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xenn [34]
4 years ago
10

find three consecutive positive odd integers such that the product of the first and the third is 4 less than 7 times the second

Mathematics
1 answer:
Digiron [165]4 years ago
7 0
X+x+1+x+3=4<7x
once you solve this you will plug in the x's an multiply the second number by 7
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Shayna's school is selling tickets to a play. On the first day of ticket sales the school sold 13 senior citizen tickets and 13
jasenka [17]

The price of one senior citizen ticket is 8$ and one student ticket is 12$.

<h3>What is the equation?</h3>

The equation is defined as mathematical statements that have a minimum of two terms containing variables or numbers that are equal.

Let the price of one senior citizen ticket = x

And the price of one student ticket = y

Given that, on the first day of ticket sales, the school sold 13 senior citizen tickets and 13 student tickets for a total of $260

The school took in $212 on the second day by selling 13 senior citizen tickets and 9 student tickets.

13x +13y = 260

13x + 9y = 212

Subtract the equation  from first

13x +13y - (13x + 9y) = 260 - 212

4y = 48

y = 48/4

y = 12

Substitute the value of y in the equation 13x + 9y = 212

13x + 9(12) = 212

13x + 108 = 212

13x = 212 - 108

13x = 104

x = 104/13

x = 8

Hence, the price of one senior citizen ticket is 8$ and one student ticket is 12$.

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brainly.com/question/10413253

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6 0
2 years ago
Tan(60)x 20<br> please hurry!!
lana [24]

Answer:

6.4008

Step-by-step explanation:

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How many times larger is 9 x 10-4 than 3 x 10-10?
Tatiana [17]

Answer:

It is 3*10^6 times bigger

Step-by-step explanation:

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Use the definition of the derivative to differentiate f(x)= In x
WINSTONCH [101]

By def. of the derivative, we have for y = ln(x),

\displaystyle \frac{dy}{dx} = \lim_{h\to0} \frac{\ln(x+h)-\ln(x)}{h}

\displaystyle \frac{dy}{dx} = \lim_{h\to0} \frac1h \ln\left(\frac{x+h}{x}\right)

\displaystyle \frac{dy}{dx} = \lim_{h\to0} \ln\left(1+\frac hx\right)^{\frac1h}

Substitute y = h/x, so that as h approaches 0, so does y. We then rewrite the limit as

\displaystyle \frac{dy}{dx} = \lim_{y\to0} \ln\left(1+y\right)^{\frac1{xy}}

\displaystyle \frac{dy}{dx} = \frac1x \lim_{y\to0} \ln\left(1+y\right)^{\frac1y}

Recall that the constant e is defined by the limit,

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