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3 years ago

12

17 Ryan and two friends volunteered to deliver

1 answer:

Stella [2.4K]3 years ago

8 0

Since there are 3 people in total
18/3 = 6 flowers
Each boy delivered 6

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5+3×=6+2×what's this

Nutka1998 [239]

First combine like terms ( x's go with x's), so subtract 2x to both sides and you get:

5+ (3x-2x) = 6 + (2x-2x)

5 + x = 6

Isolate x by bringing 5 to the other side by subtraction

(5-5) + x = (6-5)

x = 1

Hope this helped!

8 0

3 years ago

A construction company completes two projects. The first project has $3,000 in labor expenses for 60 hours worked, while the sec

d1i1m1o1n [39]

Answer: 2,100 = 0.02(42) + b

2,100 = 50(42) + b

Step-by-step explanation: good day man

3 0

2 years ago

Kelly needs to order lunch for 6 people at a business meeting. Her menu choices are chicken salad for a cost of $5 per person an

Sophie [7]

4*6 = 24. (4 dollars to spare)

(5*2) + (4*4) = 26 (2 dollars to spare)
^ ^
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6 0

4 years ago

to prove that the triangles are congruent by asa which statement and reason could br used as part if the proof

Zielflug [23.3K]

Problem Example :

Given: HF || JK; HG ≅ JG

Prove: FHG ≅ KJG

Answer: A. ∠FGH ≅∠KGJ because vertical angles are congruent.

8 0

3 years ago

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

3 years ago