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vazorg [7]
2 years ago
8

Jason considered two similar televisions at a local electronics store. The generic version was based off the brand name and was

2/3 the size of the brand name. If the generic television set is 12 inches by 24 inches, what are the dimensions of the brand name television?
Mathematics
2 answers:
FinnZ [79.3K]2 years ago
5 0

Answer:

Width of brand name television = 18 inches

Height of brand name television = 36 inches

Step-by-step explanation:

Given,  

Width of generic TV = 12 inches  

Height of generic TV = 24 inches

Let the width of brand name television = W

Let the height of brand name television = H

As given in question,  

Size of generic television = 2/3  of the size of the brand name television  

Thus,  

2/3 W = 12 inches  

2/3 H = 24 inches  

On solving the given equations, we get –  

W = (12 x 3)/2 = 18 inches

H = (24 x 3)/2 = 36 inches

Width of brand name television = 18 inches

Height of brand name television = 36 inches

slavikrds [6]2 years ago
3 0

Keywords:

<em>Variables, televisions, generic version, TV brand, dimensions </em>

For this case we have two televisions, one generic version and one brand. We know that the generic version represents \frac {2} {3}the size of the brand. We must define two variables that represent the dimensions of the brand TV, so we have:

  • x: Brand TV length
  • y: Brand TV width

Dimensions of the generic TV:

Length = 12\\Width = 24

So:

\frac {2} {3} x = 12

\frac {2} {3} y = 24

By clearing the variables we have:

x = 12 \frac {3} {2} = 18\\y = 24 \frac {3} {2} = 36

Thus, the dimensions of the brand TV are 18 inches by 36 inches

Answer:

 The dimensions of the brand TV are 18 inches by 36 inches

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seropon [69]

Answer:

BD = <u>1</u> unit

AD = <u>1</u> unit

AB = <u>1.6</u> units

AC = <u>1.6</u> units

Step-by-step explanation:

In the picture attached, the triangle ABC is shown.

Given that triangle ABC is isosceles, then ∠B = ∠C

∠A + ∠B + ∠C = 180°

36° + 2∠B = 180°

∠B = (180° - 36°)/2

∠B = ∠C  = 72°

From Law of Sines:

sin(∠A)/BC = sin(∠B)/AC = sin(∠C)/AB

(Remember that BC is 1 unit long)

AB = AC = sin(72°)/sin(36°) = 1.6

In triangle ABD, ∠B = 72°/2 = 36°, then:

∠A + ∠B + ∠D = 180°

36° + 36° + ∠D = 180°

∠D =  180° - 36° - 36° = 108°

From Law of Sines:

sin(∠A)/BD = sin(∠B)/AD = sin(∠D)/AB

(now ∠A = ∠B)

BD = AD = sin(∠A)*AB/sin(∠D)

BD = AD = sin(36°)*1.6/sin(108°) = 1

3 0
3 years ago
Find an equation that models the path of a satellite if its path is a hyperbola, a = 55,000 km, and c= 81,000 km. Assume the cen
elena-14-01-66 [18.8K]

Answer:

\frac{x^2}{55000^2} - \frac{y^2}{59464^2} =1

Step-by-step explanation:

the transverse axis is horizontal.

so its a horizontal hyperbola

Center is the origin so center is (0,0)

Equation of horizontal hyperbola is

\frac{x^2}{a^2} - \frac{y^2}{b^2} =1

Given a= 55000 and c= 81000

c^2 = a^2 + b^2

81000^2 = 55000^2 + b^2

subtract 55000^2 on both sides

b  = sqrt(81000^2 - 55000^2)= 59464.27

now plug in the values

\frac{x^2}{55000^2} - \frac{y^2}{59464^2} =1

7 0
2 years ago
If r(X)=2-x^2 and w(X)=X-2, what is the range of (w*r)(X)
slega [8]
The given function are
r(x) = 2 - x²    and     w(x) = x - 2
<span>(w*r)(x) can be obtained by multiplying the both function together
</span>
So, <span>(w*r)(x) = w(x) * r(x) = (x-2)*(2-x²)</span>
<span>(w*r)(x) = x (2-x²) - 2(2-x²)</span>
            = 2x - x³ - 4 + 2x²

∴ <span>(w*r)(x) = -x³ + 2x² + 2x - 4
</span>

<span>It is a polynomial function with a domain equal to R
</span>
The range of <span>(w*r)(x) can be obtained by graphing the function
</span>
To graph (w*r)(x), we need to make a table between x and (w*r)(x)

See the attached figure which represents the table and the graph of <span>(w*r)(x)
</span>

As shown in the graph the range of <span>(w*r)(x) is (-∞,∞)
</span>

8 0
3 years ago
Read 2 more answers
PLEASE HELP QUESTION IN PICTURE.
nika2105 [10]

Answer:

D

Step-by-step explanation:

5 0
2 years ago
Read 2 more answers
Look at this cylinder:
Sedbober [7]
  • Height=h=8cm
  • Radius=r=4cm

We know

\boxed{\sf \star TSA_{(Cylinder)}=2\pi r(h+r)}

\\ \sf\longmapsto TSA_{(Old\:Cylinder)}=2\times \dfrac{22}{7}\times 4(8+4)

\\ \sf\longmapsto TSA_{(Old\:Cylinder)}=\dfrac{176}{7}(12)

\\ \sf\longmapsto TSA_{(Old\:Cylinder)}=\dfrac{2112}{7}

\\ \sf\longmapsto TSA_{(Old\:Cylinder)}=301.7cm^2

Now

  • New Radius=2(4)=8cm
  • New Height=2(8)=16cm

\\ \sf\longmapsto TSA_{(New\:Cylinder)}=2\times \dfrac{22}{7}\times 8(16+8)

\\ \sf\longmapsto TSA_{(New\:Cylinder)}=\dfrac{352}{7}(24)

\\ \sf\longmapsto TSA_{(New\:Cylinder)}=\dfrac{8448}{7}

\\ \sf\longmapsto TSA_{(New\:Cylinder)}=1204.7cm^2

So

\\ \sf\longmapsto \dfrac{TSA_{(New\:Cylinder)}}{TSA_{(Old\:Cylinder)}}=\dfrac{1204.7}{301.7}

\\ \sf\longmapsto \dfrac{TSA_{(New\:Cylinder)}}{TSA_{(Old\:Cylinder)}}=\dfrac{4}{1}

\\ \sf\longmapsto\underline{\boxed{\bf{ {TSA_{(New\:Cylinder)}}:{TSA_{(Old\:Cylinder)}}=4:1}}}

Hence our correct option is Option C

6 0
2 years ago
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