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3 years ago

a car covers a distance of 490 km with a speed of 70km and then covers a speed of 90km calculate its average speed

1 answer:

3 0

Average speed = total distance divided by total time.

In the first part, the car covered 490 km in 7 hours, but I think you mistyped the second part (“then covers a speed of 90km”) and a critical piece is missing to find the total distance and time.

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Solve for x ~ <img src="https://tex.z-dn.net/?f=4x%20-%2016%20%3D%2064" id="TexFormula1" title="4x - 16 = 64" alt="4x - 16 =

Naddika [18.5K]

$\qquad\qquad\huge\underline{{\sf Answer}}♨$

Let's solve for x ~

$\qquad \sf \dashrightarrow \:4x - 16 = 64$

$\qquad \sf \dashrightarrow \:4(x - 4) = 64$

$\qquad \sf \dashrightarrow \:(x - 4) = 64 \div 4$

$\qquad \sf \dashrightarrow \:x - 4 = 16$

$\qquad \sf \dashrightarrow \:x = 16 + 4$

$\qquad \sf \dashrightarrow \:x =20$

6 0

2 years ago

Read 2 more answers

Sarah is a computer engineer and manager and works for a software company. She receives a

daser333 [38]

Answer:

a) Number of projects in the first year = 90

b) Earnings in the twelfth year = $116500

Total money earned in 12 years = $969000

Step-by-step explanation:

Given that:

Number of projects done in fourth year = 129

Number of projects done in tenth year = 207

There is a fixed increase every year.

a) To find:

Number of projects done in the first year.

This problem is nothing but a case of arithmetic progression.

Let the first term i.e. number of projects done in first year = $a$

Given that:

$a_4=129\\a_{10}=207$

Formula for $n^{th}$ term of an Arithmetic Progression is given as:

$a_n=a+(n-1)d$

Where $d$ will represent the number of projects increased every year.

and $n$ is the year number.

$a_4=129=a+(4-1)d \\\Rightarrow 129=a+3d .....(1)\\a_{10}=207=a+(10-1)d \\\Rightarrow 207=a+9d .....(2)$

Subtracting (2) from (1):

$78 = 6d\\\Rightarrow d =13$

By equation (1):

$129 =a+3\times 13\\\Rightarrow a =129-39\\\Rightarrow a =90$

Number of projects in the first year = 90

b)

Number of projects in the twelfth year =

$a_{12} = a+11d\\\Rightarrow a_{12} = 90+11\times 13 =233$

Each project pays $500

Earnings in the twelfth year = 233 $\times$ 500 = $116500

Sum of an AP is given as:

$S_n=\dfrac{n}{2}(2a+(n-1)d)\\\Rightarrow S_{12}=\dfrac{12}{2}(2\times 90+(12-1)\times 13)\\\Rightarrow S_{12}=6\times 323\\\Rightarrow S_{12}=1938$

It gives us the total number of projects done in 12 years = 1938

Total money earned in 12 years = 500 $\times$ 1938 = $969000

8 0

2 years ago

The quotient of any two integers is an intiger? True or false

Serggg [28]

The correct answer is true since an integer is simply just a number

4 0

3 years ago

Reka Maharis drove her vehicle to and from work last year. Her records show a total of $1,560 for fixed costs and $3,740 for var

amid [387]

Answer: 5,300

Step-by-step explanation:

5 0

2 years ago

Write the equation of the line in fully simplified slope-intercept form.

Nastasia [14]

Answer:

y= $\frac{5}{6} x$ + 4

Step-by-step explanation:

Pick two points: (-6,-10) and (0,4)

Then solve for m (slope): I picked (0,4)

$m= \frac{4-(-1)}{0-(-6)} =\frac{5}{6}$

Then put in slope intercept form:

4= $\frac{5}{6} (0)$ + b

4=b

Then put in final form:

y= $\frac{5}{6} x$ + 4

Hope this helps!

7 0

3 years ago