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3 years ago

F(x) = x^3- 12x^2 + 35x – 24

Mathematics

1 answer:

tatyana61 [14]3 years ago

6 0

Answer:

To find the x-intercept, substitute in 0 for y and solve for x . To find the y-intercept, substitute in 0 for x and solve for y . x-intercept(s): ( 1 , 0 ) , ( 8 , 0 ) , ( 3 , 0 ) y-intercept(s): ( 0 , − 24 )

Step-by-step explanation:

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HELP ME PLSSSSS I NEED HELP PLS PLS PLS ITS PYTHAGOREAN THEOREM

sergij07 [2.7K]

H=9ft
B=6ft

Using Pythagorean thereon

$\\ \Large\sf\longmapsto P^2=H^2-B^2$

$\\ \Large\sf\longmapsto P^2=9^2-6^2$

$\\ \Large\sf\longmapsto P^2=81-36$

$\\ \Large\sf\longmapsto P^2=45$

$\\ \Large\sf\longmapsto P=\sqrt{45}$

$\\ \Large\sf\longmapsto P=6.3ft$

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3 years ago

Tamira invests $5,000 in an account that pays 4% annual interest. How much will there be in the account after 3 years if the int

Talja [164]

Answer:

There will be $5624.32 in the account after 3 years if the interest is compounded annually.

There will be $5630.812 in the account after 3 years if the interest is compounded semi-annually.

There will be $5634.125 in the account after 3 years if the interest is compounded quarterly.

There will be $5636.359 in the account after 3 years if the interest is compounded monthly

Step-by-step explanation:

Tamira invests $5,000 in an account

Rate of interest = 4%

Time = 3 years

Case 1:

Principal = 5000

Rate of interest = 4%

Time = 3 years

No. of compounds per year = 1

Formula : $A=P(1+r)^t$

$A=5000(1+0.04)^3$

A=5624.32

There will be $5624.32 in the account after 3 years if the interest is compounded annually.

Case 2:

Principal = 5000

Rate of interest = 4%

Time = 3 years

No. of compounds per year = 2

Formula : $A=P(1+\frac{r}{n})^{nt}$

$A=5000(1+\frac{0.04}{2})^{2 \times 3}$

A=5630.812

There will be $5630.812 in the account after 3 years if the interest is compounded semi-annually.

Case 3:

Principal = 5000

Rate of interest = 4%

Time = 3 years

No. of compounds per year = 4

Formula : $A=P(1+\frac{r}{n})^{nt}$

$A=5000(1+\frac{0.04}{4})^{4 \times 3}$

A=5634.125

There will be $5634.125 in the account after 3 years if the interest is compounded quarterly.

Case 4:

Principal = 5000

Rate of interest = 4%

Time = 3 years

No. of compounds per year = 4

Formula : $A=P(1+\frac{r}{n})^{nt}$

$A=5000(1+\frac{0.04}{12})^{12 \times 3}$

A=5636.359

There will be $5636.359 in the account after 3 years if the interest is compounded monthly

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