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Mathematics

1 answer:

miss Akunina [59]2 years ago

8 0

Finding slope?
5. Slope is 5/3
6. Slope is 7/3
7. Slope is -3
8. Slope is 7/3

Just do rise/run

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Solve the equation for P: 3k=7Q+6p

iren [92.7K]

3k=7Q+6p

Solve for P which means we have to make 'p' alone

$3k=7Q+6p\\\\ Add 7Q on both sides \\\\ 3k - 7Q = 7Q - 7Q + 6p\\\\ \left ( 3k - 7Q \right ) = 6p\\\\ Divide both side by 6 to make p alone \\\\ \frac{\left ( 3k - 7Q \right )}{6} = \frac{6p}{6}\\\\ p = \frac{\left ( 3k - 7Q \right )}{6}$

7 0

2 years ago

Read 2 more answers

How do I solve this question ?

Julli [10]

Step-by-step explanation:

here you go!

so basically x = 9 cars

4 0

1 year ago

In New York a crowed subway has 375 passengers distributed between 5 cars. What is the unit rate for each subway car?

Andrews [41]

<h2>Answer:</h2><h2 /><h2>The unit rate is 75/1.</h2><h2 /><h2>Step-by-step explanation:</h2><h2 /><h2>To figure out this problem you have to divide. </h2><h2>375 divided by 5 is equal to 75. </h2><h2 /><h2>Therefore the unit rate is 75/1.</h2><h2 /><h2>Hope this helps and have a great day!!!</h2><h2 /><h2>:)</h2>

7 0

2 years ago

The summit charges $4 fee and $3 per hour to rent a kayak. Extreme Paddling charges a $5 fee and $2 per hour to rent a kayak. Wr

djverab [1.8K]

Answer:

The answer would be S=4x+3y and E=5x+2y

7 0

3 years ago

Solve the equation on the interval [0,2π]

WARRIOR [948]

$\bf 16sin^5(x)+2sin(x)=12sin^3(x)
\\\\\\
16sin^5(x)+2sin(x)-12sin^3(x)=0
\\\\\\
\stackrel{common~factor}{2sin(x)}[8sin^4(x)+1-6sin^2(x)]=0\\\\
-------------------------------\\\\
2sin(x)=0\implies sin(x)=0\implies \measuredangle x=sin^{-1}(0)\implies \measuredangle x=
\begin{cases}
0\\
\pi \\
2\pi 
\end{cases}\\\\
-------------------------------$

$\bf 8sin^4(x)+1-6sin^2(x)=0\implies 8sin^4(x)-6sin^2(x)+1=0$

now, this is a quadratic equation, but the roots do not come out as integers, however it does have them, the discriminant, b² - 4ac, is positive, so it has 2 roots, so we'll plug it in the quadratic formula,

$\bf 8sin^4(x)-6sin^2(x)+1=0\implies 8[~[sin(x)]^2~]^2-6[sin(x)]^2+1=0
\\\\\\
~~~~~~~~~~~~\textit{quadratic formula}
\\\\
\begin{array}{lcccl}
& 8 sin^4& -6 sin^2(x)& +1\\
&\uparrow &\uparrow &\uparrow \\
&a&b&c
\end{array} 
\qquad \qquad 
sin(x)= \cfrac{ - b \pm \sqrt { b^2 -4 a c}}{2 a}
\\\\\\
sin(x)=\cfrac{-(-6)\pm\sqrt{(-6)^2-4(8)(1)}}{2(8)}\implies sin(x)=\cfrac{6\pm\sqrt{4}}{16}
\\\\\\
sin(x)=\cfrac{6\pm 2}{16}\implies sin(x)=
\begin{cases}
\frac{1}{2}\\\\
\frac{1}{4}
\end{cases}$

$\bf \measuredangle x=
\begin{cases}
sin^{-1}\left( \frac{1}{2} \right)
sin^{-1}\left( \frac{1}{4} \right)
\end{cases}\implies \measuredangle x=
\begin{cases}
\frac{\pi }{6}~,~\frac{5\pi }{6}\\
----------\\
\approx~0.252680~radians\\
\qquad or\\
\approx~14.47751~de grees\\
----------\\
\pi -0.252680\\
\approx 2.88891~radians\\
\qquad or\\
180-14.47751\\
\approx 165.52249~de grees
\end{cases}$

3 0

3 years ago