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2 years ago

(c+ d4)2 – (c + d), if c= -1 and d = 2

Mathematics

1 answer:

Citrus2011 [14]2 years ago

8 0

Answer:

224

Step-by-step explanation:

If c = 1 and d = 2:

(-1³ + 2⁴)² - (-1 + 2)³

(-1 + 16)² -(1)³

(15)² -1

225 -1

224

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Karen is riding her bike at 4 miles per hour she wants to show this on a graph what should she draw

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Graph needs to be labeled. I labeled it as shown on picture. It also needs a title. The title can be up to you. :) you can title it “Karen riding her bike”

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3 years ago

Katie walks 1/2 mile in 1/6 hour. what is her unit rate of speed

Bond [772]

Katie travels at 0.05 mi/ min

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3 years ago

Read 2 more answers

The greatest integer that satisfies the inequality 1.6−(3−2y)<5

Cerrena [4.2K]

Solve the inequality 1.6-(3-2y)<5.

1. Rewrite this inequality without brackets:

1.6-3+2y<5.

2. Separate terms with y and without y in different sides of inequality:

2y<5-1.6+3,

2y<6.4.

3. Divide this inequality by 2:

y<3.2

4. The greatest integer that satisfies this inequality is 3.

Answer: 3.

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3 years ago

A roast turkey is taken from an oven when its temperature has reached 185 degrees F and is placed on a table in a room where the

Vedmedyk [2.9K]

(a) Using Newton's Law of Cooling, $\dfrac{dT}{dt} = k(T - T_s)$ , we have $\dfrac{dT}{dt} = k(T - 75)$ where T is temperature after T minutes.
Separate by dividing both sides by T - 75 to get $\dfrac{dT}{T - 75} = k dt$ . Integrate both sides to get $\ln|T - 75| = kt + C$ .

Since $T(0) = 185$ , we solve for C:
$|185 - 75| = k(0) + C\ \Rightarrow\ C = \ln 110$
So we get $\ln|T - 75| = kt + \ln 110$ . Use T(30) = 150 to solve for k:
$\ln| 150 - 75 | = 30k + \ln 110\ \Rightarrow\ \ln 75 - \ln 110= 30k \Rightarrow \\ k= \frac{1}{30}\ln (75/110) = \frac{1}{30}\ln(15/22)$

So

$\ln|T - 75| = kt + \ln 110 \Rightarrow |T - 75| = e^{kt + \ln110} \Rightarrow \\ \\
|T - 75| = 110e^{kt} \Rightarrow T - 75 = \pm110e^{(1/30)\ln(15/22)t} \Rightarrow \\
T = 75 \pm110e^{(1/30)\ln(15/22)t}$

But choose Positive because T > 75. Temp of turkey can't go under.

$T(t) = 75 + 110e^{(1/30)\ln(15/22)t} \\
T(45) = 75 + 110e^{(1/30)\ln(15/22)(45)} = 136.929 \approx 137{}^{\circ}F$

(b)

$T(t) = 75 + 110e^{(1/30)\ln(15/22)t} = 75 + 110(15/22)^{t/30} \\
100 = 75 + 110(15/22)^{t/30} \\
25 = 110(15/22)^{t/30} 
\frac{25}{110} = (15/22)^{t/30} \\
\ln(25/110) / ln(15/22) = t/30 \\
t = 30\ln(25/110) / ln(15/22) \approx 116\ \mathrm{min}$

Dogs of the AMS.

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3 years ago

6. All angles in the figure below are right angles. What is the area of the figure? (1 point)

lilavasa [31]

Answer:

Option (4)

Step-by-step explanation:

Area of the figure given in the picture = Area of large rectangle - Area of rectangle A

Area of rectangle A = Length × Width

= [9 - (2 + 4)] × 3

= 3 × 3

= 9

Area of the large rectangle = Length × Width

= 9 × 5

= 45

Therefore, area of the given figure = 45 - 9

= 36

Option (4) will be the correct option.

5 0

3 years ago