 elements (they are 10, 10, 10, 11, 11, 12 from the table)

The probability for event A will be

$\begin{gathered} p(A)=\frac{\text{required outcome}}{\text{total outcome}} \\ p(A)=\frac{6}{36} \\ p(A)=\frac{1}{6} \end{gathered}$

P(A) = 1/6

Event B: The sum is an even number.

We will look at the table and count the number of elements that are even

There are 18 elements (notice that there are 3 even number on each of the 6 rows of the table)

The probability for event B will be

$\begin{gathered} p(B)=\frac{\text{required outcome}}{\text{total outcome}} \\ p(B)=\frac{18}{36} \\ p(B)=\frac{1}{2} \end{gathered}$

p(B) = 1/2

5 0

6 months ago

In ABC,if sin A=4/5 and tan A=4/3, then what I s cos A?

Svet_ta [14]

Cos(A) will be 3/5 Cos(A) is adjacent/hypotenuse. So here we have sin(A)=opp/hyp which is 4/5 and tan(A)=opp/adj which is 4/3. So cos(A) is adj./hyp and knowing the previous 2 terms of sin(A) and cos(A), you can go ahead and find Cos(A). So cos(A) is 3/5

5 0

2 years ago