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lys-0071 [83]
3 years ago
11

A radio station is giving away a $100 bill to every 30th caller and movie tickets to every 40th caller. Which call will be the f

irst to win both prizes
Mathematics
1 answer:
Ymorist [56]3 years ago
3 0

Answer:

120th caller

Step-by-step explanation:

According to the problem, calculation of the given data are as follows,

$100 bill = 30th caller

Movie tickets = 40th caller

So, to find out first to win both prizes, we will use LCM method,

LCM of 30 and 40 =       10 | 30 , 40

                                            | 3    , 4

LCM of 30 and 40 = 10 × 3 × 4 = 120

Hence, 120th caller will be the first to win both prizes.

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Use the properties of real numbers to rewrite the expression. 2/5 * (-7) * 5/2
Mama L [17]

For this case we have the following expression:

\frac{2}{5}*(- 7)*\frac{5}{2}

Using the associative property we can rewrite the expression in the following way:

(\frac{2}{5}*\frac{5}{2})*(-7)

Finally, simplifying we have:

(1) (- 7)

-7

Answer:

Rewriting the expression we have that the result is:

d. -7

4 0
3 years ago
Find the length of chord QS <br><br> 1)5<br> 2)33<br> 3)21<br> 4)32
Harrizon [31]

Answer:

Length of Chord QS = 33

Step-by-step explanation:

<u>Length of Chord QS</u>:

QW X WS = PW = WR

12(4x + 1) = 14(3x + 3)

48x + 12 = 42x + 42

48x - 42x = 42 - 12

6x = 30

x = \frac{30}{6} = 5

∴ Length of Chord QS = 12 + 4(5) + 1 = 13 + 20 = 33

The intersecting chords theorem or just The chord theorem is a statement in elementary geometry that describes a relation of the four line segments created by two intersecting chords within a circle. It states that the products of the lengths of the line segments on each chord are equal.  Each chord is cut into two segments at the point of where they intersect. One chord is cut into two line segments A and B. The other into the segments C and D.  This theorem states that A×B is always equal to C×D no matter where the chords are.

8 0
3 years ago
Please help.<br> Is algebra.<br> PLEASE HELP NO LINKS OR FILES.<br> I don't want links.
alexandr1967 [171]

9514 1404 393

Answer:

  9n^4

Step-by-step explanation:

The divisor and quotient can be interchanged to find the divisor:

  \dfrac{18n^6+27n^5-36n^4}{2n^2+3n-4}=\boxed{\text{Blank 1}}

Such division is carried out by first finding the quotient of the highest-degree terms:

  \dfrac{18n^6}{2n^2}=\dfrac{18}{2}n^{6-2}=9n^4

This value is used to multiply the denominator and subtract that product from the numerator to find the new numerator. The new numerator is zero, so the value that goes in Blank 1 is ...

  9n^4

_____

The attachment shows the long division.

7 0
3 years ago
Helppppppppppppppppppppppppppppppppppppppppppppppppppppp
DiKsa [7]

Answer:

8.40049 × 10^7

Step-by-step explanation:

4 0
3 years ago
I NEED HELP PLEASE, THANKS! :)
Alenkasestr [34]

Answer:

Option A

Step-by-step explanation:

This is a great question!

The first thing we want to do here is to graph the system of " inequalities " -

\begin{bmatrix}2x+9y\le \:100\\ 9x+y\le \:54\end{bmatrix}

As x is ≥ 0, respectively y ≥ 0, it makes things a lot simpler, as it kind of restricts us to quadrant 1, but not entirely.

First change each inequality to " slope - intercept " form, as such -

2x + 10y \leq 100,\\10y \leq 100 - 2x,\\y \leq - 1 / 5x + 10\\----------\\9x + y \leq 54,\\y \leq - 9x + 54

The graphed solutions would be in the attachment. I have colored the region with which the two intersect, and the fact that we are limited to quadrant 1. In this colored region there are 3 major points, ( 5, 9 ), ( 0, 10 ), and ( 6, 0 ). We have to determine which of these are are maximum points, as the minimum are the same. Substitute these values ( ( 5, 9 ), ( 0, 10 ), and ( 6, 0 ) ) into the equation f( x, y ) = 10x + 4y,

f ( x, y ) = 10( 5 ) + 4( 9 ),\\f ( x, y ) = 86 -\\f ( x, y ) = 10( 0 ) + 4( 10 ),\\f ( x, y ) = 40 -\\f ( x, y ) = 10( 6 ) + 4( 0 ),\\f ( x, y ) = 60

( 5, 9 ) resulted in the greatest amount, 86. That would make it our maximum point -

<u><em>Solution = Option A</em></u>

3 0
4 years ago
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