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3 years ago

Help please!!!!!!!!!!!!

Mathematics

1 answer:

Tpy6a [65]3 years ago

3 0

Answer:

no solution

Step-by-step explanation:

if you plug in y=8-3x to 5y,

5(8-3x) the x's cancel out

40=20

^no real solution

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which equation represents an exponential function that passes through the point (2, 80)? f(x) = 4(x)5 f(x) = 5(x)4 f(x) = 4(5)x

pav-90 [236]

we know that

if the exponential function passes through the given point, then the point must satisfy the equation of the exponential function

we proceed to verify each case if the point $(2,80)$ satisfied the exponential function

case A $f(x)=4(x^{5})$

For $x=2$ calculate the value of y in the equation and then compare with the y-coordinate of the point

$f(2)=4(2^{5})=128$

$128\neq 80$

therefore

the exponential function $f(x)=4(x^{5})$ not passes through the point $(2,80)$

case B $f(x)=5(x^{4})$

For $x=2$ calculate the value of y in the equation and then compare with the y-coordinate of the point

$f(2)=5(2^{4})=80$

$80=80$

therefore

the exponential function $f(x)=5(x^{4})$ passes through the point $(2,80)$

case C $f(x)=4(5^{x})$

For $x=2$ calculate the value of y in the equation and then compare with the y-coordinate of the point

$f(2)=4(5^{2})=100$

$100\neq 80$

therefore

the exponential function $f(x)=4(5^{x})$ not passes through the point $(2,80)$

case D $f(x)=5(4^{x})$

For $x=2$ calculate the value of y in the equation and then compare with the y-coordinate of the point

$f(2)=5(4^{2})=80$

$80=80$

therefore

the exponential function $f(x)=5(4^{x})$ passes through the point $(2,80)$

therefore

the answer is

$f(x)=5(x^{4})$

$f(x)=5(4^{x})$

7 0

3 years ago

Find the volume of the largest circular cone that can beinscribed in a shpere of radius 3.

Aleksandr [31]

Answer:

$V =\dfrac{32}{3}\pi$

Step-by-step explanation:

given,

radius of sphere = 3

volume of cone:

$V = \dfrac{1}{3}\pi r^2h$

r is the radius of circular base

h is the height of the cone

here r = x and h = 3 + y

now, volume in term of x and y

$V = \dfrac{1}{3}\pi x^2(3+y)$

Applying Pythagoras theorem

x² + y² = 3²

$x = \sqrt{9-y^2}$

$V = \dfrac{1}{3}\pi ( \sqrt{9-y^2})^2(3+y)$

$V = \dfrac{1}{3}\pi ( 9-y^2)(3+y)$

$V = \dfrac{1}{3}\pi (27 + 9 y - 3 y^2-y^3)$

differentiating both side

$\dfrac{dV}{dy} =\dfrac{1}{3}\pi ( 9-6y- 3y^2)$

for maxima $\dfrac{dV}{dy} = 0$

$\pi ( 3-2 y - y^2)=0$

y² + 2 y - 3 = 0

(y+3)(y-1)=0

y = 1,-3

y cannot be negative so, volume at y = 1

$V = \dfrac{1}{3}\pi (27 + 9 (1)- 3(1)^2-(1)^3)$

$V =\dfrac{32}{3}\pi$

Hence, the largest cone which can be inscribed in the spheres of the radius 3 has volume $V =\dfrac{32}{3}\pi$

5 0

3 years ago

Anyone who understand this for 16 point s plz

Zielflug [23.3K]

This is pretty open-ended. How about:

$\sqrt 2$ and $\sqrt 3$ are similar because they're both irrational, meaning neither is the ratio of two integers.

$\sqrt 2$ and $\sqrt 3$ are different because $\sqrt{2} < 1.5}$ and $\sqrt{3} > 1.5$ .

5 0

3 years ago

The drama club was selling ticketsto the school play. Adult ticketscost $8.00 each, and studenttickets cost $5.00 each. The litt

BlackZzzverrR [31]

We are given a problem that can be solved using a system of linear equations. Let A, be the number of adults, and S the number of students. Since there are in total 142 people and there were two days, this means that the sum of the number of adults and the number of students must be 284, which can be written mathematically as follows:

$A+S=284,(1)$

This is our first equation. The second equation is found using the total sales of $1948. Since the ticket per adult is $8 and per student is $5, we have the following equations: