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stepladder [879]
3 years ago
5

a web music store offers two versions of a popular song. the size of the standard version is 2.9 megabytes (MB). the size of the

high-quality version is 4.8 MB. yesterday, the high-quality version was downloaded four times as often as the standard version. The total size downloaded for the two versions was 6188 MB. How many downloads of the standard version were there?​
Mathematics
1 answer:
kenny6666 [7]3 years ago
8 0

Answer:

280 downloads

Step-by-step explanation:

-let x denote the number of standard version downloads.

-Given that the high-quality version was downloaded four times as often as the standard version it is denoted as 4x

-This can then be expressed as:

x(2.9)+4x(4.8)=6188\\\\2.9x+19.2x=6188\\\\22.10x=6188\\\\x=280

Hence, the number of standard version downloads is 280

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The following polynomial is defined on the interval [-2, 2]:
Bess [88]

Answer:

Using Matlab code for Fourier series to calculate for the function, see the attached

Step-by-step explanation:

Go through the picture step by step.

6 0
3 years ago
If 1 meter = 100cm then what is the relationship between m3 and cm3?
ddd [48]
If the 1m turns into 3m then he cm increases also by 3.The 3 meters would equal 300 cm.
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3 years ago
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The area of the triangle formed by x− and y− intercepts of the parabola y=0.5(x−3)(x+k) is equal to 1.5 square units. Find all p
Juliette [100K]

Check the picture below.


based on the equation, if we set y = 0, we'd end up with 0 = 0.5(x-3)(x-k).

and that will give us two x-intercepts, at x = 3 and x = k.

since the triangle is made by the x-intercepts and y-intercepts, then the parabola most likely has another x-intercept on the negative side of the x-axis, as you see in the picture, so chances are "k" is a negative value.

now, notice the picture, those intercepts make a triangle with a base = 3 + k, and height = y, where "y" is on the negative side.

let's find the y-intercept by setting x = 0 now,


\bf y=0.5(x-3)(x+k)\implies y=\cfrac{1}{2}(x-3)(x+k)\implies \stackrel{\textit{setting x = 0}}{y=\cfrac{1}{2}(0-3)(0+k)} \\\\\\ y=\cfrac{1}{2}(-3)(k)\implies \boxed{y=-\cfrac{3k}{2}} \\\\[-0.35em] ~\dotfill\\\\ \stackrel{\textit{area of a triangle}}{A=\cfrac{1}{2}bh}~~ \begin{cases} b=3+k\\ h=y\\ \quad -\frac{3k}{2}\\ A=1.5\\ \qquad \frac{3}{2} \end{cases}\implies \cfrac{3}{2}=\cfrac{1}{2}(3+k)\left(-\cfrac{3k}{2} \right)


\bf \cfrac{3}{2}=\cfrac{3+k}{2}\left( -\cfrac{3k}{2} \right)\implies \stackrel{\textit{multiplying by }\stackrel{LCD}{2}}{3=\cfrac{(3+k)(-3k)}{2}}\implies 6=-9k-3k^2 \\\\\\ 6=-3(3k+k^2)\implies \cfrac{6}{-3}=3k+k^2\implies -2=3k+k^2 \\\\\\ 0=k^2+3k+2\implies 0=(k+2)(k+1)\implies k= \begin{cases} -2\\ -1 \end{cases}


now, we can plug those values on A = (1/2)bh,


\bf \stackrel{\textit{using k = -2}}{A=\cfrac{1}{2}(3+k)\left(-\cfrac{3k}{2} \right)}\implies A=\cfrac{1}{2}(3-2)\left(-\cfrac{3(-2)}{2} \right)\implies A=\cfrac{1}{2}(1)(3) \\\\\\ A=\cfrac{3}{2}\implies A=1.5 \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ \stackrel{\textit{using k = -1}}{A=\cfrac{1}{2}(3+k)\left(-\cfrac{3k}{2} \right)}\implies A=\cfrac{1}{2}(3-1)\left(-\cfrac{3(-1)}{2} \right) \\\\\\ A=\cfrac{1}{2}(2)\left( \cfrac{3}{2} \right)\implies A=\cfrac{3}{2}\implies A=1.5

7 0
3 years ago
Two rectangular prisms have a combined volume of 432 cubic feet. Prism A has half the volume of Prism B.
dexar [7]

Answer:

a. 144 cubic foot

b. 12 ft

c. 18 ft

Step-by-step explanation:

Let the volume of prism A be V_a

and that of prism b be V_b

ATQ,   V_a  +  V_b= 432

also V_a= 0.5 V_b

⇒1.5 V_b= 432

= V_b= 432/1.5= 288 cubic feet

therefore V_a= 144  cubic feet

volume of prism= area of base×height = V_a

24×h = 144

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A_b= 2/3×24

⇒base area of prism B= 16 sq.ft

now 16×h_b= 288⇒h_b= 288/16= 18 ft

3 0
3 years ago
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Tesla Battery Recharge Time.The electric­ vehicle manufacturing company Tesla esti­mates that a driver who commutes 50 miles per
madam [21]

Answer:

a.) f(x) = \frac{1}{30} where 90 < x < 120

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Step-by-step explanation:

Let

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⇒X ≈ ∪ ( 90, 120 )

a.)

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b.)

P(x < 110) = \int\limits^{110}_{90} {\frac{1}{30} } \, dx

               = \frac{1}{30}[x]\limits^{110}_{90}  = \frac{1}{30} [ 110 - 90 ] = \frac{1}{30} [ 20] = \frac{2}{3}

c.)

P(x > 100 ) = \int\limits^{120}_{100} {\frac{1}{30} } \, dx

                 = \frac{1}{30}[x]\limits^{120}_{100}  = \frac{1}{30} [ 120 - 100 ] = \frac{1}{30} [ 20] = \frac{2}{3}

d.)

P(95 < x< 110)  = \int\limits^{110}_{95} {\frac{1}{30} } \, dx

                       = \frac{1}{30}[x]\limits^{110}_{95}  = \frac{1}{30} [ 110 - 95 ] = \frac{1}{30} [ 15] = \frac{1}{2}

7 0
3 years ago
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