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Cerrena [4.2K]
4 years ago
13

George is comparing two different phones. The screen of phone A has a width of 5.3 cm. The width of the screen on phone B

Mathematics
2 answers:
bija089 [108]4 years ago
8 0

Answer:

To answer this question, you need to makes the unit of the smartphone screen same. In this case, phone A is using decimal and phone B is using a fraction. You can use which was easier.

Let's try to use decimal. Then you need to convert phone B fraction width into decimal width. The calculation would be: 5cm + 1/3cm= 5 cm + 0.333cm= 5.333cm

From here it clear that phone B has a wider screen than phone A.

Step-by-step explanation:

liq [111]4 years ago
3 0

Answer:

The screen of phone B is wider because 5\frac{1}{3}  is greater than 5.3.

Step-by-step explanation:

We are given that

The width of screen of phone A=5.3 cm

The width of screen of phone B=5\frac{1}{3} cm

We have to find the statement explains which phone has the wider screen.

The width of screen of phone B=5+\frac{1}{3} cm

The width of screen of phone B=5+0.33 cm

The width of screen of phone B=5.33 cm

Now, we make first like decimal number and then  we will compare.

Now, write zero at end of number in 5.3

Then, the number becomes 5.30 .

The hundredth part 3 in  5.33 is greater than hundredth part 0 of 5.30.

Therefore, 5.33 >5.30

Hence, the screen of phone B is wider because 5\frac{1}{3} >5.3.

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arsen [322]

Hi There!

Step-by-step explanation and Answer:

21 * 0.2 = $4.2 (Discount)

21 - 4.2 = $16.8 (Sale Price)

Hope This Helps :)

5 0
4 years ago
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The data table represents the distance between a well-known lighthouse and a cruise ship over time. The cruise ship is travellin
Trava [24]
Find the velocity first, the "slope".

m=(y2-y1)/(x2-x1)

m=(95.5-53)/(4-2)

m=21.25  ?  Your equations are wrong, I'll double check with another set of points...

m=(350.5-308)/(16-14)

m=21.25  LOL, yep none of your lines has the correct slope.

Anyway...

y=21.25x+b, using any point we can solve for the y-intercept, or "b", (2, 53)

53=21.25(2)+b

53=42.5+b

10.5=b so the distance to the lighthouse as a function of time is:

y=21.25x+10.5

So essentially, you should fire your teacher or burn that book you are using :)

6 0
4 years ago
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16. Multiply the polynomials: (x + 3)(x2 – 5x + 12)
Dvinal [7]
The answer would be -3x^2+3 x + 36
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3 years ago
(c). It is well known that the rate of flow can be found by measuring the volume of blood that flows past a point in a given tim
aleksklad [387]

(i) Given that

V(R) = \displaystyle \int_0^R 2\pi K(R^2r-r^3) \, dr

when R = 0.30 cm and v = (0.30 - 3.33r²) cm/s (which additionally tells us to take K = 1), then

V(0.30) = \displaystyle \int_0^{0.30} 2\pi \left(0.30-3.33r^2\right)r \, dr \approx \boxed{0.0425}

and this is a volume so it must be reported with units of cm³.

In Mathematica, you can first define the velocity function with

v[r_] := 0.30 - 3.33r^2

and additionally define the volume function with

V[R_] := Integrate[2 Pi v[r] r, {r, 0, R}]

Then get the desired volume by running V[0.30].

(ii) In full, the volume function is

\displaystyle \int_0^R 2\pi K(R^2-r^2)r \, dr

Compute the integral:

V(R) = \displaystyle \int_0^R 2\pi K(R^2-r^2)r \, dr

V(R) = \displaystyle 2\pi K \int_0^R (R^2r-r^3) \, dr

V(R) = \displaystyle 2\pi K \left(\frac12 R^2r^2 - \frac14 r^4\right)\bigg_0^R

V(R) = \displaystyle 2\pi K \left(\frac{R^4}2- \frac{R^4}4\right)

V(R) = \displaystyle \boxed{\frac{\pi KR^4}2}

In M, redefine the velocity function as

v[r_] := k*(R^2 - r^2)

(you can't use capital K because it's reserved for a built-in function)

Then run

Integrate[2 Pi v[r] r, {r, 0, R}]

This may take a little longer to compute than expected because M tries to generate a result to cover all cases (it doesn't automatically know that R is a real number, for instance). You can make it run faster by including the Assumptions option, as with

Integrate[2 Pi v[r] r, {r, 0, R}, Assumptions -> R > 0]

which ensures that R is positive, and moreover a real number.

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Pepsi [2]

Answer:

Vertical Asymptote:

x=1

Horizontal asymptote:

it does not exist

Step-by-step explanation:

we are given

f(x)=log_3(x-1)

Vertical asymptote:

we know that vertical asymptotes are values of x where f(x) becomes +inf or -inf

we know that any log becomes -inf when value inside log is zero

so, we can set value inside log to zero

and then we can solve for x

x-1=0

we get

x=1

Horizontal asymptote:

we know that

horizontal asymptote is a value of y when x is +inf or -inf

For finding horizontal asymptote , we find lim x-->inf or -inf

\lim_{x \to \infty}  f(x)= \lim_{x \to \infty}log_3(x-1)

\lim_{x \to \infty}  f(x)=log_3(\infty-1)

\lim_{x \to \infty}  f(x)=undefined

so, it does not exist

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3 years ago
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