Which of the following best describes the slope of the line below?

The number of bacteria in a refrigerated food product is given by N ( T ) = 27 T 2 − 180 T + 100 N(T)=27T2-180T+100, 7 < T &l

NeX [460]

Answer:

$N(T(t))=432t^2-374.4t-118.88$

The number of Bactria after 5.8 hours is 12242.

Step-by-step explanation:

The number of bacteria in a refrigerated food product is given by

$N(T)=27T^2-180T+100$

where, T is the temperature of the food.

When the food is removed from the refrigerator, then the temperature is given by

$T(t)=4t+1.6$

We need to find the composite function N(T(t)).

$N(T(t))=N(4t+1.6)$

$N(T(t))=27(4t+1.6)^2-180(4t+1.6)+100$

$N(T(t))=432t^2+345.6t+69.12-720t-288+100$

$N(T(t))=432t^2-374.4t-118.88$

where N(T(t)) is the number of bacteria after t hours.

Substitute t=5.8 in the above function.

$N(T(5.8))=432(5.8)^2-374.4(5.8)-118.88$

$N(T(5.8))=14532.48-2290.4$

$N(T(5.8))=12242.08$

$N(T(5.8))\approx 12242$

Therefore, the number of Bactria after 5.8 hours is 12242.

7 0

3 years ago

What is 2(10) + 2(x – 4) simplified

Digiron [165]

20+2x-8

Answer: 12+2x

5 0

1 year ago

Why did the scientist create an exact duplicate of himself

galben [10]

To do his work that he didn't want to do.

3 0

4 years ago

Find all points on the x-axis that are 14 units from the point (6,-7) All points on the x-axis that are 14 units from the point

Maksim231197 [3]

Answer: $(6+7\sqrt{3},0)\text{ and }(6-7\sqrt{3},0)$ are the required points.

or (18.124,0) and ( -6.124,0) are the required points.

Step-by-step explanation:

Let (x,0) be the point on x -axis that are 14 units from the point (6,-7) .

Then by distance formula , we have

$\sqrt{(x-6)^2+(0-(-7))^2}=14\ \ \ [\ \because distance=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}]$

Taking square on both the sides , we get

$(x-6)^2+7^2=14^2\\\\\Rightarrow\ x^2+6^2-2(6)x+49=196\\\\\Rightarrow\ x^2+36-12x=147\\\\\Rightarrow\ x^2-12x=111\\\\\Rightarrow\ x^2-12x-111=0$

Using quadratic formula : $x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}$

$x=\dfrac{12\pm\sqrt{(-12)^2-4(1)(-111)}}{2}\\\\\Rightarrow\ x=\dfrac{12\pm\sqrt{144+444}}{2}\\\\\Rightarrow\ x=\dfrac{12\pm\sqrt{588}}{2}\\\\\Rightarrow\ x=\dfrac{12\pm\sqrt{2^2\times7^2\times3}}{2}\\\\\Rightarrow\ x=\dfrac{12\pm14\sqrt{3}}{2}\\\\\Rightarrow\ x=6\pm7\sqrt{3}$

so, $(6+7\sqrt{3},0)\text{ and }(6-7\sqrt{3},0)$ are the required points.

since $\sqrt{3}=1.732$

so, $(6+7(1.732),0)\text{ and }(6-7(1.732),0)$ are the required points.

i.e. (18.124,0) and ( -6.124,0) are the required points.

3 0

3 years ago

George buys some first class and second class stamps to put on some letters. Each first class stamp costs 67p, and each second c

Dafna1 [17]

Answer:

£191.36

Step-by-step explanation:

Sum the parts of the ratio, 1 + 4 = 5 parts

Divide 320 by 5 to find the value of one part of the ratio.

320 ÷ 5 = 64 ← value of 1 part of the ratio , thus

4 parts = 4 × 64 = 256

Thus he bought 64 first class and 256 second class.

Cost = (64 × £0.67) + (256 × £0.58)

= £42.88 + 148.48

= £191.36

3 0

3 years ago