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statuscvo [17]
2 years ago
12

How do you convert fraction to a decimal?​

Mathematics
1 answer:
Romashka-Z-Leto [24]2 years ago
7 0

\huge\text{Hey there!}

\large\text{Some people usually convert fractions into decimals by DIVIDING}\\\large\text{the TOP  number) by the DENOMINATOR (the BOTTOM number)}.

\large\text{Here is an EXAMPLE  of what I meant!}

\large\text{Fraction: }\mathsf{\dfrac{5}{6}}

\large\text{Equation: } \mathsf{5\div6}

\large\text{Answer: } \mathsf{0.833333}

\text{Good luck on your assignment and enjoy your day!}

~\frak{Amphitrite1040:)}

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If the sin of angle x is 4 over 5 and the triangle was dilated to be two times as big as the original, what would be the value o
beks73 [17]

Answer:

Sin of x does not change

Step-by-step explanation:

Whenever a triangle is dilated, the angle remains the same as well as the ratio for sides of triangle. For smshapes with dimensions, when shapes are dilated the dimensions has increment with common factor.

From trigonometry,

Sin(x)=opposite/hypotenose

Where x=4/5

Sin(4/5)= opposite/hypotenose

But we were given the scale factor of 2 which means the dilation is to two times big.

Then we have

Sin(x)=(2×opposite)/(2×hypotenose)

Then,if we divide by 2 the numerator and denominator we still have

Sin(x)=opposite/hypotenose

Which means the two in numerator and denominator is cancelled out.

Then we still have the same sin of x. as sin(4/5)

Hence,Sin of x does not change

4 0
2 years ago
A soft drink machine outputs a mean of 27 ounces per cup. The machine's output is normally distributed with a standard deviation
Rzqust [24]

Answer:  0.6065

Step-by-step explanation:

Given : The machine's output is normally distributed with

\mu=27\text{ ounces per cup}

\sigma=3\text{ ounces per cup}

Let x be the random variable that represents the output of machine .

z-score : z=\dfrac{x-\mu}{\sigma}

For x= 21 ounces

z=\dfrac{21-27}{3}\approx-2

For x= 28 ounces

z=\dfrac{28-27}{3}\approx0.33

Using the standard normal distribution table , we have

The p-value : P(21

P(z

Hence, the probability of filling a cup between 21 and 28 ounces=  0.6065

8 0
3 years ago
2/9 h = 8<br><br> h=<br><br> Pls tell me what h = !
Wewaii [24]

Answer:

h=36

.......................

3 0
3 years ago
Solve for x round to the nearest tenth.
lesantik [10]

Answer:

x ≈ 15.9

Step-by-step explanation:

Using Pythagoras' identity in the right triangle

x² + 6² = 17²

x² + 36 = 289 ( subtract 36 from both sides )

x² = 253 ( take the square root of both sides )

x = \sqrt{253} ≈ 15.9 ( to the nearest tenth )

3 0
2 years ago
Read 2 more answers
For each of the following vector fields F , decide whether it is conservative or not by computing curl F . Type in a potential f
Phantasy [73]

The key idea is that, if a vector field is conservative, then it has curl 0. Equivalently, if the curl is not 0, then the field is not conservative. But if we find that the curl is 0, that on its own doesn't mean the field is conservative.

1.

\mathrm{curl}\vec F=\dfrac{\partial(5x+10y)}{\partial x}-\dfrac{\partial(-6x+5y)}{\partial y}=5-5=0

We want to find f such that \nabla f=\vec F. This means

\dfrac{\partial f}{\partial x}=-6x+5y\implies f(x,y)=-3x^2+5xy+g(y)

\dfrac{\partial f}{\partial y}=5x+10y=5x+\dfrac{\mathrm dg}{\mathrm dy}\implies\dfrac{\mathrm dg}{\mathrm dy}=10y\implies g(y)=5y^2+C

\implies\boxed{f(x,y)=-3x^2+5xy+5y^2+C}

so \vec F is conservative.

2.

\mathrm{curl}\vec F=\left(\dfrac{\partial(-2y)}{\partial z}-\dfrac{\partial(1)}{\partial y}\right)\vec\imath+\left(\dfrac{\partial(-3x)}{\partial z}-\dfrac{\partial(1)}{\partial z}\right)\vec\jmath+\left(\dfrac{\partial(-2y)}{\partial x}-\dfrac{\partial(-3x)}{\partial y}\right)\vec k=\vec0

Then

\dfrac{\partial f}{\partial x}=-3x\implies f(x,y,z)=-\dfrac32x^2+g(y,z)

\dfrac{\partial f}{\partial y}=-2y=\dfrac{\partial g}{\partial y}\implies g(y,z)=-y^2+h(y)

\dfrac{\partial f}{\partial z}=1=\dfrac{\mathrm dh}{\mathrm dz}\implies h(z)=z+C

\implies\boxed{f(x,y,z)=-\dfrac32x^2-y^2+z+C}

so \vec F is conservative.

3.

\mathrm{curl}\vec F=\dfrac{\partial(10y-3x\cos y)}{\partial x}-\dfrac{\partial(-\sin y)}{\partial y}=-3\cos y+\cos y=-2\cos y\neq0

so \vec F is not conservative.

4.

\mathrm{curl}\vec F=\left(\dfrac{\partial(5y^2)}{\partial z}-\dfrac{\partial(5z^2)}{\partial y}\right)\vec\imath+\left(\dfrac{\partial(-3x^2)}{\partial z}-\dfrac{\partial(5z^2)}{\partial x}\right)\vec\jmath+\left(\dfrac{\partial(5y^2)}{\partial x}-\dfrac{\partial(-3x^2)}{\partial y}\right)\vec k=\vec0

Then

\dfrac{\partial f}{\partial x}=-3x^2\implies f(x,y,z)=-x^3+g(y,z)

\dfrac{\partial f}{\partial y}=5y^2=\dfrac{\partial g}{\partial y}\implies g(y,z)=\dfrac53y^3+h(z)

\dfrac{\partial f}{\partial z}=5z^2=\dfrac{\mathrm dh}{\mathrm dz}\implies h(z)=\dfrac53z^3+C

\implies\boxed{f(x,y,z)=-x^3+\dfrac53y^3+\dfrac53z^3+C}

so \vec F is conservative.

4 0
3 years ago
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