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2 years ago

Find the value of x.

Mathematics

2 answers:

Vilka [71]2 years ago

7 0

Answer:

Generally, the algebraic expression should be any one of the forms such as addition, subtraction, multiplication and division. To find the value of x, bring the variable to the left side and bring all the remaining values to the right side. Simplify the values to find the result.

Ksju [112]2 years ago

5 0

I don’t know dawg sorry g

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How do I find a2 and a3 for the following geometric sequence? 54, a2, a3, 128

vlada-n [284]

The formula for the nth term of a geometric sequence:
$a_n=a_1 \times r^{n-1}$
a₁ - the first term, r - the common ratio

$54, a_2, a_3, 128 \\ \\
a_1=54 \\
a_4=128 \\ \\
a_n=a_1 \times r^{n-1} \\
a_4=a_1 \times r^3 \\
128=54 \times r^3 \\
\frac{128}{54}=r^3 \\ \frac{128 \div 2}{54 \div 2}=r^3 \\
\frac{64}{27}=r^3 \\
\sqrt[3]{\frac{64}{27}}=\sqrt[3]{r^3} \\
\frac{\sqrt[3]{64}}{\sqrt[3]{27}}=r \\
r=\frac{4}{3}$

$a_2=a_1 \times r= 54 \times \frac{4}{3}=18 \times 4=72 \\
a_3=a_2 \times r=72 \times \frac{4}{3}=24 \times 4=96 \\ \\
\boxed{a_2=72, a_3=96}$

7 0

3 years ago

Read 2 more answers

you can buy 5 cans of green beans at the village market for $3 you can buy 10 of the same cans of beans at Sams Club for $6.10 w

ioda

Each can of green beans costs ($3.00/5) = 60¢ at the
village market, and ($6.10/10) = 61¢ at Sam's Club.

If you're watching every penny, then that MIGHT be enough
of a difference to make you decide to buy your beans at the
village market, but not necessarily.

If, say, the village market is farther away from you, or if there are
other things you're going to get from Sam's anyway, then you
should buy your beans there too.

6 0

3 years ago

Can someone tell me how to do this? with steps?

MrRa [10]

Answer:

Step-by-step explanation:

In Δ AFB,

∠AFB + ∠ABF + ∠A = 180 {Angle sum property of triangle}

90 + 48 + ∠1 = 180

138 + ∠1 = 180

∠1 = 180 - 138

∠1 = 42°

FC // ED and FD is transversal

So, ∠CFD ≅∠EDF {Alternate interior angles are congruent}

∠2 = 39°

In ΔFCD,

∠2 + ∠3 + ∠FCD = 180

39 + ∠3 + 90 = 180

129 +∠3 = 180

∠3 = 180- 129

∠3 = 51°

3 0

2 years ago

Solve the following System of Three Equations:<br> x−3y+z=−15<br> 2x+y−z=−2<br> x+y+2z=1

SashulF [63]

Answer:

x = -3 , y = 4 , z = 0

Step-by-step explanation:

Solve the following system:

{x - 3 y + z = -15

2 x + y - z = -2

x + y + 2 z = 1

Hint: | Choose an equation and a variable to solve for.

In the first equation, look to solve for z:

{x - 3 y + z = -15

2 x + y - z = -2

x + y + 2 z = 1

Hint: | Solve for z.

Subtract x - 3 y from both sides:

{z = 3 y + (-x - 15)

2 x + y - z = -2

x + y + 2 z = 1

Hint: | Perform a substitution.

Substitute z = -15 - x + 3 y into the second and third equations:

{z = -15 - x + 3 y

15 + 3 x - 2 y = -2

x + y + 2 (-15 - x + 3 y) = 1

Hint: | Expand the left hand side of the equation x + y + 2 (-15 - x + 3 y) = 1.

x + y + 2 (-15 - x + 3 y) = x + y + (-30 - 2 x + 6 y) = -30 - x + 7 y:

{z = -15 - x + 3 y

15 + 3 x - 2 y = -2

-30 - x + 7 y = 1

Hint: | Choose an equation and a variable to solve for.

In the second equation, look to solve for x:

{z = -15 - x + 3 y

15 + 3 x - 2 y = -2

-30 - x + 7 y = 1

Hint: | Isolate terms with x to the left hand side.

Subtract 15 - 2 y from both sides:

{z = -15 - x + 3 y

3 x = 2 y - 17

-30 - x + 7 y = 1

Hint: | Solve for x.

Divide both sides by 3:

{z = -15 - x + 3 y

x = (2 y)/3 - 17/3

-30 - x + 7 y = 1

Hint: | Perform a substitution.

Substitute x = (2 y)/3 - 17/3 into the third equation:

{z = -15 - x + 3 y

x = (2 y)/3 - 17/3

(19 y)/3 - 73/3 = 1

Hint: | Choose an equation and a variable to solve for.

In the third equation, look to solve for y:

{z = -15 - x + 3 y

x = (2 y)/3 - 17/3

(19 y)/3 - 73/3 = 1

Hint: | Isolate terms with y to the left hand side.

Add 73/3 to both sides:

{z = -15 - x + 3 y

x = (2 y)/3 - 17/3

(19 y)/3 = 76/3

Hint: | Solve for y.

Multiply both sides by 3/19:

{z = -15 - x + 3 y

x = (2 y)/3 - 17/3

y = 4

Hint: | Perform a back substitution.

Substitute y = 4 into the first and second equations:

{z = -x - 3

x = -3

y = 4

Hint: | Perform a back substitution.

Substitute x = -3 into the first equation:

{z = 0

x = -3

y = 4

Hint: | Sort results.

Collect results in alphabetical order:

Answer: {x = -3 , y = 4 , z = 0

4 0

3 years ago

Diagram 5 shows a right cylinder with a diameter of 2xcm. Given that the total surface area of the cylinder is 96cm³.Find the ma

Paraphin [41]

Given:

The diameter of the right cylinder is 2x cm.

The total surface area is 96 cm cube.

The radius is calculated as,

$\begin{gathered} r=\frac{d}{2} \\ r=\frac{2x}{2} \\ r=x\text{ cm} \end{gathered}$

The total surface area is,

$\begin{gathered} S=2\pi rh+2\pi(r)^2 \\ 96=2\pi xh+2\pi(x^2) \\ h=\frac{96-2\pi(x^2)}{2\pi x} \end{gathered}$

Volume is,

$\begin{gathered} V=\pi(r)^2h \\ =\pi(x^2)\frac{96-2\pi(x^2)}{2\pi x} \\ =\frac{x(96-2\pi(x^2)}{2} \end{gathered}$

Now, differentiate with respect to x,

$\begin{gathered} \frac{dV}{dx}^{}=\frac{d}{dx}(\frac{x(96-2\pi(x^2)}{2}) \\ =\frac{d}{dx}\mleft(x\mleft(-\pi x^2+48\mright)\mright) \\ =\frac{d}{dx}\mleft(x\mright)\mleft(-\pi x^2+48\mright)+\frac{d}{dx}\mleft(-\pi x^2+48\mright)x \\ =1\cdot\mleft(-\pi x^2+48\mright)+\mleft(-2\pi x\mright)x \\ =84-3\pi(x^2)\ldots\ldots\ldots\ldots\text{.}(1) \end{gathered}$

Now,

$\begin{gathered} \frac{dV}{dx}=0 \\ 84-3\pi(x^2)=0 \\ x^2=\frac{16}{\pi} \\ x=\sqrt[]{\frac{16}{\pi}} \end{gathered}$

Now, differentiate (1) with respect to x again,

$\begin{gathered} \frac{d^2V}{dx^2}=\frac{d}{dx}(84-3\pi(x^2)) \\ =-6\pi x \\ At\text{ x=}\sqrt[]{\frac{16}{\pi}} \\ \frac{d^2V}{dx^2}=-6\pi\sqrt[]{\frac{16}{\pi}}$

Since, the double derivative is negative.

$So,\text{ the volume is maximum at }\sqrt[]{\frac{16}{\pi}}$

So, the volume becomes,

$\begin{gathered} V=\pi(x^2)h \\ V=\pi(\sqrt[]{\frac{16}{\pi}})^2h \\ V=\frac{16h}{\pi} \end{gathered}$

Answer: maximum volume of the cylinder is,

6 0

1 year ago