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3 years ago

The exact value of sin 5pi/12 is:

Mathematics

2 answers:

monitta3 years ago

6 0

Answer: A. SQRT2 (SQRT3 -1) / 4

explanation:

Just took the quiz on A P E X.

larisa86 [58]3 years ago

3 0

Answer:

this kinda math isn't my strong suit my best guess would be 0.02

You might be interested in

Which graph represents the function f(x)=x4+4x3−12x2−32x+64?

sp2606 [1]

$f(x)=x^4+4x^3-12x^2-32x+64$

The degree of f(x) is 4. Also the leading coefficient is 1 and it is positive

So as x approaches infinity then y approaches infinity

as x approaches -infinity then y approaches infinity

The first and fourth graph goes up and it satisfies the above . so we ignore the second and third graph.

Now we check the x intercepts of the first graph

x intercepts of first graph is -4 and 2

Plug in -4 for x in f(x) and check whether we get 0

$f(x)=x^4+4x^3-12x^2-32x+64$

$f(x)=(-4)^4+4(-4)^3-12(-4)^2-32(-4)+64=0$

Now plug in 2 for x and check

$f(x)=(2)^4+4(2)^3-12(2)^2-32(2)+64=0$

So -4 and 2 are the x intercepts that satisfies f(x)

Hence first option is the graph of $f(x)=x^4+4x^3-12x^2-32x+64$

5 0

3 years ago

Read 2 more answers

An experiment was performed to compare the abrasive wear of two different laminated materials. Twelve pieces of material 1 were

vladimir1956 [14]

Answer:

The calculated value Z= 4.8389> 1.96 at 0.05 level of significance.

The null hypothesis is rejected.

There is significance difference between that the abrasive wear of material 1 not exceeds that of material 2 by more than 2 units

Step-by-step explanation:

Step:-(1)

Given data the samples of material 1 gave an average (coded) wear of 85 units with a sample standard deviation of 4

Mean of the first sample x₁⁻ =85

standard deviation of the first sample S₁ = 4

Given data the samples of material 2 gave an average of 81 and a sample standard deviation of 5.

Mean of the first sample x₂⁻ =81

standard deviation of the first sample S₂ = 5

Step :-2

Null hypothesis: H₀: there is no significance difference between that the abrasive wear of material 1 exceeds that of material 2 by more than 2 units

Alternative hypothesis :H₁: there is significance difference between that the abrasive wear of material 1 exceeds that of material 2 by more than 2 units

Assume the populations to be approximately normal with equal variances.σ₁² =σ₂²

The test statistic

$Z= \frac{x_{1} -x_{2} }{\sqrt{\frac{S^2_{1} }{n_{1} } +\frac{S^2_{2} }{n_{2} } } }$

Given n₁=n₂=60.

$Z= \frac{85-81 }{\sqrt{\frac{(4)^2 }{60 } +\frac{5^2 }{60} } }$

On calculation, we get

Z = $\frac{4}{\sqrt{0.6833} }$

z = 4.8389

The tabulated value Z =1.96 at 0.05 level of significance.

The calculated value Z= 4.8389> 1.96 at 0.05 level of significance.

The null hypothesis is rejected.

Conclusion:-

there is significance difference between that the abrasive wear of material 1 not exceeds that of material 2 by more than 2 units.

3 0

3 years ago

Consider the following system of equations:

Umnica [9.8K]

X=3 and the work to get the answer is attached.

4 0

3 years ago

The graph of the function C(x) = −0.34x2 + 12x + 62 is shown. The function models the production cost, C, in thousands of dollar

Virty [35]

Answer:

0 ≤ x < 1.12 and 34.18 < x ≤ 39.87

Step-by-step explanation:

Let

x ----> is the number of tires produced, in thousands

C(x) ---> the production cost, in thousands of dollars

we have

$C(x)=-0.34x^{2} +12x+62$

This is a vertical parabola open downward (the leading coefficient is negative)

The vertex represent a maximum

The graph in the attached figure

we know that

Looking at the graph

For the interval [0,1.12) -----> $0\leq x$

The value of C(x) ----> $C(x) < 75$

That means ----> The production cost is under $75,000

For the interval (34.18,39.87] -----> $34.18 < x\leq 39.87$

The value of C(x) ----> $C(x) < 75$

That means ----> The production cost is under $75,000

Remember that the variable x (number of tires) cannot be a negative number

therefore

If the company wants to keep its production costs under $75,000 a reasonable domain for the constraint x is

0 ≤ x < 1.12 and 34.18 < x ≤ 39.87

7 0

4 years ago

Help help , Please help! Brainliest if correct! What was the equation of the graph below before it was shifted to the left 1.5 u

DENIUS [597]

Answer:

A. G(x) = (x -3)^3 -(x -3)

Step-by-step explanation:

The graph before it was shifted left will be a right-shift of the equation shown. That is accomplished by replacing x with (x-1.5). Then the right-shifted equation is ...

G(x) = ((x-1.5) -1.5)^3 -((x -1.5) -1.5)

G(x) = (x -3)^3 -(x -3) . . . . matches choice A

3 0

3 years ago