BRAINLIEST TO THE BEST ANSWER !!!!

400 passengers go on a coach trip.

Nataliya [291]

Go be to if Derek ID do oh next 37

7 0

3 years ago

Sandy is upgrading her Internet service. Fast Internet charges $45 for installation and $49.45 per month.

butalik [34]

Answer:

x = 6 months.

The equation is given by $45 + ($49.45 × x) = ($56.95 × x).

Step-by-step explanation:

i) Let x be the number of months of Internet Service purchased till the Fast

Internet charges and Quick Internet charges become the same.

ii) Charges for Fast Internet for x months is given by $45 + ($49.45 × x)

iii) Charges for Quick Internet for x months is given by ($56.95 × x)

iv) According to the first statement we will now equate the equations in ii)

and iii) and solve for x.

Therefore, $45 + ($49.45 × x) = ($56.95 × x)

45 + 49.45 x = 56.95 x

Therefore (56.95 - 49.45) x = 45

7.50 × x = 45

Therefore x = 45 ÷ 7.5 = 6

6 0

3 years ago

+(7p-8q+6pq)+(q-2p+pq)-(10pq-p-4q)

lilavasa [31]

Answer:

Step-by-step explanation:

3 0

3 years ago

Maggie is standing on a platform that is 8 feet above the ground. She throws a ball in the air that hits the ground after 2 seco

Paha777 [63]

The equation that offers the best approximation to this result is: $h = -16\cdot t^{2}+28\cdot t + 8$ . (Choice D)

<h3>How to find the free fall formula for a given scenario</h3>

An object experiments a free fall when it is solely accelerated by gravity on the assumption of an uniform acceleration. The formula is described below:

$h =h_{o} + v_{o}\cdot t + \frac{1}{2}\cdot g\cdot t^{2}$ (1)

Where:

$h_{o}$ - Initial height, in feet.
$v_{o}$ - Initial speed, in feet per second.
$t$ - Time, in seconds.
$g$ - Gravitational acceleration, in feet per square second.

If we know that $t = 2\,s$ , $h_{o} = 8\,ft$ , $h = 0\,ft$ , $g = -32.174\,\frac{ft}{s^{2}}$ , then the height formula is:

$0 = 8 +2\cdot v_{o} - \frac{1}{2}\cdot (32.174)\cdot (2)^{2}$

$0 = -56.348+2\cdot v_{o}$

$v_{o} = 28.174\,\frac{ft}{s}$

The equation that offers the best approximation to this result is: $h = -16\cdot t^{2}+28\cdot t + 8$ . (Choice D)

To learn more on free fall, we kindly invite to check this verified question: brainly.com/question/13796105

5 0

3 years ago

Which of the following graphs shows the solution set for the inequality below? 3|x + 1| < 9

Bas_tet [7]

Step-by-step explanation:

The absolute value function is a well known piecewise function (a function defined by multiple subfunctions) that is described mathematically as

$f(x) \ = \ |x| \ = \ \left\{\left\begin{array}{ccc}x, \ \text{if} \ x \ \geq \ 0 \\ \\ -x, \ \text{if} \ x \ < \ 0\end{array}\right\}$ .

This definition of the absolute function can be explained geometrically to be similar to the straight line $\textbf{\textit{y}} \ = \ \textbf{\textit{x}}$ , however, when the value of x is negative, the range of the function remains positive. In other words, the segment of the line $\textbf{\textit{y}} \ = \ \textbf{\textit{x}}$ where $\textbf{\textit{x}} \ < \ 0$ (shown as the orange dotted line), the segment of the line is reflected across the x-axis.

First, we simplify the expression.

$3\left|x \ + \ 1 \right| \ < \ 9 \\ \\ \\\-\hspace{0.2cm} \left|x \ + \ 1 \right| \ < \ 3$ .

We, now, can simply visualise the straight line, $y \ = \ x \ + \ 1$ , as a line having its y-intercept at the point $(0, \ 1)$ and its x-intercept at the point $(-1, \ 0)$ . Then, imagine that the segment of the line where $x \ < \ 0$ to be reflected along the x-axis, and you get the graph of the absolute function $y \ = \ \left|x \ + \ 1 \right|$ .

Consider the inequality

$\left|x \ + \ 1 \right| \ < \ 3$ ,

this statement can actually be conceptualise as the question

$``\text{For what \textbf{values of \textit{x}} will the absolute function \textbf{be less than 3}}"$ .

Algebraically, we can solve this inequality by breaking the function into two different subfunctions (according to the definition above).

Case 1 (when $x \ \geq \ 0$ )

$x \ + \ 1 \ < \ 3 \\ \\ \\ \-\hspace{0.9cm} x \ < \ 3 \ - \ 1 \\ \\ \\ \-\hspace{0.9cm} x \ < \ 2$

Case 2 (when $x \ < \ 0$ )

$-(x \ + \ 1) \ < \ 3 \\ \\ \\ \-\hspace{0.15cm} -x \ - \ 1 \ < \ 3 \\ \\ \\ \-\hspace{1cm} -x \ < \ 3 \ + \ 1 \\ \\ \\ \-\hspace{1cm} -x \ < \ 4 \\ \\ \\ \-\hspace{1.5cm} x \ > \ -4$

*remember to flip the inequality sign when multiplying or dividing by

negative numbers on both sides of the statement.

Therefore, the values of x that satisfy this inequality lie within the interval

$-4 \ < \ x \ < \ 2$ .

Similarly, on the real number line, the interval is shown below.

The use of open circles (as in the graph) indicates that the interval highlighted on the number line does not include its boundary value (-4 and 2) since the inequality is expressed as "less than", but not "less than or equal to". Contrastingly, close circles (circles that are coloured) show the inclusivity of the boundary values of the inequality.

3 0

2 years ago