2 years ago

Jim uses ribbon to make bookmarks. Jim has 9 feet of ribbon, He uses 1/3 foot of ribbon to make each bookmark.

1 answer:

7 0

The answer to this is 27 ribbons bc first u have to convert 9 feet to inches which is 12*9=108in. Then since he uses 1/3 foot of ribbon to make each book mark ( u would know that 1/3 of a feet is 4 inches ) so u divided 108 by 4 which is 27 ribbons!!!! : )

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Georgia [21]

Answer:

If s(x) = x – 7 and t(x) = 4x2 – x + 3, which expression is equivalent to (t*s)(x)? 4(x – 7)2 – x – 7 + 3 4(x – 7)2 – (x – 7) + 3 (4x2 – x + 3) – 7 (4x2 – x + 3)(x – 7)

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Juli2301 [7.4K]

A = (1/2) * b * h
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2 years ago

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Jet001 [13]

Answer:

The three correct answers are B "The sine function increases on (0°, 90°) and (270°, 360°)." , E "Both the cosine and sine functions have a maximum value of 1.", and F "Both the cosine and sine functions are periodic."

Step-by-step explanation:

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2 years ago

(1 point) Let P(t) be the performance level of someone learning a skill as a function of the training time t. The derivative dPd

monitta

Answer:

$P(t)=M+Ce^{-kt}$

Step-by-step explanation:

Given the differential model

$\dfrac{dP}{dt}=k[M-P(t)]$

We are required to solve the equation for P(t).

$\dfrac{dP}{dt}=kM-kP(t)\\$Add kP(t) to both sides\\\dfrac{dP}{dt}+kP(t)=kM\\$Taking the integrating factor\\e^{\int k dt} =e^{kt}\\$Multiply all through by the integrating factor\\\dfrac{dP}{dt}e^{kt}+kP(t)e^{kt}=kMe^{kt}\\\dfrac{dP}{dt}e^{kt}=kMe^{kt}\\(Pe^{kt})'=kMe^{kt} dt\\$Take the integral of both sides with respect to t\\\int (Pe^{kt})'=\int kMe^{kt} dt\\Pe^{kt}=kM \int e^{kt} dt\\Pe^{kt}=\dfrac{kM}{k} e^{kt} + C_0, C_0$ a constant of integration$

$Pe^{kt}=Me^{kt} + C\\$Divide both side by e^{kt}\\P(t)=M+Ce^{-kt}\\P(t)=M+Ce^{-kt}\\$Therefore:\\P(t)=M+Ce^{-kt}$

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1 year ago

Can you Solve this?<br> IF AT=4 CAT=6 CROW=8 BRAIN=10<br><br> TWISTER = ?

Artyom0805 [142]

The answer is 14 there just adding an extra letter each time twister has 7 letters

3 0

1 year ago