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3 years ago

A rectangle has a perimeter of 68 inches. If the width of the rectangle is 10 inches, what is the length of the rectangle

Mathematics

2 answers:

Gwar [14]3 years ago

7 0

I believe it’s 6.8 if I’m wrong I’m sorry

patriot [66]3 years ago

3 0

The length of the rectangle is 6.8.

You have to divide 68 by 10 = 6.8 =) =)

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Can anyone answer this I need it ASAP plz

Pani-rosa [81]

Answer: x = 13.4

Step-by-step explanation:

1. Subtract 90 degrees and 33 degrees to get angle X, which is 57 degrees.

2. Using Laws of Sine, plug in the known elements of the triangle:

(16) / (sin(90)) = (x) / (sin(57))

3. Cross multiply, this will leave 16sin(57) = xsin(90)

4. Divide sin(90) to the other side, leaving x = (16sin(57)) / (sin(90))

5. Plug this into a calculator and your final answer should be 13.4

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1 year ago

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Hakeem puts $53 from each paycheck into his savings account. He gets paid every two weeks or 26 times per year. His goal is to s

garik1379 [7]

Step-by-step explanation:

53×26=1378. hence if hakeem saves $1300 he still has about $78.

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1 year ago

At which values of x does the function F(x) have a vertical asymptote? Check all that apply. F(x)=2/3x(x-1)(x 5).

Alekssandra [29.7K]

The function F(x) have a vertical asymptote at x =0 and x =1.

Given that

The given Function;

$\rm F(x)=\dfrac{2}{3x(x-1)(x^ 5)}$

We have to determine

At which value the function F(x) has a vertical asymptote.

According to the question

The given Function;

$\rm F(x)=\dfrac{2}{3x(x-1)(x^ 5)}$

The function is undefined when the denominator becomes 0. Find values of x for which the denominator is 0:

Then,

$\rm 3x (x-1) (x^5)= 0$

$\rm x^5 =0, \ \ x=0\\\\ x-1 = 0, \ \ x =1\\$

The value of x is 0 and 1.

Hence, the function F(x) have a vertical asymptote at x =0 and x =1.

To know more about Parabola click the link given below.

brainly.com/question/12868913

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1 year 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

Evaluate the expression: v ⋅ w

dexar [7]

V ⋅ w = <2, -8, -8> ⋅ <-2, 6, -5> = 2(-2) + -8(6) + -8(-5) = -4 + -48 + 40 = -12

v ⋅ w = -12

4 0

2 years ago

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