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bulgar [2K]
3 years ago
13

The school that Brenda goes to is selling tickets to the annual talent show. On the first day of the ticket sales the school sol

d 2 adult tickets and 14 child tickets for a total of $96. The school took in $120 on the second day by selling 12 adult tickets and 8 child tickets. find the price of an adult ticket and the price of a child ticket.
Mathematics
1 answer:
KiRa [710]3 years ago
7 0
A=cost er adult ticket
c=cost per child ticket

2a+14c=96
12a+8c=120

solve

simplify by dividing firs equation by 2 and 2nd by 4

a+7c=48
3a+2c=30

multiply first equaiton by -3 and add to second
-3a-21c=-144
<u>3a+2c=30 +</u>
0a-19c=-114

-19c=-114
divide both sides by -19
x=6

sub back

a+7c=48
6+7c=48
minus 6
7c=42
divide 7
c=6

adult ticket=$6
child ticekt=$6
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Answer:

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Step-by-step explanation:

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2 years ago
If 5 + 20 times 2 Superscript 2 minus 3 x Baseline = 10 times 2 Superscript negative 2 x Baseline + 5, what is the value of x? –
sp2606 [1]

Answer:

<h3>Option D) 3 is correct</h3><h3>Therefore the value of x is 3</h3>

Step-by-step explanation:

Given equation is 5+20\times (2)^{2-3x}=10\times (2)^{-2x}+5

<h3>To find the value of x :</h3>

First solving the given equation we have,

5+20\times (2)^{2-3x}=10\times (2)^{-2x}+5

5+20\times (2)^{2-3x}-5=10\times (2)^{-2x}+5-5

20\times (2)^{2-3x}=10\times (2)^{-2x}

20\times (2)^2.(2)^{-3x}=10\times (2)^{-2x}

20\times 4.(2)^{-3x}=10\times (2)^{-2x}

80(2)^{-3x}=10\times (2)^{-2x}

\frac{80}{10}(2)^{-3x}=\frac{10\times (2)^{-2x}}{10}

8(2)^{-3x}=(2)^{-2x}

\frac{(2)^{-3x}}{(2)^{-2x}}=\frac{1}{8}

(2)^{-3x}.(2)^{2x}=\frac{1}{2^3} ( by using the property \frac{1}{a^{-m}}=a^m )

2^{-3x+2x}=\frac{1}{2^3} ( by using the property a^m.a^n=a^{m+n} )

2^{-x}=\frac{1}{2^3}

\frac{1}{2^x}=\frac{1}{2^3} ( by using the property a^{-m}=\frac{1}{a^m} )

Since bases are same so powers are same

Therefore we can equate the powers we get x=3

<h3>Therefore the value of x is 3</h3><h3>Option D) 3 is correct</h3>
5 0
3 years ago
Read 2 more answers
Please help logarithms!
nlexa [21]

Given:

\log_34\approx 1.262

\log_37\approx 1.771

To find:

The value of \log_3\left(\dfrac{4}{49}\right).

Solution:

We have,

\log_34\approx 1.262

\log_37\approx 1.771

Using properties of log, we get

\log_3\left(\dfrac{4}{49}\right)=\log_34-\log_349      \left[\because \log_a\dfrac{m}{n}=\log_am-\log_an\right]

\log_3\left(\dfrac{4}{49}\right)=\log_34-\log_37^2      

\log_3\left(\dfrac{4}{49}\right)=\log_34-2\log_37          [\log x^n=n\log x]

Substitute \log_34\approx 1.262 and \log_37\approx 1.771.

\log_3\left(\dfrac{4}{49}\right)=1.262-2(1.771)

\log_3\left(\dfrac{4}{49}\right)=1.262-3.542

\log_3\left(\dfrac{4}{49}\right)=-2.28

Therefore, the value of \log_3\left(\dfrac{4}{49}\right) is -2.28.

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3 years ago
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Answer:

A≈78.54

Step-by-step explanation:

A=πr2

I think this is right

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4 years ago
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Darina [25.2K]

ECH is an acute angle.

FCG is also an acute angle.

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3 years ago
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