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3 years ago

Classify the equation 64x - 16 = 16(4x - 1) as

Mathematics

2 answers:

Maksim231197 [3]3 years ago

7 0

Answer:

there is no solution 64x-16=64x-16

Step-by-step explanation:

if you distrubute then you will see the numbers are the same on both sides

kicyunya [14]3 years ago

6 0

Wait no... it’s infinitely many solutions both sides have the same exact line. The solution of 2 lines is where the lines intersect. The lines intersect on every single point of the line. Therefore it has infinitely many solutions.

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HELP

adelina 88 [10]

Answer:

$(x-6)^2=8$

Step-by-step explanation:

$x^2-12x+32=4$

$x^2-12x+32+4=4+4$

$x^2-12x+36=8$

$(x-6)^2=8$

Basically, you need to add a number to both sides so that you can reduce the left side to a square. In this case, you add 4.

8 0

2 years ago

Solve the equation sin38=x/10

sashaice [31]

Answer:

6.1566

Step-by-step explanation

Easy:

Sin 38 = x/10

Multiply 10 on both sides to cancel out the 10 under the x

This will equal to:

10(sin 38) = x

Simply enter 10(sin 38) in the calculator like desmos scientific.

Your answer is about 6.1566

5 0

3 years ago

Read 2 more answers

An important problem in industry is shipment damage. A pottery producing company ships its product by truck and determines that

Veronika [31]

Answer:

e) (10.94, 11.66)

Step-by-step explanation:

We have a small sample size of n = 15 departing truck loads (we suppose that the sample comes from a normal distribution), the average number of damaged items per truck load is calculated to be $\bar{x} = 11.3$ with a calculated sample variance $s^2$ = 0.64. The 90% confidence interval for the true mean of damaged items is given by $\bar{x}\pm t_{\alpha/2}(\frac{s}{\sqrt{n}})$ where $t_{\alpha/2}$ is the $\alpha/2$ th quantile of the t distribution with n - 1 = 14 degrees of freedom.

As we want the 90% confidence interval, we have that $\alpha = 0.1$ and the confidence interval is $11.3\pm t_{0.05}(\frac{0.8}{\sqrt{15}})$ where $t_{0.05} = -1.7613$ . Then, we have $11.3\pm (-1.7613)(\frac{0.8}{\sqrt{15}})$ and the 90% confidence interval is given by (10.94, 11.66)