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2 years ago

Draw a diagram to show 3 pizzas shared equally among 6 friends

Mathematics

2 answers:

Yuliya22 [10]2 years ago

8 0

They each get two slices

Reil [10]2 years ago

4 0

Easy just eat more rice

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PLS HELP W THIS QUESTION ASAP

Mashcka [7]

The total cost will be $20.21 because if u do $0.44×16 it will be $7.26 then, you will add $7.26+$12.95 together and get $20.21.

7 0

2 years ago

Read 2 more answers

Q15. Ken earned x dollars at his part-time job on Friday. His wife earned $12 more

mafiozo [28]

Ken earns 54 dollars at his part-time job on Friday

Given that Total amount earned by ken and his wife at their part-time job on Friday= $174

Money earned by ken= x dollars

Money earned by ken's wife = (2x+12) dollars

Total amount earned by ken and his wife at their part-time job on Friday= x dollars + (2x+12) dollars

(x + 2x + 12 ) dollars = $ 174 ( addition of ken and ken's wife income and total amount earned by ken and his wife at their part-time job on Friday is $174)

(3x + 12 ) dollars = $ 174

3x=$162

x= $\frac{162}{3}$

x= 54 dollars

Therefore,54 dollars were earned by ken at his part-time job on Friday

Hence,Ken earns 54 dollars at his part-time job on Friday

Learn more about dollars here:

brainly.com/question/14982791

#SPJ9

4 0

1 year ago

Evaluate the line integral, where c is the given curve. (x + 9y) dx + x2 dy, c c consists of line segments from (0, 0) to (9, 1)

viktelen [127]

$\displaystyle\int_C(x+9y)\,\mathrm dx+x^2\,\mathrm dy=\int_C\langle x+9y,x^2\rangle\cdot\underbrace{\langle\mathrm dx,\mathrm dy\rangle}_{\mathrm d\mathbf r}$

The first line segment can be parameterized by $\mathbf r_1(t)=\langle0,0\rangle(1-t)+\langle9,1\rangle t=\langle9t,t\rangle$ with $0\le t\le1$ . Denote this first segment by $C_1$ . Then

$\displaystyle\int_{C_1}\langle x+9y,x^2\rangle\cdot\mathbf dr_1=\int_{t=0}^{t=1}\langle9t+9t,81t^2\rangle\cdot\langle9,1\rangle\,\mathrm dt$
$=\displaystyle\int_0^1(162t+81t^2)\,\mathrm dt$
$=108$

The second line segment ( $C_2$ ) can be described by $\mathbf r_2(t)=\langle9,1\rangle(1-t)+\langle10,0\rangle t=\langle9+t,1-t\rangle$ , again with $0\le t\le1$ . Then

$\displaystyle\int_{C_2}\langle x+9y,x^2\rangle\cdot\mathrm d\mathbf r_2=\int_{t=0}^{t=1}\langle9+t+9-9t,(9+t)^2\rangle\cdot\langle1,-1\rangle\,\mathrm dt$
$=\displaystyle\int_0^1(18-8t-(9+t)^2)\,\mathrm dt$
$=-\dfrac{229}3$

Finally,

$\displaystyle\int_C(x+9y)\,\mathrm dx+x^2\,\mathrm dy=108-\dfrac{229}3=\dfrac{95}3$

5 0

3 years ago

Please help thanks<br> Xxxxx

Stells [14]

Answer:

Step-by-step explanation:

5 0

3 years ago

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Simplify <br> 2/5(a+b)+3/5(a+c)

Sergeu [11.5K]

Distribute 2/5 and 3/5 into the ():
2/5(a+b)+3/5(a+c)

2/5 a+ 2/5 b+3/5 a+ 3/5 c

combine the like terms:
2/5 a+3/5 a= 5/5 a --> 1a --> a

new simplified equation:
a+2/5 b+3/5 c

3 0

3 years ago

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