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3 years ago

8

What is the height of a container if the volume is 56.52 in.³ in the radius is 1.5 inches

1 answer:

iren2701 [21]3 years ago

8 0

Djdkcjkjoeeefifhefjiej

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Write an equation for the value of y in terms of x. Then solve the equation for x.

UkoKoshka [18]

Answer:

x=y-48, y=x+48

Step-by-step explanation:

3 0

3 years ago

crimeas [40]

Answer:

The zeros are:

$x =4, x=2, x = 5$

The function has three distinct real zeros.

Hence, option (B) is true.

Step-by-step explanation:

Given the expression

$h\left(x\right)=\left(x-4\right)^2\left(x^2-7x+\:10\right)$

Let us determine the zeros of the function by putting h(x) = 0 and solving the expression

$0=\left(x-4\right)^2\left(x^2-7x+10\right)$

switch sides

$\left(x-4\right)^2\left(x^2-7x+10\right)=0$

as

$x^2-7x+\:10=\left(x-2\right)\left(x-5\right)$

so

$\left(x-4\right)^2\left(x-2\right)\left(x-5\right)=0$

Using the zero factor principle

$\mathrm{If}\:ab=0\:\mathrm{then}\:a=0\:\mathrm{or}\:b=0\:\left(\mathrm{or\:both}\:a=0\:\mathrm{and}\:b=0\right)$

so

$x-4=0\quad \mathrm{or}\quad \:x-2=0\quad \mathrm{or}\quad \:x-5=0$

$x =4, x=2, x = 5$

Thus, the zeros are:

$x =4, x=2, x = 5$

It is clear that there are three zeros and all the zeros are distinct real numbers.

Therefore,

The function has three distinct real zeros.

Hence, option (B) is true.

4 0

3 years ago

!TWELVE POINTS + BRAINLIEST ANSWER! Write the equation for an ellipse with y intercepts (0,-3) and (0, 3) and major axis of leng

Mandarinka [93]

Answer:

x²/25 + y²/9 = 1 or 9x² + 25y² = 225

Step-by-step explanation:

We have two points which is y intercepts (0,-3) and (0,3).

We know that the major axis is 2a and secondary axis is 2b

These two points are vertical top of the ellipse which they give us value of half secondary axis b = 3

The length of major axis is 2a = 10 => a = 10/2 = 5 => a = 5

The equation of the ellipse is:

x²/a² + y²/b² = 1 or b²x² + a²y² = a²b²

When we replace value of a and b we get:

x²/5² + y²/3² = 1 or 3²x² + 5²y² = 5² · 3² and finally

x²/25 + y²/9 = 1 or 9x² + 25y² = 225

God with you!!!

8 0

3 years ago

Q3: Identify the graph of the equation and write and equation of the translated or rotated graph in general form. (Picture Provi

natta225 [31]

Answer:

b. circle; $2(x')^2+2(y')^2-5x'-5\sqrt{3}y'-6 =0$

Step-by-step explanation:

The given conic has equation;

$x^2-5x+y^2=3$

We complete the square to obtain;

$(x-\frac{5}{2})^2+(y-0)^2=\frac{37}{4}$

This is a circle with center;

$(\frac{5}{2},0)$

This implies that;

$x=\frac{5}{2},y=0$

When the circle is rotated through an angle of $\theta=\frac{\pi}{3}$ ,

The new center is obtained using;

$x'=x\cos(\theta)+y\sin(\theta)$ and $y'=-x\sin(\theta)+y\cos(\theta)$

We plug in the given angle with x and y values to get;

$x'=(\frac{5}{2})\cos(\frac{\pi}{3})+(0)\sin(\frac{\pi}{3})$ and $y'=--(\frac{5}{2})\sin(\frac{\pi}{3})+(0)\cos(\frac{\pi}{3})$

This gives us;

$x'=\frac{5}{4} ,y'=\frac{5\sqrt{3} }{4}$

The equation of the rotated circle is;

$(x'-\frac{5}{4})^2+(y'-\frac{5\sqrt{3} }{4})^2=\frac{37}{4}$

Expand;

$(x')^2+(y')^2-\frac{5\sqrt{3} }{2}y'-\frac{5}{2}x'+\frac{25}{4} =\frac{37}{4}$

Multiply through by 4; to get

$4(x')^2+4(y')^2-10\sqrt{3}y'-10x'+25 =37$

Write in general form;

$4(x')^2+4(y')^2-10x'-10\sqrt{3}y'-12 =0$

Divide through by 2.

$2(x')^2+2(y')^2-5x'-5\sqrt{3}y'-6 =0$

8 0

3 years ago

Of all customers purchasing automatic garage-door openers, 75% purchase Swedish model. Let X = the number among the next 15 purc

Lelechka [254]

Answer:

a)

$P(X=k) = {15 \choose k} * 0.75^{k}*0.25^{15-k}$

For any integer k between 0 and 15, and 0 for other values of k.

b)

$P(X>10) = 0.2252+ 0.2252+ 0.1559+0.0668+0.0134 = 0.6865$

c) P(6 ≤ X ≤ 10) = 0.2737

d) μ = 15*0.75 = 11.25. σ² = 11.25*0.25 = 2.8125

Step-by-step explanation:

X is a binomial random variable with parameters n = 15, p = 0.75. Therefore

a)

$P(X=k) = {15 \choose k} * 0.75^{k}*0.25^{15-k}$

For any integer k between 0 and 15, and 0 for other values of k.

b)

P(X>10) = P(X=11) + P(X=12)+ P(X=13)+P(X=14)+P(x=15)

$P(X=11) = {15 \choose 11} * 0.75^{11} * 0.25^4 = 0.2252$

$P(X=12) = {15 \choose 12} * 0.75^{12} * 0.25^3 = 0.2252$

$P(X=13) = {15 \choose 13} * 0.75^{13} * 0.25^2 = 0.1559$

$P(X=14) = {15 \choose 14} * 0.75^{14} * 0.25 = 0.0668$

$P(X=15) = {15 \choose 15} * 0.75^{15} = 0.0134$

Thus,

$P(X>10) = 0.2252+ 0.2252+ 0.1559+0.0668+0.0134 = 0.6865$

c) P(6 ≤ X ≤ 10) = P(X = 6) + P(X = 7) + P(X = 8) + P(X=9) + P(X=10)

$P(X=6) = {15 \choose 6} * 0.75^{6} * 0.25^9 = 0.0034$

$P(X=7) = {15 \choose 7} * 0.75^{7} * 0.25^8 = 0.0131$

$P(X=8) = {15 \choose 8} * 0.75^{8} * 0.25^7 = 0.0393$

$P(X=9) = {15 \choose 9} * 0.75^{9} * 0.25^6 = 0.0918$

$P(X=10) = {15 \choose 10} * 0.75^{10} * 0.25^{5} = 0.1652$

Thereofre,

$P(6 \leq X \leq 10) = 0.0034 + 0.0134 + 0.0393 + 0.0918 + 0.1652 = 0.2737$

d) μ = n*p = 15*0.75 = 11.25

σ² = np(1-p) = 11.25*0.25 = 2.8125

3 0

3 years ago