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3 years ago

Is √16/49 rational ?

Mathematics

2 answers:

Effectus [21]3 years ago

8 0

Answer:

no it is not rational

Ierofanga [76]3 years ago

8 0

Answer:

It is rational

Step-by-step explanation:

A rational number is any integer, fraction, terminating decimal, or repeating decimal.

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What is the correct answer for this?

My name is Ann [436]

Answer:

Tan (X) = 72/65. = 1.1

Step-by-step explanation:

tan x = (opposite side) / (adjacent side)

From angle X the adjancent is the closest leg which has the value of 65 and the opposite is the furthest leg with a value of 72. (the line that is connecting the legs which is the biggest line is the hypotenuse and has a value of 97). You will not need the hypotenuse I just gave you extra information for future purposes.

so, since tan x = (opposite side) / (adjacent side):

Tan (X) = 72/65.

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2 years ago

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1. The cost of 2 pounds of potatoes is $3.78. What is

Kobotan [32]

Answer:

3.78$/2=1.89

y/x=1.89

y=1.89*x

Step-by-step explanation:

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3 years ago

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Suppose you invest 1,600 at an annual interest rate of 7.9% compounded continuously. How much will you have in the account after

Elina [12.6K]

$\bf ~~~~~~ \textit{Continuously Compounding Interest Earned Amount} \\\\ A=Pe^{rt}\qquad \begin{cases} A=\textit{accumulated amount}\\ P=\textit{original amount deposited}\to& \$1600\\ r=rate\to 7.9\%\to \frac{7.9}{100}\to &0.079\\ t=years\to &10 \end{cases} \\\\\\ A=1600e^{0.079\cdot 10}\implies A=1600e^{0.79}\implies A\approx 3525.434$

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3 years ago

What is the derivative of 2te^-t

Sveta_85 [38]

$(2te^{-t})'=(2t)'e^{-t}+2t(e^{-t})'=2e^{-t}+2t(-e^{-t})=2e^{-t}-2te^{-t}$

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3 years ago

Solve 0 = (x – 4)2 – 1 by graphing the related function.

fomenos

Answer:

Therefore, the solutions of the quadratic equations are:

$x=5,\:x=3$

The graph is also attached.

Step-by-step explanation:

The solution of the graph could be obtained by finding the x-intercept.

$y=\left(x-\:4\right)^2-1$

Finding the x-intercept by substituting the value y = 0

$y=\left(x-\:4\right)^2-1$

$\:0\:=\:\left(x\:-\:4\right)^2\:-\:1$ ∵ y = 0

$\left(x-4\right)^2-1=0$

$\left(x-4\right)^2-1+1=0+1$

$\left(x-4\right)^2=1$

$\mathrm{For\:}\left(g\left(x\right)\right)^2=f\left(a\right)\mathrm{\:the\:solutions\:are\:}g\left(x\right)=\sqrt{f\left(a\right)},\:\:-\sqrt{f\left(a\right)}$

$\mathrm{Solve\:}\:x-4=\sqrt{1}$

$\mathrm{Apply\:rule}\:\sqrt{1}=1$

$x-4=1$

$x=5$

$\mathrm{Solve\:}\:x-4=-\sqrt{1}$

$\mathrm{Apply\:rule}\:\sqrt{1}=1$

$x-4=-1$

$x=3$

So, when y = 0, then x values are 3, and 5.

Therefore, the solutions of the quadratic equations are:

$x=5,\:x=3$

The graph is also attached. As the graph is a Parabola. It is visible from the graph that the values of y = 0 at x = 5 and x = 3. As the graph is a Parabola.

8 0

3 years ago