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2 years ago

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A TV show had 4,500,000 viewers for their first episode and 8,500,000 viewers for their second episode. How many viewers did the

y have overall? Ima give alot of points for this because its for a quiz

1 answer:

Nataly [62]2 years ago

3 0

Answer:

13,000,000

Step-by-step explanation:

You might be interested in

The question is below the screen

sesenic [268]

9514 1404 393

Answer:

x = 12

Step-by-step explanation:

The square of the length of the tangent segment is equal to the product of lengths from the external common point to the two circle intersections of the secant.

x^2 = (2)(70+2)

x^2 = 144

x = √144

x = 12

4 0

3 years ago

B) The price of an electric fan is fixed 20% above its cost price. When it is sold allowing

kenny6666 [7]

Answer:

$\boxed{ \boxed{ \sf {Marked \: price \: = Rs \: 1500}}}$

$\boxed{ \bold{ \boxed{ \sf{Selling \: price = \: Rs \: 1230}}}}$

Step-by-step explanation:

Let Cost price ( C.P ) be x

<u>Finding </u><u>the </u><u>Marked </u><u>price </u><u>and </u><u>selling </u><u>price </u>

Marked price = $\sf{x + 20\% \: of \: x}$

⇒ $\sf{x + \frac{20}{100} \times x}$

⇒ $\sf{ \frac{x \times 100 + 20x}{100} }$

⇒ $\sf{ \frac{120x}{100} }$ ⇒ ( i )

Selling price = $\sf{marked \: price \: - 18\% \: of \: marked \: price}$

⇒ $\sf{ \frac{120x}{100} - \frac{18}{100} \times \frac{120x}{100} }$

⇒ $\sf{ \frac{120x}{100} - \frac{54x}{250} }$

⇒ $\sf{ \frac{120x \times 5 - 54x \times 2}{500} }$

⇒ $\sf{ \frac{600x - 108x}{500} }$

⇒ $\sf{ \frac{492x}{500} }$ ⇒ ( ii )

<u>Finding </u><u>the </u><u>value </u><u>of </u><u>x </u><u>(</u><u> </u><u>Cost </u><u>price </u><u>)</u>

$\sf{loss = cost \: price - selling \: price}$

⇒ $\sf{20 = x - \frac{492x}{500} }$

⇒ $\sf{20 = \frac{x \times 500 - 492x}{500} }$

⇒ $\sf{20 = \frac{8x}{500} }$

⇒ $\sf{8x = 10000}$

⇒ $\sf{x = \frac{10000}{8} }$

⇒ $\sf{x = \: Rs \: 1250}$

Value of x ( cost price ) = Rs 1250

<u>Now</u><u>,</u><u> </u><u>Replacing </u><u>the </u><u>value </u><u>of </u><u>x </u><u>in </u><u>(</u><u> </u><u>i </u><u>)</u><u> </u><u>in </u><u>order </u><u>to </u><u>find </u><u>the </u><u>value </u><u>of </u><u>marked </u><u>price</u>

$\sf{marked \: price = \frac{120x}{100} }$

⇒ $\sf{ \frac{120 \times 1250}{100} }$

⇒ $\sf{ \: Rs \: 1500}$

<u>Replacing </u><u>value </u><u>of </u><u>x </u><u>in </u><u>(</u><u> </u><u>ii </u><u>)</u><u> </u><u>in </u><u>order </u><u>to </u><u>find </u><u>the </u><u>value </u><u>of </u><u>selling </u><u>price</u>

$\sf{selling \: price = \frac{492 \: x}{500} }$

⇒ $\sf{ \frac{615000}{500} }$

⇒ $\sf{ \: Rs \: 1230}$

Thus , Marked price of the fan = Rs 1500

Selling price of the fan = Rs 1230

Hope I helped!

Best regards!!

6 0

3 years ago

A triangle with a height of 5cm and a base length of 12cm what is the perimeter

Lisa [10]

Answer: 5+5+12+12

Step-by-step explanation:

7 0

3 years ago

Giving 45 points! I need help!

Monica [59]

where is the question?

8 0

3 years ago

Looking at the two quadratic functions below (1 & 2), answer the following questions.

algol [13]

Part A:

Given parabola (1) to be $f(x) =-(x+12)^2 -6$ , and parabola (2) to be $f(x)=13(x-4)^2+1$

Notice that parabola (2) is stretched horizontally by a factor of 13 which is greater than 1. This means that parabola (2) is further away from the x-axis than parabola (1). (i.e. parabola (2) is more 'vertical' than parabola (1).

Therefore, parabola (1) is wider than parabola (2).

Part B:

A parabola open up when the coefficient of the quadratic term (the squared term) is positive and opens down when the coefficient of the quadratic term is negative.

Given parabola (1) to be $f(x) =-(x+12)^2 -6$ , and parabola (2) to be $f(x)=13(x-4)^2+1$

Notice that the coefficient of the quadratic term is positive for parabola (2) and negative for parabola (1), therefore, the parabola that will open down is parabola (1).

Part C:

For any function, f(x), the graph of the function is moved p places to the left when p is added to x (i.e. f(x + p)) and moves p places to the right when p is subtracted from x (i.e. f(x - p)).

Given parabola (1) to be $f(x) =-(x+12)^2 -6$ , and parabola (2) to be $f(x)=13(x-4)^2+1$

Notice that in parabola (1), 12 is added to x, which means that the graph of the parent function is shifted 12 places to the left while in parabola (2), 4 is subtracted from x, which means that the graph of the parent function is shifted 4 places to the right.

Therefore, the parabola that would be furthest left on the x-axis is parabola (1).

Part D:

For any function, f(x), the graph of the function is moved q places up when q is added to the function (i.e. f(x) + q) and moves q places down when q is subtracted from the function (i.e. f(x) - q).

Given parabola (1) to be $f(x) =-(x+12)^2 -6$ , and parabola (2) to be $f(x)=13(x-4)^2+1$

Notice that in parabola (2), 1 is added to the function, which means that the graph of the parent function is shifted 1 place up while in parabola (1), 6 is subtracted from the function, which means that the graph of the parent function is shifted 6 places down.

Therefore, the parabola that would be highest on the y-axis is parabola (2).

7 0

3 years ago