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3 years ago

What is the value of x if x = 64?

Mathematics

2 answers:

mrs_skeptik [129]3 years ago

5 0

Answer: 8x8, 2x32, or 4x16

Step-by-step explanation:

The value of x is either one of these multiplications. All of these equal 64, so choose one and use the multiplication that fits you best.

weeeeeb [17]3 years ago

4 0

I am pretty sure it is 64

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Polynomial division (4x^2 +7x +6) ÷(x+ 2)

jolli1 [7]

Answer:

4x2 + 7x - 6

————————————

x - 2

Step-by-step explanation:

8 0

3 years ago

Read 2 more answers

Find the general solution of the differential equation and check the result by differentiation. (Use C for the constant of integ

atroni [7]

Answer: $y=Ce^(^3^t^{^9}^)$

Step-by-step explanation:

Beginning with the first differential equation:

$\frac{dy}{dt} =27t^8y$

This differential equation is denoted as a separable differential equation due to us having the ability to separate the variables. Divide both sides by 'y' to get:

$\frac{1}{y} \frac{dy}{dt} =27t^8$

Multiply both sides by 'dt' to get:

$\frac{1}{y}dy =27t^8dt$

Integrate both sides. Both sides will produce an integration constant, but I will merge them together into a single integration constant on the right side:

$\int\limits {\frac{1}{y} } \, dy=\int\limits {27t^8} \, dt$

$ln(y)=27(\frac{1}{9} t^9)+C$

$ln(y)=3t^9+C$

We want to cancel the natural log in order to isolate our function 'y'. We can do this by using 'e' since it is the inverse of the natural log:

$e^l^n^(^y^)=e^(^3^t^{^9} ^+^C^)$

$y=e^(^3^t^{^9} ^+^C^)$

We can take out the 'C' of the exponential using a rule of exponents. Addition in an exponent can be broken up into a product of their bases:

$y=e^(^3^t^{^9}^)e^C$

The term e^C is just another constant, so with impunity, I can absorb everything into a single constant:

$y=Ce^(^3^t^{^9}^)$

To check the answer by differentiation, you require the chain rule. Differentiating an exponential gives back the exponential, but you must multiply by the derivative of the inside. We get:

$\frac{d}{dx} (y)=\frac{d}{dx}(Ce^(^3^t^{^9}^))$

$\frac{dy}{dx} =(Ce^(^3^t^{^9}^))*\frac{d}{dx}(3t^9)$

$\frac{dy}{dx} =(Ce^(^3^t^{^9}^))*27t^8$

Now check if the derivative equals the right side of the original differential equation:

$(Ce^(^3^t^{^9}^))*27t^8=27t^8*y(t)$

$Ce^(^3^t^{^9}^)*27t^8=27t^8*Ce^(^3^t^{^9}^)$

QED

I unfortunately do not have enough room for your second question. It is the exact same type of differential equation as the one solved above. The only difference is the fractional exponent, which would make the problem slightly more involved. If you ask your second question again on a different problem, I'd be glad to help you solve it.

7 0

2 years ago

Can anyone help me solve this

yawa3891 [41]

Answer:

a. $45.54; b. $51.42; c. See below

Step-by-step explanation:

a. 193 min

Service charge = $40.00

1st 100 min @ 0.021 = 2.10

Next 93 min @0.037 = 3.44

Total = $45.54

b. 317 min

Service charge = $40.00

1st 100 min @ 0.021 = 2.10

2nd 100 min @ 0.037 = 3.70

117 min @ 0.048 = 5.62

Total = $51.42

c. Piecewise function

The charge is

$40.00 + 0.021t if t ≤ 100
$42.10 + 0.037(t - 100) if 100 < t ≤ 200
$45.80 + 0.048 (t - 200) if t > 200

which we can write like this:

$f(t) =\begin{cases}40.00 + 0.021t & t \leq 100\\42.10 + 0.037(t-100) & 100 < t \leq200\\45.80 + 0.048(t-200) & t > 200\end{cases}$

5 0

3 years ago

Write 0.05 as a fraction. A) 1 12 B) 1 15 Eliminate C) 1 18 D) 1 9

zheka24 [161]

$x=0.0\overline{5}\\\\x=0.0555...\qquad\text{multiply both sides by 10}\\\\10x=0.555...\qquad\text{multiply both sides by 10}\\\\100x=5.555...\\\\\text{Make the difference:}\\\\100x-10x=5.555...-0.555...\\\\90x=5\qquad\text{divide both sides by 90}\\\\x=\dfrac{5}{90}\\\\x=\dfrac{5:5}{90:5}\\\\x=\dfrac{1}{18}\\\\Answer:\ \boxed{0.0\overline{5}=\dfrac{1}{18}}\to\boxed{C)}$

6 0

3 years ago

Prove AED is equal to DEB. Do not use vertical angle theorem

Nastasia [14]

Answer:

Step-by-step explanation:

From the figure attached,

Given:

Lines AB and CD are intersecting each other at a point E.

To prove:

∠ACB ≅ ∠DEB

Statements Reasons

1). ∠AEC + ∠BEC = 180° 1). Linear pair theorem

2). ∠DEB + ∠BEC = 180° 2). Linear pair theorem

3). ∠AEC + ∠BEC = ∠DEB + ∠BEC 3). Transitive property of equality

4). ∠AEC = ∠DEB 4). Subtraction property of equality

6 0

3 years ago