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3 years ago

PLS HELP WILL MARK BRAINLIEST!!!

Mathematics

2 answers:

Anton [14]3 years ago

8 0

Answer:

8 3/10, 8.304, 167/20, 8.92

Step-by-step explanation:

Turn all of them into decimals

8 3/10 = 8.30

8.304

167/20 = 8.350

8.92

GrogVix [38]3 years ago

4 0

Answer:

The answer is 8.304 then 8.92 then 167/20 then 8 3/10. :)

Step-by-step explanation:

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Please help, im totally lost.

Korvikt [17]

B) multiplication of improper fractions: 4/3 x 4/3
expressed as improper fraction: 16/3
product expressed as mixed number: 5 1/3

why: 4 x 4 = 16, keeping the denominator (3) is 16/3. 16 isn't a multiple of 3, the closest is 15, and 15 / 3 is 5, but we since we have 16, we have one left, so it'd be 5 1/3

c) multiplication of improper fractions: 7/3 x 3/2
expressed as improper fraction: 21/6
product expressed as mixed number: 3 3/6 = 3 1/2

why: 7 x 3 = 21, 2 x 3 is 6, combine it, 21/6. multiple of 6 that is closest to 21 is 18. 18 / 6 = 3. we have 3 numbers left over. (21 - 18 = 3) so it's the 3 that we get, plus the 3 that we have left, is 3 3/6

d) multiplication of improper fractions: 11/6 x 7/5
expressed as improper fraction: 77/30
product expressed as mixed number: 2 17/30

why: 11 x 7 = 77, 6 x 5 = 30, combine it, 77/30. 77 goes into 30 twice. (30 x 6 = 60, can't add another 30, because its more than 77) we have 17 numbers left over, (77 - 60 = 17) so it's 2 17/30. can't reduce 17/30 because 17 doesn't have any common factors with 30. hope i helped!

6 0

3 years ago

For what side length(s) is the area of an equilateral triangle equal to 30 cm?? Only enter the number, in centimeters, rounded t

xenn [34]

Answer: The sides length are 8.32 cm

Step-by-step explanation:

An equilateral triangle has all his sides of the same lenght, so we assume that the triangle has an L lenght in his sides.

The area of a triangle is $Area = \frac{base * height}{2}$ where the base is L, the Area is 30 and an unknown height.

To determine the height, we cut the triangle in half and take one side. By simetry, one side has a base of $\frac{L}{2}$ , a hypotenuse of L and a the unknown height.

Then we apply the <em>Pythagoras theorem</em>, this states that <em>in a right triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides</em>, or, $hypotenuse = \sqrt{c^{2} + c^{2} }$ Where one c is $\frac{L}{2}$ and the other is the height.

Then we find one of the c of the equation wich will be the height.

$height = \sqrt{hypotenuse^{2}-base^{2} }$

$height = \sqrt{ L^{2} -\frac{L}{4} ^{2}}\\height = \sqrt{\frac{ 3L^{2}}{4} } \\\\height = \frac{\sqrt{3}L }{2}$

Finally, we use the triangle area mentioned before an find the value of L.

$30 = \frac{L*\frac{\sqrt{3}L }{2} }{2} \\\\L = \sqrt{\frac{120}{\sqrt{3} } } \\\\L = 8.32 cm$

6 0

3 years ago

Let S be the surface z^2+x^2=1 with -17<=y<=7. Find a parameteriztion of the surface using the polar coordinates transform

julsineya [31]

The parameterization is already given to you, you just need to determine the domain for each parameter. $z^2+x^2=1$ is a cylinder that lies on the $y$ -axis with radius 1. You need to take $0\le v\le2\pi$ to get the entire lateral face, and $-17\le u\le7$ .

6 0

4 years ago

(1 point) Find the length traced out along the parametric curve x=cos(cos(4t))x=cos⁡(cos⁡(4t)), y=sin(cos(4t))y=sin⁡(cos⁡(4t)) a

Mazyrski [523]

The length of a curve $C$ given parametrically by $(x(t),y(t))$ over some domain $t\in[a,b]$ is

$\displaystyle\int_C\mathrm ds=\int_a^b\sqrt{\left(\frac{\mathrm dx}{\mathrm dt}\right)^2+\left(\frac{\mathrm dy}{\mathrm dt}\right)^2}\,\mathrm dt$

In this case,

$x(t)=\cos(\cos4t)\implies\dfrac{\mathrm dx}{\mathrm dt}=-\sin(\cos4t)(-\sin4t)(4)=4\sin4t\sin(\cos4t)$

$y(t)=\sin(\cos4t)\implies\dfrac{\mathrm dy}{\mathrm dt}=\cos(\cos4t)(-\sin4t)(4)=-4\sin4t\cos(\cos4t)$

So we have

$\displaystyle\left(\frac{\mathrm dx}{\mathrm dt}\right)^2+\left(\frac{\mathrm dy}{\mathrm dt}\right)^2=16\sin^24t\sin^2(\cos4t)+16\sin^24t\cos^2(\cos4t)=16\sin^24t$

and the arc length is

$\displaystyle\int_0^1\sqrt{16\sin^24t}\,\mathrm dt=4\int_0^1|\sin4t|\,\mathrm dt$

We have

$\sin(4t)=0\implies4t=n\pi\implies t=\dfrac{n\pi}4$

where $n$ is any integer; this tells us $\sin(4t)\ge0$ on the interval $\left[0,\frac\pi4\right]$ and $\sin(4t)$ on $\left[\frac\pi4,1\right]$ . So the arc length is

$=\displaystyle4\left(\int_0^{\pi/4}\sin4t\,\mathrm dt-\int_{\pi/4}^1\sin4t\,\mathrm dt\right)$

$=-\cos(4t)\bigg_0^{\pi/4}-\left(-\cos(4t)\bigg_{\pi/4}^1\right)$

$=(\cos0-\cos\pi)+(\cos4-\cos\pi)=\boxed{3+\cos4}$

7 0

3 years ago

How do I solve this using cosine or sine ratio?

skad [1K]

You would use the simple format SOH-CAH-TOA
SOH: sine = opposite/hypotenuse
CAH: cosine = adjacent/hypotenuse
TOA: tangent = opposite/adjacent
----
So for this problem to find c you would use this equation:
sin42= c/7
7sin42=c
c=4.68

Then for d you would use this equation:
tan48=d/4.68
4.68tan48=d
=5.197

Hope this helps :)

6 0

4 years ago