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3 years ago

What are the numbers called in a multiplication

Mathematics

2 answers:

igomit [66]3 years ago

4 0

The two numbers you multiply are called factors and the answer is called the product

mart [117]3 years ago

3 0

The numbers that are called in multiplication are called factors. Multiplying factors together will create a product.

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KIM [24]

Answer:

The answer to 17 is -8. 2+ = 10 - 12 = -2*4 = -8.

The answer to 18 is 28. 4^2 = 16, 7 + 1 = 9 + 3 = 12, 16 + 12 = 28.

Step-by-step explanation:

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svetoff [14.1K]

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Step-by-step explanation:

(20x15)0.5

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(-1,-2) and (3,10)?<br> what’s the equation

Ahat [919]

Answer:

Step-by-step explanation:

If we have two coordinates on a line (x1,y1 =1,2) and (x2, y2 =3,6) we can solve for m as follows. (x2,y2) 6=m3+c - (x1,y1) 2=m1+c

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3 years ago

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NemiM [27]

9514 1404 393

Answer:

162 degrees

Step-by-step explanation:

Angles P and F are supplementary, so ...

3x^2 +12x = 180

x^2 +4x = 60

x^2 +4x -60 = 0

(x +10)(x -6) = 0

The useful value of x that makes this true is x=6.

The measure of angle Q is 3x, so is 3·6 = 18 degrees. The measure of angle S is the supplement to that:

∠S = 180° -18° = 162°

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3 years ago

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