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3 years ago

PLzzzzz helppp meee!!!

Mathematics

1 answer:

Tanzania [10]3 years ago

7 0

Download the app "photomath" it's available on apple and android devices it's an app where you take a picture of the problem and it gives you the answer :) ♡ ☆

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3x + y = 6 4x + 2y = 8

Lilit [14]

Answer:

x=2 y = 0

Step-by-step explanation:

good luck

3 0

3 years ago

the sum of two numbers is 158.if we add 25 to each one of them,then the first becomes the triple of the other.what are these num

IRISSAK [1]

Answer:

Step-by-step explanation:

8 0

3 years ago

Determine the mean and variance of the random variable with the following probability mass function. f(x) = (216/43)(1/6)^x, x =

anzhelika [568]

Answer:

The mean of function provided is 1.186.

The variance of the provided f(x) is 0.198

Step-by-step explanation:

It is provided that the probability mass function is,

f(x)= (214/43)×(1/6)ˣ; x=1,2,3

The mean is calculated as,

E(X)=∑ x × f(x)

=1×(216/43)×(1/6)¹ + 2 × (216/43)×(1/6)² × 3 × (216/43)×(1/6)³

=36/43 + 12/43 +3/43

=1.186

The mean of function provided is 1.186

Explanation | Common mistakes | Hint for next step

The expected value of the probability mass function,f(x)= (216/43×(1/6)ˣ

is 1.1861.186 .

Step 2 of 2

To calculate the variance, first calculate E(X²)=∑ x² × f(x)

= 1² ×(216/43) × (1/6)¹ + 2² × (216/43) × (1/6)² × 3² × (216/43) ×(1/6)³

=36/43 +24/43 +9/43

=1.605

The variance is calculated as,

V(X) =E(X²) - [E(X)]²

=1.605 -(1.186)²

= 0.198

The variance of the provided f(x) is 0.198

Explanation | Common mistakes

The variance of function f(x)=(216/43) × (1/6)ˣ ; x =1,2,3 is 0.198

3 0

3 years ago

Solve the equation in the interval [0,2π]. If there is more than one solution write them separated by commas.

Sedaia [141]

$\large\begin{array}{l} \textsf{Solve the equation for x:}\\\\ 
\mathsf{(tan\,x)^2+2\,tan\,x-4.76=0}\\\\\\ \textsf{Substitute}\\\\ 
\mathsf{tan\,x=t\qquad(t\in \mathbb{R})}\\\\\\ \textsf{so the equation 
becomes}\\\\ \mathsf{t^2+2t-4.76=0}\quad\Rightarrow\quad\begin{cases} 
\mathsf{a=1}\\\mathsf{b=2}\\\mathsf{c=-4.76} \end{cases} 
\end{array}$

$\large\begin{array}{l} \textsf{Using 
the quadratic formula:}\\\\ \mathsf{\Delta=b^2-4ac}\\\\ 
\mathsf{\Delta=2^2-4\cdot 1\cdot (-4.76)}\\\\ 
\mathsf{\Delta=4+19.04}\\\\ \mathsf{\Delta=23.04}\\\\ 
\mathsf{\Delta=\dfrac{2\,304}{100}}\\\\ 
\mathsf{\Delta=\dfrac{\diagup\!\!\!\! 4\cdot 576}{\diagup\!\!\!\! 4\cdot
 25}}\\\\ \mathsf{\Delta=\dfrac{24^2}{5^2}} \end{array}$

$\large\begin{array}{l}
 \mathsf{\Delta=\left(\dfrac{24}{5}\right)^{\!2}}\\\\ 
\mathsf{\Delta=(4.8)^2}\\\\\\ 
\mathsf{t=\dfrac{-b\pm\sqrt{\Delta}}{2a}}\\\\ 
\mathsf{t=\dfrac{-2\pm\sqrt{(4.8)^2}}{2\cdot 1}}\\\\ 
\mathsf{t=\dfrac{-2\pm 4.8}{2}}\\\\ \mathsf{t=\dfrac{\diagup\!\!\!\! 
2\cdot (-1\pm 2.4)}{\diagup\!\!\!\! 2}}\\\\\mathsf{t=-1\pm 2.4} 
\end{array}$

$\large\begin{array}{l} \begin{array}{rcl} 
\mathsf{t=-1-2.4}&~\textsf{ or }~&\mathsf{t=-1+2.4}\\\\ 
\mathsf{t=-3.4}&~\textsf{ or }~&\mathsf{t=1.4} \end{array} 
\end{array}$

$\large\begin{array}{l} \textsf{Both 
are valid values for t. Substitute back for }\mathsf{t=tan\,x:}\\\\ 
\begin{array}{rcl} \mathsf{tan\,x=-3.4}&~\textsf{ or 
}~&\mathsf{tan\,x=1.4} \end{array}\\\\\\ \textsf{Take the inverse 
tangent function:}\\\\ \begin{array}{rcl} 
\mathsf{x=tan^{-1}(-3.4)+k\cdot \pi}&~\textsf{ or 
}~&\mathsf{x=tan^{-1}(1.4)+k\cdot \pi}\\\\ 
\mathsf{x=-tan^{-1}(3.4)+k\cdot \pi}&~\textsf{ or 
}~&\mathsf{x=tan^{-1}(1.4)+k\cdot \pi} \end{array}\\\\\\ 
\textsf{where k in an integer.} \end{array}$

__________

$\large\begin{array}{l}
 \textsf{Now, restrict x values to the interval 
}\mathsf{[0,\,2\pi]:}\\\\ \bullet~~\textsf{For }\mathsf{k=0:}\\\\ 
\begin{array}{rcl} 
\mathsf{x=-tan^{-1}(3.4)$

$\large\begin{array}{l}
 \bullet~~\textsf{For }\mathsf{k=1:}\\\\ \begin{array}{rcl} 
\mathsf{x=-tan^{-1}(3.4)+\pi}&~\textsf{ or 
}~&\mathsf{x=tan^{-1}(1.4)+\pi} \end{array}\\\\\\ 
\boxed{\begin{array}{c}\mathsf{x=-tan^{-1}(3.4)+\pi} 
\end{array}}\textsf{ is in the 2}^{\mathsf{nd}}\textsf{ quadrant.}\\\\ 
\mathsf{x\approx 1.86~rad~~(106.39^\circ)}\\\\\\ 
\boxed{\begin{array}{c}\mathsf{x=tan^{-1}(1.4)+\pi} \end{array}}\textsf{
 is in the 3}^{\mathsf{rd}}\textsf{ quadrant.}\\\\ \mathsf{x\approx 
4.09~rad~~(234.46^\circ)}\\\\\\ \end{array}$

$\large\begin{array}{l}
 \bullet~~\textsf{For }\mathsf{k=2:}\\\\ \begin{array}{rcl} 
\mathsf{x=-tan^{-1}(3.4)+2\pi}&~\textsf{ or 
}~&\mathsf{x=tan^{-1}(1.4)+2\pi>2\pi~~\textsf{(discard)}} 
\end{array}\\\\\\ \boxed{\begin{array}{c}\mathsf{x=-tan^{-1}(3.4)+2\pi} 
\end{array}}\textsf{ is in the 4}^{\mathsf{th}}\textsf{ quadrant.}\\\\ 
\mathsf{x\approx 5.00~rad~~(286.39^\circ)} \end{array}$

$\large\begin{array}{l}
 \textsf{Solution set:}\\\\ 
\mathsf{S=\left\{tan^{-1}(1.4);\,-tan^{-1}(3.4)+\pi;\,tan^{-1}(1.4)+\pi;\,-tan^{-1}(3.4)+2\pi\right\}}
 \end{array}$

If you're having problems understanding this answer, try seeing it through your browser: brainly.com/question/2071152

$\large\textsf{I hope it helps.}$

Tags: trigonometric trig quadratic equation tangent tan solve inverse symmetry parity odd function

6 0

3 years ago

Cube divided into 4 square pyramids with a height of 2h. The image above can be used to help construct an informal argument for

zubka84 [21]

A.
because the volume of the pyramids added together would be the cubes total volume.

8 0

3 years ago

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