All categories

3 years ago

2. Evaluate the polynomial two ways: by substituting in the given

Mathematics

2 answers:

leonid [27]3 years ago

8 0

P(-3) = -2 i think that’s what it’s keeps giving me

natali 33 [55]3 years ago

7 0

Answer:

P(-3) = -2

Step-by-step explanation:

<u>Substitution</u>

P(-3) = 4(-3)^2 +10(-3) -8 = 36 -30 -8

P(-3) = -2

<u>Long Division</u>

See the working below. The remainder is -2, so ...

P(-3) = -2

You might be interested in

I need help on this may you help

HACTEHA [7]

Answer:

should be 63

Step-by-step explanation:

6 0

1 year ago

-6(7 + x) plzzzz help

adelina 88 [10]

Answer:

-6x-42

Step-by-step explanation:

Open parenthesis.

-42-6x is what you get once opened.

7 0

3 years ago

Read 2 more answers

3 bags of chips cost 14.70 what is the price per bag

yawa3891 [41]

Answer:

4.90

Step-by-step explanation:

p/a=x

price/amount=x

14.70/3=4.90

7 0

3 years ago

Read 2 more answers

Aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa

ipn [44]

Answer:

The answer is C.

Step-by-step explanation:

The slow cooling of intrusive igneous rock can cause large crystals to form. Crystals also form in metamorphic rock, but due the pressure, are often smaller.

Hope this helped!

7 0

2 years ago

Read 2 more answers

A metal beam was brought from the outside cold into a machine shop where the temperature was held at 65degreesF. After 5 min, t

ivolga24 [154]

Answer:

The beam initial temperature is 5 °F.

Step-by-step explanation:

If T(t) is the temperature of the beam after t minutes, then we know, by Newton’s Law of Cooling, that

$T(t)=T_a+(T_0-T_a)e^{-kt}$

where $T_a$ is the ambient temperature, $T_0$ is the initial temperature, $t$ is the time and $k$ is a constant yet to be determined.

The goal is to determine the initial temperature of the beam, which is to say $T_0$

We know that the ambient temperature is $T_a=65$ , so

$T(t)=65+(T_0-65)e^{-kt}$

We also know that when $t=5 \:min$ the temperature is $T(5)=35$ and when $t=10 \:min$ the temperature is $T(10)=50$ which gives:

$T(5)=65+(T_0-65)e^{k5}\\35=65+(T_0-65)e^{-k5}$

$T(10)=65+(T_0-65)e^{k10}\\50=65+(T_0-65)e^{-k10}$

Rearranging,

$35=65+(T_0-65)e^{-k5}\\35-65=(T_0-65)e^{-k5}\\-30=(T_0-65)e^{-k5}$

$50=65+(T_0-65)e^{-k10}\\50-65=(T_0-65)e^{-k10}\\-15=(T_0-65)e^{-k10}$

If we divide these two equations we get

$\frac{-30}{-15}=\frac{(T_0-65)e^{-k5}}{(T_0-65)e^{-k10}}$

$\frac{-30}{-15}=\frac{e^{-k5}}{e^{-k10}}\\2=e^{5k}\\\ln \left(2\right)=\ln \left(e^{5k}\right)\\\ln \left(2\right)=5k\ln \left(e\right)\\\ln \left(2\right)=5k\\k=\frac{\ln \left(2\right)}{5}$

Now, that we know the value of $k$ we can use it to find the initial temperature of the beam,

$35=65+(T_0-65)e^{-(\frac{\ln \left(2\right)}{5})5}\\\\65+\left(T_0-65\right)e^{-\left(\frac{\ln \left(2\right)}{5}\right)\cdot \:5}=35\\\\65+\frac{T_0-65}{e^{\ln \left(2\right)}}=35\\\\\frac{T_0-65}{e^{\ln \left(2\right)}}=-30\\\\\frac{\left(T_0-65\right)e^{\ln \left(2\right)}}{e^{\ln \left(2\right)}}=\left(-30\right)e^{\ln \left(2\right)}\\\\T_0=5$

so the beam started out at 5 °F.

6 0

3 years ago