Determine whether these two triangles are similar, and if so what is the ratio of their sides?

Yes, ratio = 3/4
No, these triangles are not similar
Yes, ratio=3/8
Yes, ratio=4/5

Answer:

A

Step-by-step explanation:

[tex]\sqrt[3]{3x}[/tex] + 7 = 4 ( subtract 7 from both sides )

[tex]\sqrt[3]{3x}[/tex] = - 3 ( cube both sides to clear the radical )

3x = (- 3)³ = - 27 ( divide both sides by 3 )

x = - 9

Answer:

Let x = cube root of 400

Using a calculator x = 7.368 to 3 decimal places

So: 7 < x < 8

Answer:

5 should be the correct answer

Answer:

Step-by-step explanation:

Given

Remember :

Solving

Answer:

[tex]\huge\boxed{\bf\:\frac{w^{6}}{9}}[/tex]

Step-by-step explanation:

[tex]\frac{ 3 ^ { -2 } \times k ^ { 0 } \times w ^ { 0 } }{ w ^ { -6 } }[/tex]

Any number with 0 as its exponent will be equal to 1. So,

[tex]\frac{3^{-2} \times 1 \times 1}{w^{-6}}\\= \frac{3^{-2}}{w^{6}}[/tex]

Now,

[tex]\frac{3^{-2}}{w^{6}}\\= 3^{-2}w^{6}[/tex]

Then,

[tex]3^{-2}w^{6}\\= \frac{1}{3^{2}}w^{6}\\= \frac{1}{9}w^{6}\\= \boxed{\bf\:\frac{w^{6}}{9}}[/tex]

[tex]\rule{150pt}{2pt}[/tex]

Answer:

there are none (sorry if I got it wrong)

Step-by-step explanation:

also, there is none  because there other numbers are really close to the other number

outliers are a number that is really small in the pack of numbers and outliers can also be the number that is really big for example I have the numbers 3 24 28 32 43 and then 100. 3 and 100 are outliers because one number (3) is really small from the other numbers are bigger and one number (100) is really big from the other number but sometimes there are no outliers

Answer:

The ratio of the new distance to the old distance is  (11/2) .

The intensity is inversely proportional to the square of the distance,

so the new intensity will be  (2/11)²  times the old intensity.

Intensity =                 8,000 (2/11)² =

                              32,000 / 121  =  262.463  units

  Rounded to the nearest whole unit:  262 units

Step-by-step explanation:

264 units of intensity came from a distance of 11 miles option (c) 264 units is correct.

It is given that the intensity is 8000 units at a distance of 2 miles.

It is required to find the intensity at a distance of 2 miles.

Fraction number consists of two parts one is the top of the fraction number which is called the numerator and the second is the bottom of the fraction number which is called the denominator.

The intensity of a radio signal from the radio station varies inversely as the square of the distance from the station.

Suppose the Intensity of the signal is I and the distance is d, then:

[tex]\rm I \propto\frac{1}{d^2}[/tex]

Intensity from the station [tex]\rm I_1=8000 \ units[/tex]

[tex]\rm I_1[/tex] intensity at distance [tex]\rm d_1=2 \ miles[/tex]

Intensity from the station [tex]=\rm I_2[/tex]

[tex]\rm I_1[/tex] intensity at distance [tex]\rm d_2=11 \ miles[/tex]

[tex]\rm \frac{I_2}{I_1} = \frac{d_1^2}{d_2^2}[/tex]       (From the proportional relation)

[tex]\rm \frac{I_2}{8000} = \frac{2^2}{11^2}[/tex]

[tex]\rm I_2 =8000\times\frac{4}{121} \\\\\rm I_2 = 264.46 \ units[/tex] ≈ 264 units

Thus, the 264 units of intensity came from a distance of 11 miles.

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Answer:

  a. 885,920

  b. year 8

Step-by-step explanation:

The population at a given time is described by the exponential formula ...

  P(t) = P0·e^(-0.04t)

where P0 is given as 1,000,000.

We are asked for the value of P(3). This ccan be found by substituting 3 for t in the equation and evaluating the numerical expression.

  P(3) = 1,000,000·e^(-0.04·3) = 1,000,000·e^(-0.12)

  P(3) ≈ 886,920

The population when t=3 is about 886,920.

__

We can put the given numbers in the equation and solve for t.

  750,000 = 1,000,000·e^(-0.04t)

  0.75 = e^(-0.04t) . . . . . divide by 1,000,000

  ln(0.75) = -0.04t . . . . . take the natural log

  -ln(0.75)/0.04 = t ≈ 7.192

The population will drop below 750,000 in year 8.

_____

Additional comment

These values can be confirmed by a graphing calculator.

Note that "year 1" is the year between t=0 and t=1. So, "year 8" is the year between t=7 and t=8.

Answer:

See below

Step-by-step explanation:

From the graph :

y = -x         -inf < x <=0

y = x             0<= x <= 2

y = 3            x > 2

Answer:

whatever number is repeating the most is the answer.

The cost of a washer is $1184 while the cost of a drier is $296

Given Data

Hence,

w+d = 1184 -------------1

we know that the cost of the washer is three times the cost of the dryer.

w = 3d -------------------2

put w = 3d in equation 1

3d + d = 1184

4d = 1184

d = 1184/4

d = 296

The cost of a drier is $296

put d = $296 in equation 2

w = 3*296

w = $888

Check:

$888+$296 = $1184

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first off let's change the mixed fractions to improper fractions, and let's Hailey's account first.

[tex]\stackrel{mixed}{9\frac{1}{4}}\implies \cfrac{9\cdot 4+1}{4}\implies \stackrel{improper}{\cfrac{37}{4}} ~\hfill \stackrel{mixed}{8\frac{7}{8}}\implies \cfrac{8\cdot 8+7}{8}\implies \stackrel{improper}{\cfrac{71}{8}} \\\\[-0.35em] ~\dotfill[/tex]

[tex]~~~~~~ \textit{Continuously Compounding Interest Earned Amount} \\\\ A=Pe^{rt}\qquad \begin{cases} A=\textit{accumulated amount}\dotfill &\stackrel{43000(2)}{\$86000}\\ P=\textit{original amount deposited}\dotfill & \$43000\\ r=rate\to \frac{71}{8}\%\to \frac{~~ \frac{71}{8}~~}{100}\dotfill &0.08875\\ t=years \end{cases} \\\\\\ 86000=43000e^{0.08875\cdot t}\implies \cfrac{86000}{43000}=e^{0.08875t}\implies 2=e^{0.08875t}[/tex]

[tex]\ln(2)=\ln(e^{0.08875t})\implies \log_e(2)=\log_e(e^{0.08875t})\implies \ln(2)=0.08875t \\\\\\ \cfrac{\ln(2)}{0.08875}=t\implies 7.81\approx t[/tex]

ok, now we know how long it takes for Hailey's money to double, how much money does Aiden have by then?

[tex]~~~~~~ \textit{Compound Interest Earned Amount} \\\\ A=P\left(1+\frac{r}{n}\right)^{nt} \quad \begin{cases} A=\textit{accumulated amount}\\ P=\textit{original amount deposited}\dotfill &\$43000\\ r=rate\to \frac{37}{4}\%\to \frac{~~ \frac{37}{4}~~}{100}\dotfill &0.0925\\ n= \begin{array}{llll} \textit{times it compounds per year}\\ \textit{monthly, thus twelve} \end{array}\dotfill &12\\ t=years\dotfill &\frac{\ln(2)}{0.08875} \end{cases}[/tex]

[tex]A=43000\left(1+\frac{0.0925}{12}\right)^{12\cdot \frac{\ln(2)}{0.08875}}\implies A=43000\left( \frac{4837}{4800} \right)^{\frac{12\ln(2)}{0.08875}}\implies \boxed{A\approx 88311}[/tex]

notice, in Hailey's amount we used the logarithmic value for "t", just to avoid any rounding issues.

Given: f(x)=2x(cube)-6x+x

reaminder of f(x) when (x+1) diveded

to find value of k

solution,

f(x)=2x(cube)-6x+k

f(-1)=7

2(-1)(cube)+6+k=7

4+x=7

k=7

Hence,k=7

Answer:

i belive the answer is 73

Step-by-step explanation:

45+62=107

180-107=73

how many pieces of tape can be cutted.

[tex]5 \frac{1}{3} \div \frac{2}{3} [/tex]

[tex] = (5 + 0) + ( \frac{1}{3} + \frac{2}{3} )[/tex]

[tex] = 5 + \frac{1 + 2}{3} [/tex]

[tex] = 5 + \frac{3}{3} [/tex]

[tex] = 5 + \frac{3 \div 3}{3 \div 3} [/tex]

[tex] = 5 + \frac{1}{1} [/tex]

[tex] = 6[/tex]

therefore, 6 pieces of tape can be cutted.

Answer:

[tex]\sqrt{X-5}[/tex], [tex]-\sqrt{X-5}[/tex]

Step-by-step explanation:

Replace X with Y

solve for Y; X = Y^2 + 5

subtract 5 from both sides

X - 5 = Y^2

take the square root

Y = plus or minus sqrt(X-5)

sqrt(X-5); -sqrt(X-5)

Answer:

The numbers inside |x| is always positive

Step-by-step explanation:

|x+1| --->  x+1
|x-2| --->  x+2
               = 3

Answer:

infinte

Step-by-step explanation:

The answer is anything greater than 12 so it would be infinte

Answer:

ok it's a good question... so answer to that is... I also don't know..

Answer:

20,000×17 gives 340,000

340,000×68 which gives 23,120,000

Answer:

The option b) is correct  (x = 7.8 and x = -12.8)

For the verification picture is below

I hope my answer helps you.

Let first consider the equations one by one and will be solving one by one ;

[tex]{:\implies \quad \sf x+\dfrac{1}{y}=4}[/tex]

Multiplying both sides by y will lead ;

[tex]{:\implies \quad \sf xy+1=4y}[/tex]

[tex]{:\implies \quad \boxed{\sf xy=4y-1\quad \cdots \cdots(i)}}[/tex]

Now, consider the second equation which is ;

[tex]{:\implies \quad \sf y+\dfrac{1}{x}=\dfrac14}[/tex]

Multiplying both sides by x will yield

[tex]{:\implies \quad \sf xy+1=\dfrac{x}{4}}[/tex]

[tex]{:\implies \quad \sf xy=\dfrac{x}{4}-1}[/tex]

[tex]{:\implies \quad \boxed{\sf xy=\dfrac{x-4}{4}\quad \cdots \cdots(ii)}}[/tex]

As LHS of both equations (i) and (ii) are same, so equating both will yield;

[tex]{:\implies \quad \sf 4y-1=\dfrac{x-4}{4}}[/tex]

Multiplying both sides by 4 will yield

[tex]{:\implies \quad \sf 16y-4=x-4}[/tex]

[tex]{:\implies \quad \sf 16y=x}[/tex]

Dividing both sides by y will yield :

[tex]{:\implies \quad \boxed{\bf{\dfrac{x}{y}=16}}}[/tex]

Hence, the required answer is 16

First, we have "at least 120 bagels"

So we write ≥120 bagels.

Now, we are given that:-

Mrs. Trevino already has 2 dozen bagels.

Remember, a dozen is equal to 12, so if we have 2 dozen, then we have

[tex]\sf{2*12=24}\longleftarrow\sf{Mrs.Trevino~already~has~24~bagels}[/tex]

Now, she'll buy the remaining bagels in packages of 6, so we write

6x (packages of 6)

So the inequality looks like so:-

[tex]\sf{6x+24\geq 120}[/tex]

Now, let's solve the inequality :)

First, subtract 24 from both sides:-

[tex]\sf{6x\geq 120-24}[/tex]

[tex]\sf{6x\geq 96}[/tex]

Divide both sides by 6:-

[tex]\bigstar{\boxed{\pmb{x\geq 16}}}\longleftarrow\sf{number~of~packages~she~should~buy}[/tex]

Hope everything is clear :)

Answer:

The information on the graph shows a positive nonlinear correlation and the weight of the dinosaur tends to increase according to its length.

A discrete graph is also known as a scatter plot that shows the discontinuous and the nonlinear data point on the graph.

From the graph, we can see the increasing weight of the Dinosaur size vertical y-axis and the increasing length of the feet on the horizontal x-axis.

Therefore, we can conclude that the data points show an increasing nonlinear weight on the graph as the length increases.

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Answer:

372

Step-by-step explanation:

area of trapezium = a+b/2 * h

a is the length on top

b is the length on the bottom

h is the height

7+24/2*24=

=372

Answer:

the graph will be an exponential function that crosses the y-axis at about (0, -4).

Step-by-step explanation:

Answer:

The answer is the first graph.

Step-by-step explanation:

I just did the quiz

Answer:

1935 were supporting the home team

Step-by-step explanation:

Since there are 4,500 people attending the game, we must find 43% of that. Doing this, we need to multiply 43% or .43 by 4,500.

After solving, you should get 1,935. There were 1,935 people supporting the home team.

Using the vertex of a quadratic equation, it is found that P is at a maximum for l = -0.5 thousand foot-candles.

A quadratic equation is modeled by:

[tex]y = ax^2 + bx + c[/tex]

The vertex is given by:

[tex](x_v, y_v)[/tex]

In which:

[tex]x_v = -\frac{b}{2a}[/tex]

[tex]y_v = -\frac{b^2 - 4ac}{4a}[/tex]

Considering the coefficient a, we have that:

In this problem, the function for P is given by:

[tex]P(l) = \frac{110l}{l^2 + l + 4}[/tex]

The function will be at a maximum when the denominator is at a minimum. The denominator is a quadratic function with coefficients a = 1, b = 1, c = 4, hence:

[tex]l_v = -\frac{b}{2a} = -\frac{1}{2} = -0.5[/tex]

P is at a maximum for l = -0.5 thousand foot-candles.

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Answer:

See below ↓↓

Step-by-step explanation:

Taking the cos ratio of the angle

⇒ The angle formed is safe

Finding furthest possible distance

Answer:

Yes, the angle formed between the ground and the ladder is safe

6.84 ft (nearest hundredth)

Step-by-step explanation:

We can use the cos trig ratio to determine the angle made between the 20ft ladder and 3ft from the base of the house (see first attached image for sketch).

[tex]\sf \cos(\theta)=\dfrac{A}{H}[/tex]

where:

Given:

[tex]\sf \implies \cos(\theta)=\dfrac{3}{20}[/tex]

[tex]\sf \implies \theta=\cos^{-1}\left(\dfrac{3}{20}\right)[/tex]

[tex]\sf \implies \theta=81.37307344..^{\circ}[/tex]

Therefore, as the ladder is forming an 81.37° angle and 81.37...° > 70° the angle formed between the ground and the ladder is safe.

To find the furthest possible distance the ladder can be placed, set [tex]\theta[/tex] to 70° (max safe angle) and the hypotenuse (ladder length) to 20, then solve for A (see second attache image for sketch):

[tex]\sf \implies \cos(70^{\circ})=\dfrac{A}{20}[/tex]

[tex]\sf \implies A=20\cos(70^{\circ})[/tex]

[tex]\sf \implies A=6.840402867...\: \sf ft[/tex]

So the furthest possible distance the ladder can be placed to maintain a safe angle is 6.84 ft (nearest hundredth).

The marginal utility is the extra satisfaction derived from the consumption of a product.

Your information isn't well written. Therefore, an overview of utility will be given. The total utility is the total amount of satisfaction that a consumer derives from a product.

The marginal utility simply means the extra satisfaction that's gotten from a product when an additional unit is consumed.

The formula for marginal utility will be:

= Total utility difference/Quantity of goods difference

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