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3 years ago

A 15- ft tree casts an 18- ft shadow at the same time a 24- ft tree casts a shadow. How long is the shadow of the 24- ft tree

Mathematics

1 answer:

anzhelika [568]3 years ago

4 0

Answer:

28.8 ft

Step-by-step explanation:

We can use proportion in calculating this.

Let X= height of the shadow of 24- ft tree.

The proportion can be set up as

(15- ft tree) /( 18- ft shadow )=(24- ft tree) / X

15/18 = 24/X

Making X subject of the formula

15X = 18 × 24

X= (18×24)/15

X=28.8 ft

Hence, the shadow of the 24- ft tree is 28.8 feet

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Tickets for a carnival cost $8 for children 12 and under and $12 for anyone over the age of 12. The carnival sold 233 tickets an

jenyasd209 [6]

Answer:

They sold 160 children's tickets.

Step-by-step explanation:

With the information provided, yoou can write the following equations:

x+y=233 (1)

8x+12y=2,156 (2), where:

x is the number of tickets for children 12 and under

y is the number of tickets for anyone over the age of 12

Now, you can solve for x in (1):

x=233-y (3)

Then, you have to replace (3) in (2) and solve for y:

8(233-y)+12y=2,156

1,864-8y+12y=2,156

4y=2,156-1,864

4y=292

y=292/4

y=73

Finally, you can replace the value of y in (3) to find x:

x=233-73

x=160

According to this, the answer is that they sold 160 children's tickets.

5 0

3 years ago

What are the values of x in the equation 4x2 + 4x – 3 = 0

Oksi-84 [34.3K]

Answer:

x = -3/2, 1/2

Step-by-step explanation:

(2x + 3)(2x - 1) = 0

2x + 3 = 0, 2x = -3, x = -3/2

2x - 1 = 0, 2x = 1, x = 1/2

5 0

3 years ago

3) In a paddling pool there are 30 floating ducks. Each duck is marked with a number on the underside. 15 are marked with the nu

Darina [25.2K]

Answer: 1/5

Step-by-step explanation:

Given the following :

Total number of ducks in pool = 30

Mark 1 = 15 ducks

Mark 2 = 9 ducks

Mark 3 = 6 ducks

Probability of picking a duck with Mark 3:

Probability = (number of required outcomes / total possible outcomes)

Number of required outcomes = number of ducks with mark 3 = 6 ducks

P(picking a duck with Mark 3) = 6/30

6/30 = 1/5

= 1/5

4 0

4 years ago

A tank contains 100 gal of water and 50 oz of salt. Water containing a salt concentration of ¼ (1 + ½ sin t) oz/gal flows ito th

Likurg_2 [28]

Answer:

Part a)

$y\left(t\right)=\frac{-1250cost+25sint}{5002}+25+\frac{63150}{2501}\frac{1}{e^{\frac{t}{50}}}$

Part b)

Check the attached figure to see the ultimate behavior of the graph.

Part c)

The level = 25, Amplitude = 0.2499

Step-by-step Solution:

Part a)

Given:

$Q(0)=50$

Rate in:

$\frac{1}{4}\left(1+\frac{1}{2}sint\right)\cdot 2\:=\:\frac{1}{2}\left(1+\frac{1}{2}sint\right)$

Rate out:

$\frac{Q}{100}\cdot 2=\frac{Q}{50}$

So, the differential equation would become:

$\frac{dQ}{dt}=\frac{1}{2}\left(1+\frac{1}{2}sint\right)-\frac{Q}{50}$

Rewriting the equation:

$\frac{dQ}{dt}+\frac{Q}{50}=\frac{1}{2}\left(1+\frac{1}{2}sint\right)$

As $p(x)$ is the coefficient of $y$ , while $q(x)$ is the constant term in the right side of the equation:

$p\left(x\right)=\frac{1}{50}$

$q\left(x\right)=\frac{1}{2}\left(1+\frac{1}{2}sint\right)$

First it is important to determine the function $\mu$ :

$\mu \left(t\right)=e^{\int \:p\left(t\right)dt}$

$=e^{\int \:\left(\frac{1}{50}\right)dt}$

$=e^{\frac{t}{50}}$

The general solution then would become:

$y\left(t\right)=\frac{1}{\mu \left(t\right)}\left(\int \mu \left(t\right)q\left(t\right)dt+c\:\right)$

$=\frac{1}{e^{\frac{t}{50}}}\int e^{\frac{t}{50}}\:\frac{1}{2}\left(1+\frac{1}{2}sint\right)dt+\frac{1}{e^{\frac{t}{50}}}c$

$=\frac{1}{e^{\frac{t}{50}}}\left(\frac{-25e^{\frac{t}{50}}\left(50cost-sint\right)}{5002}+25e^{\frac{t}{50}}\right)+\frac{1}{e^{\frac{t}{50}}}c$

$=\frac{\left-1250cost+25sint\right}{5002}+25+\frac{1}{e^{\frac{t}{50}}}c$

Evaluate at $t=0$

$50=y\left(0\right)=\frac{\left(-1250cos0+25sin0\right)}{5002}+25+\frac{1}{e^{\frac{0}{50}}}c$

Solve to c:

$c=25+\frac{1250}{5002}$

$\mathrm{Cancel\:}\frac{1250}{5002}:\quad \frac{625}{2501}$

$c=25+\frac{625}{2501}$

$\mathrm{Convert\:element\:to\:fraction}:\quad \:25=\frac{25\cdot \:2501}{2501}$

$c=\frac{25\cdot \:2501}{2501}+\frac{625}{2501}$

$\mathrm{Since\:the\:denominators\:are\:equal,\:combine\:the\:fractions}:\quad \frac{a}{c}\pm \frac{b}{c}=\frac{a\pm \:b}{c}$

$c=\frac{25\cdot \:2501+625}{2501}$

$c=\frac{63150}{2501}$

$c\approx 25.25$

Therefore, the general solution then would become:

$y\left(t\right)=\frac{-1250cost+25sint}{5002}+25+\frac{63150}{2501}\frac{1}{e^{\frac{t}{50}}}$

Part b) <em>Plot the Solution to see the ultimate behavior of the graph</em>

The graph appears to level off at about the value of $Q=25$ .

The graph is attached below.

Part c)

In the graph we note that the level is $Q=25$ .

Therefore, the level = 25

The amplitude is the (absolute value of the) coefficient of $cost\:t$ in the general solution (as the coefficient of the sine part is a lot smaller):

Therefore,

$A=\frac{1250}{5002}\:\approx 2.499$

Keywords: differential equation, word problem

Learn more about differential equation word problem from brainly.com/question/14614696

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4 0

3 years ago

Directions: Graph each function using a table of Graph each values, then identify its key characteristics.

Tom [10]

Answer:

Step-by-step explanation:

Given function is,

$y=(\frac{1}{2})^x$

In the given exponential function,

Base of the function = $\frac{1}{2}$

By the property of an exponential function,

1). If the base is between 0 and 1, function will be a growth function.

2). If the base is greater than 1, function will be a decay function.

Therefore, given function is a decay function.

Input-output table,

x -2 -1 0 1 2

y $(\frac{1}{2})^{-2}=4$ $(\frac{1}{2})^{-1}=2$ $(\frac{1}{2})^{0}=1$ $(\frac{1}{2})^1=\frac{1}{2}$ $(\frac{1}{2})^2=\frac{1}{4}$

By plotting these points we can get the graph of the function.

Domain: (-∞, ∞)

Range: (0, ∞)

y-intercept: 1

Asymptote: y = 0

7 0

3 years ago