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2 years ago

Answers to question 1, a,b,c,d,e and f

Mathematics

1 answer:

denpristay [2]2 years ago

6 0

No idea I don't know the answer

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Inverse Variation

AURORKA [14]

$(1)\\R\cdot I=a\ \ \ \ and\ \ \ 50\cdot 0.15=a\\\\ \ \Rightarrow\ \ \ a=7.5\ \ \ \Rightarrow\ \ I= \frac{7.5}{R} \ \ \ \Rightarrow\ \ \ Ans.\ A.\\\\(2)\\x\cdot y=a\ \ \ and\ \ \ 8\cdot6=a\\\\ \ \Rightarrow\ \ \ a=48\ \Rightarrow\ \ \ xy=48\ \ \ \Rightarrow\ \ \ 10y=48\ \Rightarrow\ \ \ y=4.8\ \Rightarrow\ \ \ Ans.\ D.\\\\(3)\\xy=-12\ \ \Rightarrow\ \ \ constant=-12\ \ \Rightarrow\ \ \ Ans.\ D.\\\\$

$(4)\\xy=-12\\\\e.g.:\ \ \ x=3\ \ and\ \ y=-4\ \ \ \ or\ \ \ \ x=-3\ \ and\ \ y=4\ \ \Rightarrow\ \ Ans.\ A.\\\\(5)\\V\cdot T=a\ \ \ and\ \ \ 18\cdot 3=a\\\\ \ \ \Rightarrow\ \ \ a=54\ \ \ \Rightarrow\ \ \ V\cdot T=54\ \ \ \Rightarrow\ \ \ V= \frac{54}{T} \ \ \ \Rightarrow\ \ \ Ans.\ C.$

5 0

3 years ago

ZA Round your answer to the nearest hundredth.

Ahat [919]

Answer:

A = 48.81°

Step-by-step explanation:

Angles in a Right Triangle

When the side lengths of a right triangle are known and an angle must be determined, we can use the trigonometric ratios that relate angles and sides.

The tangent ratio is defined as:

$\displaystyle \tan A=\frac{\text{opposite leg}}{\text{adjacent leg}}$

Opposite leg to angle A is 8 and adjacent leg is 7, thus:

$\displaystyle \tan A=\frac{8}{7}$

Using the inverse tangent funcion:

$\displaystyle A=\arctan\frac{8}{7}$

Calculating:

A = 48.81°

5 0

2 years ago

The volume of a cube is 1 cubic foot. What is the length of each side of the cube

Alona [7]

Answer:

Step-by-step explanation:

7 0

2 years ago

What is the volume of a block of ice measuring 3 meters long, 1.5 meters wide, and 2 meters high?

vekshin1

The volume would be 9meters³

3x1.5x2

8 0

2 years ago

Read 2 more answers

Evaluate the limit as x approaches 0 of (1 - x^(sin(x)))/(x*log(x))

e-lub [12.9K]

$sin~ x \approx x ~ ~\sf{as}~~ x \rightarrow 0$

We can replace sin x with x anywhere in the limit as long as x approaches 0.

Also,

$\large \lim_{ x \to 0 } ~ x^x = 1$

I will make the assumption that log(x)=ln(x).

The limit result can be proven if the base of log(x) is 10.

$\large \lim_{x \to 0^{+}} \frac{1- x^{\sin x} }{x \log x } \\~\\ \large = \lim_{x \to 0^{+}} \frac{1- x^{\sin x} }{ \log( x^x) } \\~\\ \large = \lim_{x \to 0^{+}} \frac{1- x^{x} }{ \log( x^x) } ~~ \normalsize{\text{ substituting x for sin x } } \\~\\ \large = \frac{\lim_{x \to 0^{+}} (1) - \lim_{x \to 0^{+}} \left( x^{x}\right) }{ \log( \lim_{x \to 0^{+}}x^x) } = \frac{1-1}{\log(1)} = \frac{0}{0}$

We get the indeterminate form 0/0, so we have to use Lhopitals rule

 $\large \lim_{x \to 0^{+}} \frac{1- x^{x} }{ \log( x^x) } =_{LH} \lim_{x \to 0^{+}} \frac{0 -x^x( 1 + \log (x)) }{1 + \log (x) } \\ = \large \lim_{x \to 0^{+}} (-x^x) = \large - \lim_{x \to 0^{+}} (x^x) = -1$

Therefore,

 $\large \lim_{x \to 0^{+}} \frac{1- x^{\sin x} }{x \log x } =\boxed{ -1}$

3 0

2 years ago