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3 years ago

What are the factors of

Mathematics

1 answer:

Mkey [24]3 years ago

8 0

Answer:

go to khan academy

Step-by-step explanation:

hope it helps

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What is 4910000 in scientific notation

n200080 [17]

4.9*10 x^{6}
i hope i helped!

5 0

3 years ago

Read 2 more answers

Given the definitions of f(x) and g(x) below, find the value of g( f(-5)).

tensa zangetsu [6.8K]

Answer:

we conclude that:

g(f(-5)) = 30

Step-by-step explanation:

Given

f(x) = x+9

g(x) = x² + 7x – 14

To determine

g(f(-5)) = ?

First, we need to find f(-5)

f(x) = x+9

substitute x = -5

f(-5) = (-5)+9

f(-5) = -5+9

f(-5) = 4

g(f(-5)) = g(4)

Now, we need to find g(4)

g(x) = x² + 7x – 14

substitute x = 4

g(4) = (4)² + 7(4) - 14

g(4) = 16 + 28 - 14

g(4) = 30

i.e.

g(f(-5)) = g(4) = 30

Therefore, we conclude that:

g(f(-5)) = 30

7 0

3 years ago

Find the circumference of the given circle given a diameter of 40 cm.

andriy [413]

The circumference of the given circle would be 125.6 cm. This is due to finding the circumference is multiplying it’s diameter by PI (3.14) giving you the total circumference of the circle.

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3 years ago

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A 100 gallon tank initially contains 100 gallons of sugar water at a concentration of 0.25 pounds of sugar per gallon suppose th

Vsevolod [243]

At the start, the tank contains

(0.25 lb/gal) * (100 gal) = 25 lb

of sugar. Let $S(t)$ be the amount of sugar in the tank at time $t$ . Then $S(0)=25$ .

Sugar is added to the tank at a rate of P lb/min, and removed at a rate of

$\left(1\frac{\rm gal}{\rm min}\right)\left(\dfrac{S(t)}{100}\dfrac{\rm lb}{\rm gal}\right)=\dfrac{S(t)}{100}\dfrac{\rm lb}{\rm min}$

and so the amount of sugar in the tank changes at a net rate according to the separable differential equation,

$\dfrac{\mathrm dS}{\mathrm dt}=P-\dfrac S{100}$

Separate variables, integrate, and solve for S.

$\dfrac{\mathrm dS}{P-\frac S{100}}=\mathrm dt$

$\displaystyle\int\dfrac{\mathrm dS}{P-\frac S{100}}=\int\mathrm dt$

$-100\ln\left|P-\dfrac S{100}\right|=t+C$

$\ln\left|P-\dfrac S{100}\right|=-100t-100C=C-100t$

$P-\dfrac S{100}=e^{C-100t}=e^Ce^{-100t}=Ce^{-100t}$

$\dfrac S{100}=P-Ce^{-100t}$

$S(t)=100P-100Ce^{-100t}=100P-Ce^{-100t}$

Use the initial value to solve for C :

$S(0)=25\implies 25=100P-C\implies C=100P-25$

$\implies S(t)=100P-(100P-25)e^{-100t}$

The solution is being drained at a constant rate of 1 gal/min; there will be 5 gal of solution remaining after time

$1000\,\mathrm{gal}+\left(-1\dfrac{\rm gal}{\rm min}\right)t=5\,\mathrm{gal}\implies t=995\,\mathrm{min}$

has passed. At this time, we want the tank to contain

(0.5 lb/gal) * (5 gal) = 2.5 lb

of sugar, so we pick P such that

$S(995)=100P-(100P-25)e^{-99,500}=2.5\implies\boxed{P\approx0.025}$

5 0

3 years ago

Mr Davis is creating a spice mixture for a recipe.2/5 of the spice mixture was oregano 1/3 of the spice mixture was basil the re

attashe74 [19]

Answer: $\frac{11}{15}$

Step-by-step explanation:

You know that:

- 2/5 of the spice mixture was oregano.

- 1/3 of the spice mixture was basil.

Then, to find the fraction of the total amount of spice mixture that was oregano and basil, you must add both fractions, as following:

- Find the least common multiply of the denominators:

$LCM=5*3=15$

- Divide the LCM by each original denominator and multiply the result by each numerator.

- Make the addition.

Then, the result is:

$\frac{(2*3)+(1*5)}{15}=\frac{6+5}{15}=\frac{11}{15}$

7 0

3 years ago