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borishaifa [10]

3 years ago

6

The table and graph below show the amount of money Mi-Ling and Daniel

1 answer:

Over [174]3 years ago

6 0

Is there a math question and can you show it please

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A printer can print 40 pages in 1.6 minutes

Rzqust [24]

Given:
40 pages in 1.6 minutes

40 / 1.6 = 25 pages per minute

25 x 5 minutes = 125 pages

40 / 1.6 = x / 5
40*5 = 1.6x
200 = 1.6x
200/1.6 = x
125 = x pages

25 : 1 = 150 : x
25x = 150
x = 150/25
x = 6 minutes

4 0

3 years ago

The length of x equals...?

Solnce55 [7]

Show us the image lol

4 0

2 years ago

Pls fast,i need it<br> ..............................

Ghella [55]

Answer:

hints to solve each. partial answer due to lack of time, hope it helps!

Step-by-step explanation:

1)

a) solve ():

5 $\sqrt{3}$ + 7 $\sqrt{3}$ - 9 $\sqrt{3}$ - 6 $\sqrt{3}$ + 12 $\sqrt{3}$ - 10 $\sqrt{3}$ =

$\sqrt{3}$ . (5 +7-9-6+12-10)

b) use $\sqrt{6\\}$ = $\sqrt{3\\}\\\\$ $\sqrt{2\\}\\\\$

c) use $\sqrt{20\\}$ = $\sqrt{4\\}\\\\$ $\sqrt{5\\}\\\\$ = 2 $\sqrt{5\\}\\\\$

use $\sqrt{8\\}$ = $\sqrt{4\\}\\\\$ $\sqrt{2\\}\\\\$ = 2 $\sqrt{2\\}\\\\$

2)

a) use $\sqrt{30\\}$ = $\sqrt{6}$ $\sqrt{5}$

use $\sqrt{30\\}$ = $\sqrt{6}$ $\sqrt{4}$ = 2 $\sqrt{6}$

6 0

2 years ago

If jerome has 3 watermelons and 4 oranges, how fast was the train to his right going

attashe74 [19]

Answer:

70 km/h

Step-by-step explanation:

6 0

2 years ago

Read 2 more answers

For the cost function c equals 0.1 q squared plus 2.1 q plus 8, how fast does c change with respect to q when q equals 11? Det

Soloha48 [4]

Answer:

Rate of change of c with respect to q is 4.3

Percentage rate of change c with respect to q is 9.95%

Step-by-step explanation:

Cost function is given as, $c=0.1\:q^{2}+2.1\:q+8$

Given that c changes with respect to q that is, $\dfrac{dc}{dq}$ . So differentiating given function,

$\dfrac{dc}{dq}=\dfrac{d}{dq}\left (0.1\:q^{2}+2.1\:q+8 \right)$

Applying sum rule of derivative,

$\dfrac{dc}{dq}=\dfrac{d}{dq}\left(0.1\:q^{2}\right)+\dfrac{d}{dq}\left(2.1\:q\right)+\dfrac{d}{dq}\left(8\right)$

Applying power rule and constant rule of derivative,

$\dfrac{dc}{dt}=0.1\left(2\:q^{2-1}\right)+2.1\left(1\right)+0$

$\dfrac{dc}{dt}=0.1\left(2\:q\right)+2.1$

$\dfrac{dc}{dt}=0.2\left(q\right)+2.1$

Substituting the value of $q=11$ ,

$\dfrac{dc}{dt}=0.2\left(11\right)+21.$

$\dfrac{dc}{dt}=2.2+2.1$

$\dfrac{dc}{dt}=4.3$

Rate of change of c with respect to q is 4.3

Formula for percentage rate of change is given as,

$Percentage\:rate\:of\:change=\dfrac{Q'\left(x\right)}{Q\left(x\right)}\times 100$

Rewriting in terms of cost C,

$Percentage\:rate\:of\:change=\dfrac{C'\left(q\right)}{C\left(q\right)}\times 100$

Calculating value of $C\left(q \right)$

$C\left(q\right)=0.1\:q^{2}+2.1\:q+8$

Substituting the value of $q=11$ ,

$C\left(q\right)=0.1\left(11\right)^{2}+2.1\left(11\right)+8$

$C\left(q\right)=0.1\left(121\right)+23.1+8$

$C\left(q\right)=12.1+23.1+8$

$C\left(q\right)=43.2$

Now using the formula for percentage,

$Percentage\:rate\:of\:change=\dfrac{4.3}{43.2}\times 100$

$Percentage\:rate\:of\:change=0.0995\times 100$

$Percentage\:rate\:of\:change=9.95%$

Percentage rate of change of c with respect to q is 9.95%

7 0

2 years ago