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professor190 [17]
2 years ago
9

In a cookie mix, we find cinnamon cookies, raisin cookies, and pinwheel cookies in a ratio of 3 : 1 : 2. If a bag of the mix con

tains 26 pinwheel cookies, how many cookies in total are there? *
Mathematics
1 answer:
lana66690 [7]2 years ago
3 0

Answer:

39 : 13 : 26 or 78 cookies total

Step-by-step explanation:

if following the rule, divide the normal amount of cookies by the total

(26 divided by 2 = 13)

then multiply that 13 with the other two numbers and thats your answer

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HELP ME PLZ!!!!! Identify the domain of the graph of y = −x2 − 6x − 13. (1 point) All real numbers x ≤ −4 x ≥ −6 x ≥ −2
vlada-n [284]

Answer:

Step-by-step explanation:

All real numbers because x can assume every possible value

5 0
2 years ago
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-5y y=-2 -3x 6y=-12 substitution
IrinaK [193]
The answer would be x=4 and y=-2.
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3 years ago
Which equation has the solution x=-4? a 2(x-4)=16 b 3x+6-2x=-2 c 1/4x+4=3 d(x+6)3=-6 need help plz
xz_007 [3.2K]

Answer:

C: 1/4x + 4 = 3

Step-by-step explanation:

1/4x + 4 = 3

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Thus, option c gives out the solution x = - 4.

4 0
3 years ago
The acceleration, in meters per second per second, of a race car is modeled by A(t)=t^3−15/2t^2+12t+10, where t is measured in s
oksian1 [2.3K]

Answer:

The maximum acceleration over that interval is A(6) = 28.

Step-by-step explanation:

The acceleration of this car is modelled as a function of the variable t.

Notice that the interval of interest 0 \le t \le 6 is closed on both ends. In other words, this interval includes both endpoints: t = 0 and t= 6. Over this interval, the value of A(t) might be maximized when t is at the following:

  • One of the two endpoints of this interval, where t = 0 or t = 6.
  • A local maximum of A(t), where A^\prime(t) = 0 (first derivative of A(t)\! is zero) and A^{\prime\prime}(t) (second derivative of \! A(t) is smaller than zero.)

Start by calculating the value of A(t) at the two endpoints:

  • A(0) = 10.
  • A(6) = 28.

Apply the power rule to find the first and second derivatives of A(t):

\begin{aligned} A^{\prime}(t) &= 3\, t^{2} - 15\, t + 12 \\ &= 3\, (t - 1) \, (t + 4)\end{aligned}.

\displaystyle A^{\prime\prime}(t) = 6\, t - 15.

Notice that both t = 1 and t = 4 are first derivatives of A^{\prime}(t) over the interval 0 \le t \le 6.

However, among these two zeros, only t = 1\! ensures that the second derivative A^{\prime\prime}(t) is smaller than zero (that is: A^{\prime\prime}(1) < 0.) If the second derivative A^{\prime\prime}(t)\! is non-negative, that zero of A^{\prime}(t) would either be an inflection point (ifA^{\prime\prime}(t) = 0) or a local minimum (if A^{\prime\prime}(t) > 0.)

Therefore \! t = 1 would be the only local maximum over the interval 0 \le t \le 6\!.

Calculate the value of A(t) at this local maximum:

  • A(1) = 15.5.

Compare these three possible maximum values of A(t) over the interval 0 \le t \le 6. Apparently, t = 6 would maximize the value of A(t)\!. That is: A(6) = 28 gives the maximum value of \! A(t) over the interval 0 \le t \le 6\!.

However, note that the maximum over this interval exists because t = 6\! is indeed part of the 0 \le t \le 6 interval. For example, the same A(t) would have no maximum over the interval 0 \le t < 6 (which does not include t = 6.)

4 0
3 years ago
14 + ( 42 ÷ 7 )+ 4^2
pochemuha

Answer:

36

Step-by-step explanation:

14+6+16

4 0
2 years ago
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