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2 years ago

HELP WITH MATH hzjdjdjdjdjd

Mathematics

2 answers:

ikadub [295]2 years ago

6 0

Answer:

9/20, 5/8, 3/4, 11/12

Step-by-step explanation:

3/4 = 9/12

5/8 = 7.5/12

9/20 = 5.4/12

Order from least to greatest:

9/20, 5/8, 3/4, 11/12

kolbaska11 [484]2 years ago

3 0

Answer:

0.916

Step-by-step explanation:

Trust me thats all you gotta do.

You might be interested in

Find the slope and y intercept (Please show work)

zepelin [54]

Answer:

slope = - 6, y- intercept = 5

Step-by-step explanation:

The equation of a line in slope- intercept form is

y = mx + c ( m is the slope and c the y- intercept )

Given

0 = 2x - $\frac{5}{3}$ + $\frac{1}{3}$ y ( multiply through by 3 to clear the fractions )

0 = 6x - 5 + y ( subtract 6x - 5 from both sides )

- 6x + 5 = y , that is

y = - 6x + 5 ← in slope- intercept form

with slope m = - 6 and y- intercept c = 5

4 0

2 years ago

<img src="https://tex.z-dn.net/?f=f%20-%20354%20%3D%201221" id="TexFormula1" title="f - 354 = 1221" alt="f - 354 = 1221" align="

alukav5142 [94]

$f-354=1221$ add 354 to both sides

$f+354-354=1221+354\\\\\boxed{f=1,575}\to\boxed{D.}$

6 0

3 years ago

in a class experiment, Sean finds that the probability that a student plays soccer is 12/25. if the school population is 300, ho

fgiga [73]

Okay, here we have this:

Considering the provided information, we are going to calculate the requested value, so we obtain the following:

Then we will substitute in the following formula:

Students who play soccer=Number of students*(Probability that they play soccer)

Replacing:

Students who play soccer=300*(12/25)

Students who play soccer=3600/25

Students who play soccer=144

Finally we obtain that 144 students would we expect to play soccer, based on Sean's experiment.

5 0

10 months ago

PKEASE HELP Based on the graph, what is one solution of the equation f(x) = g(x)? graph of function f of x equals negative x plu

Oksanka [162]

Answer:

x=1

x=-5

Step-by-step explanation:

we know that

The solution of the system of equations is the intersection point both graphs

In this problem there are two intersection points

The system has two solutions

we have

one solution is the point (-4.9, 5.4) and the other (0.92, -0.42)

see the attached figure

therefore

x=1

x=-5

Step-by-step explanation:

8 0

2 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

2 years ago