Nigel sent 644 text messages last week. write and solve an equation to find the average number of message he sent each day

2 answers:

7 0

<h2><em>Too many.</em></h2><h2><em></em></h2><h2><em></em></h2><h2><em>In all seriousness... Do 644 / 7. He Sent 92 Texts on Average every day.</em></h2>

Rom4ik [11]3 years ago

7 0

Answer:

644=7x

Since he did 644 in a week.

Since there are 7 days in a week, it would be 644=7x

Answer: 92 texts on average.

You might be interested in

1. In school lottery to win a price for "Free Domino's Pizza" it is required to select 5 numbers (in any order) out of 40 number

barxatty [35]

Answer:

1/8 ;

12 / 51

Step-by-step explanation:

To select any 5 numbers out of 40

Number of selection required / total numbers

Probability of winning = 5 /40 = 1/8

Total number of socks = 18

White, W = 8

Black, B = 6

Blue, C = 4

probability, without looking in the drawer, that you will first select and remove a black pair, then select either a blue or a white pair?

P(B) * [(P(C) + P(W)]

Probability = required outcome / Total possible outcomes

P(B) = 6 / 18 = 1/3

Without replacement :

P(W or C) = 8/17 + 4/17 = 12/17

Hence,

1/3 * 12/17

= 12 / 51

8 0

3 years ago

A firework is launched at the rate of 10 feet per second from a point on the ground 50 feet from an observer. to 2 decimal place

Kazeer [188]

The rate of change of the angle of elevation when the firework is 40 feet above the ground is 0.12 radians/second.

First we will draw a right angle triangle ΔABC, where ∠B = 90°

Lets, assume the height(AB) = h and base(BC)= x

If the angle of elevation, ∠ACB = α, then

tan(α) = $\frac{AB}{BC} = \frac{h}{x}$

Taking inverse trigonometric function, α = tan⁻¹ ( $\frac{h}{x}$ ) .............(1)

As we need to find the rate of change of the angle of elevation, so we will differentiate both sides of equation (1) with respect to time (t) :

$\frac{d\alpha}{dt}=[\frac{1}{1+ \frac{h^2}{x^2}}]*(\frac{1}{x})\frac{dh}{dt}$

Here, the firework is launched from point B at the rate of 10 feet/second and when it is 40 feet above the ground it reaches point A,

that means h = 40 feet and $\frac{dh}{dt}$ = 10 feet/second.

C is the observer's position which is 50 feet away from the point B, so x = 50 feet.

$\frac{d\alpha}{dt}= [\frac{1}{1+ \frac{40^2}{50^2}}] *\frac{1}{50} *10\\ \\ \frac{d\alpha}{dt} = [\frac{1}{1+\frac{16}{25}}] *\frac{1}{5}\\ \\ \frac{d\alpha}{dt} = [\frac{25}{41}] *\frac{1}{5}\\ \\ \frac{d\alpha}{dt}= \frac{5}{41} =0.1219512$