A=1/2h(B+b); A=70, B-6, b=1 solve for h

Mathematics

1 answer:

larisa86 [58]3 years ago

5 0

this should help alil better.

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How were the numbers 1 10 100 and 1000 written by romans?

Arturiano [62]

1 -- I
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8 0

3 years ago

BRAINEST IF YOU GET THSI RIGHT

Scorpion4ik [409]

<h3>Solution :</h3>

$\sf \dfrac{6x+4}{3} = \dfrac{5x-5}{8}$

By cross multiply, we get

$\sf : \implies (6x+4) \times 8 = (5x-5) \times 3$

$\sf : \implies 48x+32 = 15x-15$

$\sf : \implies 48x+32 - 15x + 15= 0$

$\sf : \implies 33x + 47 = 0$

$\sf : \implies 33x = -47$

$\sf : \implies x = \dfrac{- 47}{33}$

Therefore, answer is $\sf x = \dfrac{- 47}{33}$

4 0

2 years ago

Read 2 more answers

Is sin theta=5/6, what are the values of cos theta and tan theta?

mina [271]

let's bear in mind that sin(θ) in this case is positive, that happens only in the I and II Quadrants, where the cosine/adjacent are positive and negative respectively.

$\bf sin(\theta )=\cfrac{\stackrel{opposite}{5}}{\stackrel{hypotenuse}{6}}\qquad \impliedby \textit{let's find the \underline{adjacent side}} \\\\\\ \textit{using the pythagorean theorem} \\\\ c^2=a^2+b^2\implies \pm\sqrt{c^2-b^2}=a \qquad \begin{cases} c=hypotenuse\\ a=adjacent\\ b=opposite\\ \end{cases} \\\\\\ \pm\sqrt{6^2-5^2}=a\implies \pm\sqrt{36-25}\implies \pm \sqrt{11}=a \\\\[-0.35em] ~\dotfill$

$\bf cos(\theta )=\cfrac{\stackrel{adjacent}{\pm\sqrt{11}}}{\stackrel{hypotenuse}{6}} \\\\\\ tan(\theta )=\cfrac{\stackrel{opposite}{5}}{\stackrel{adjacent}{\pm\sqrt{11}}}\implies \stackrel{\textit{and rationalizing the denominator}~\hfill }{tan(\theta )=\pm\cfrac{5}{\sqrt{11}}\cdot \cfrac{\sqrt{11}}{\sqrt{11}}\implies tan(\theta )=\pm\cfrac{5\sqrt{11}}{11}}$