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Blababa [14]
3 years ago
6

Samantha is preheating her oven before using it to bake. The initial temperature of the oven is 70° and the temperature will inc

rease at a rate of 20° per minute after being turned on. What is the temperature of the oven 17 minutes after being turned on? What is the temperature of the oven t t minutes after being turned on?
Mathematics
1 answer:
spin [16.1K]3 years ago
5 0

The answer is 410°

You have to multiply 17x20 and then add 70 because that’s the initial temp.

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Need help on my math!!!!!
Firlakuza [10]

Answer:

The answer is A.

Step-by-step explanation:

Excluded values are values that will make the denominator of a fraction equal to 0. You can't divide by 0, so it's very important to find these excluded values when you're solving a rational expression.

8 0
3 years ago
For which inequality is x = 5 a solution
vesna_86 [32]

Answer:

4<x<6

Step-by-step explanation:

4 is smaller than x

and x is samller than 6

so its 5

6 0
2 years ago
For which value of x is the equation 2(1 + x) = x + 3 true?<br><br> A) 1<br> B) 2<br> C) 3<br> D) 4
amm1812

Answer:

A) 1

Step-by-step explanation:

To solve, just plug in the numbers to the equation for x until both sides are equal to each other.

Plug in 1:

2(1 + x) = x + 3

Let x = 1:

2(1 + 1) = 1 + 3

Solve. Remember to follow PEMDAS. First, add parenthesis, then multiply. On the other side, add:

2(1 + 1) = 1 + 3

2(2) = 4

4 = 4 (True)

A) 1 is your answer.

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5 0
3 years ago
Which expression is equivalent to -6(6 + 8) + 4
vazorg [7]

Answer:

-6 x 14 + 4

Step-by-step explanation:

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8 0
3 years ago
Read 2 more answers
The Laplace Transform of a function f(t), which is defined for all t &gt; 0, is denoted by L{f(t)} and is defined by the imprope
lesya692 [45]

(1) D

L_s\left\{t\right\} = \displaystyle\int_0^\infty te^{-st}\,\mathrm dt

Integrate by parts, taking

u = t \implies \mathrm du=\mathrm dt

\mathrm dv = e^{-st}\,\mathrm dt \implies v=-\dfrac1se^{-st}

Then

L_s\left\{t\right\} = \displaystyle \left[-\frac1ste^{-st}\right]\bigg|_{t=0}^{t\to\infty}+\frac1s\int_0^\infty e^{-st}\,\mathrm dt

L_s\left\{t\right\} = \displaystyle \frac1s\int_0^\infty e^{-st}\,\mathrm dt

L_s\left\{t\right\} = \displaystyle -\frac1{s^2}e^{-st}\bigg|_{t=0}^{t\to\infty}

L_s\left\{t\right\} = \displaystyle \boxed{\frac1{s^2}}

(2) A

L_s\left\{1\right\} = \displaystyle\int_0^\infty e^{-st}\,\mathrm dt

L_s\left\{1\right\} = \displaystyle\left[-\frac1se^{-st}\right]\bigg|_{t=0}^{t\to\infty}

L_s\left\{1\right\} = \displaystyle\boxed{\frac1s}

7 0
3 years ago
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