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3 years ago

13

Jackson is going to the state fair. The cost of admission is $4.00, and it costs $1.50 for each ride. If Jackson has $20.00, whi

ch equation can be used to determine how many rides Jackson can ride?

1 answer:

wel3 years ago

4 0

1.50X-$4.00=$20 is the answer

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What is the simplified form of ? 8x4 8x8 32x4 32x8

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Answer:

480

Step-by-step explanation:

8x4=32

8x8=64

32x4=128

32x8=256

256+128+64+32=480

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2 years ago

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What is a reasonable estimate for -3/2x (-5 1/4)

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To estimate, you should round some of the numbers to the closest number (rounding up or down). It will be more accurate the less you round. In this problem, we should round -5 1/4 to -5 since it is the closest. -5*-3/2 = 7.7. You will end up with -3/2x (-5 1/4) ≈ 7.5x

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2(3 – 4a) + 5(a – 7)

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Answer:

The answer is -(29+3a).

Step-by-step explanation:

=2(3-4a)+5(a-7)

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Population Growth The population P (in thousands) of Oriando,Florida from 1980 through 2009 can be modeled by p = 130e0.0205t,wh

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Answer: a) 236,000.

b) 2021

Step-by-step explanation:

Given : Population Growth The population P (in thousands) of Oriando,Florida from 1980 through 2009 can be modeled by $p = 130e^{0.0205t}$ (1)

where t = 0 corresponds to 1980.

Then , for 2009

t= 2009-1980=29

a. ⇒The population of Phoenix in 2009 = $p = 130e^{0.0205(29)}$

$p = 130e^{0.5945}$

$p =130(1.81212465608)=235.57620529\approx236$

Hence, the population of Orlando in 2009 was about 236,000.

b) Substitute p= 300 in (1) , we get

$300 = 130e^{0.0205t}$

$2.3077= e^{0.0205t}$ (Divide both sides by 130)

Taking Natural log on both sides,

$\ln(2.3077) = 0.0205t\\\\\Rightarrow\ t=\dfrac{\ln(2.3077)}{0.0205}$

$t=\dfrac{0.836251357528}{0.0205}=40.7927491477\approx41$

Hence, The year Orlando will have a population of 300,000 =1980+41 =2021

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3 years ago

What is the expansion of (3+x)^4

Vlad1618 [11]

Answer:

$\left(3+x\right)^4:\quad x^4+12x^3+54x^2+108x+81$

Step-by-step explanation:

Considering the expression

$\left(3+x\right)^4$

Lets determine the expansion of the expression

$\left(3+x\right)^4$

$\mathrm{Apply\:binomial\:theorem}:\quad \left(a+b\right)^n=\sum _{i=0}^n\binom{n}{i}a^{\left(n-i\right)}b^i$

$a=3,\:\:b=x$

$=\sum _{i=0}^4\binom{4}{i}\cdot \:3^{\left(4-i\right)}x^i$

Expanding summation

$\binom{n}{i}=\frac{n!}{i!\left(n-i\right)!}$

$i=0\quad :\quad \frac{4!}{0!\left(4-0\right)!}3^4x^0$

$i=1\quad :\quad \frac{4!}{1!\left(4-1\right)!}3^3x^1$

$i=2\quad :\quad \frac{4!}{2!\left(4-2\right)!}3^2x^2$

$i=3\quad :\quad \frac{4!}{3!\left(4-3\right)!}3^1x^3$

$i=4\quad :\quad \frac{4!}{4!\left(4-4\right)!}3^0x^4$

$=\frac{4!}{0!\left(4-0\right)!}\cdot \:3^4x^0+\frac{4!}{1!\left(4-1\right)!}\cdot \:3^3x^1+\frac{4!}{2!\left(4-2\right)!}\cdot \:3^2x^2+\frac{4!}{3!\left(4-3\right)!}\cdot \:3^1x^3+\frac{4!}{4!\left(4-4\right)!}\cdot \:3^0x^4$

$=\frac{4!}{0!\left(4-0\right)!}\cdot \:3^4x^0+\frac{4!}{1!\left(4-1\right)!}\cdot \:3^3x^1+\frac{4!}{2!\left(4-2\right)!}\cdot \:3^2x^2+\frac{4!}{3!\left(4-3\right)!}\cdot \:3^1x^3+\frac{4!}{4!\left(4-4\right)!}\cdot \:3^0x^4$

as

$\frac{4!}{0!\left(4-0\right)!}\cdot \:\:3^4x^0:\:\:\:\:\:\:81$

$\frac{4!}{1!\left(4-1\right)!}\cdot \:3^3x^1:\quad 108x$

$\frac{4!}{2!\left(4-2\right)!}\cdot \:3^2x^2:\quad 54x^2$

$\frac{4!}{3!\left(4-3\right)!}\cdot \:3^1x^3:\quad 12x^3$

$\frac{4!}{4!\left(4-4\right)!}\cdot \:3^0x^4:\quad x^4$

so equation becomes

$=81+108x+54x^2+12x^3+x^4$

$=x^4+12x^3+54x^2+108x+81$

Therefore,

$\left(3+x\right)^4:\quad x^4+12x^3+54x^2+108x+81$

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3 years ago