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3 years ago

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Help please!! Stuck on this question!! ∠ABC is adjacent to ∠CBD. If the m∠ABC=4x+23, m∠CBD=6x+7, and m∠ABD=130°, what is the mea

sure of angle ABC?

1 answer:

Ulleksa [173]3 years ago

3 0

Answer:

63 degrees

Step-by-step explanation:

add 4x + 23 and 6x + 7, then equal that to 130. When you find x, plug it into 4x+23 and get 63.

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Solve: 2cos(x)-square root 3=0 for 0 less than x less than 2 pi

Leona [35]

Answer:

The general solution of $cos x = cos(\frac{\pi }{6})$ is

x = 2nπ± $\frac{\pi }{6}$

The general solution values

$x = - \frac{\pi }{6} and x = \frac{\pi }{6}$

Step-by-step explanation:

Explanation:-

Given equation is

$2cosx-\sqrt{3} =0 for 0$

$2cosx =\sqrt{3}$

Dividing '2' on both sides, we get

$cos x =\frac{\sqrt{3} }{2}$

$cos x = cos(\frac{\pi }{6})$

<em>General solution of cos θ = cos ∝ is θ = 2nπ±∝</em>

<em>Now The general solution of </em> $cos x = cos(\frac{\pi }{6})$ <em> is </em>

<em> x = 2nπ±</em> $\frac{\pi }{6}$ <em></em>

put n=0

$x = - \frac{\pi }{6} and x = \frac{\pi }{6}$

Put n=1

$x = 2\pi +\frac{\pi }{6} = \frac{13\pi }{6}$

$x = 2\pi -\frac{\pi }{6} = \frac{11\pi }{6}$

put n=2

$x = 4\pi +\frac{\pi }{6} = \frac{25\pi }{6}$

$x = 4\pi -\frac{\pi }{6} = \frac{23\pi }{6}$

And so on

But given 0 < x< 2π

The general solution values

$x = - \frac{\pi }{6} and x = \frac{\pi }{6}$

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3 years ago

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Answer:

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Step-by-step explanation:

9+(-3) = 6

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3 years ago

Dwight has sold 22 cars and he sells 2 each week.

elixir [45]

Im thinking it is 7 weeks.

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3 years ago

Which relationships describe angles 1 and 2?

lozanna [386]

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2 years ago

Anyone have the answer for this

sergiy2304 [10]

$\huge\bold{Given:}$

Length of the base = 16 km.

Length of the hypotenuse = 34 km. $\huge\bold{To\:find:}$

✎ The length of the missing leg '' $a$ ".

$\large\mathfrak{{\pmb{\underline{\orange{Solution}}{\orange{:}}}}}$

The length of the missing leg "a" is $\boxed{30\:km}$ .

$\large\mathfrak{{\pmb{\underline{\red{Step-by-step\:explanation}}{\orange{:}}}}}$

Using Pythagoras theorem, we have

$({perpendicular})^{2} + ({base})^{2} = ({hypotenuse})^{2} \\ ⇢ {a}^{2} + ({16 \: km})^{2} = ({34 \: km})^{2} \\ ⇢ {a}^{2} + 256 \: {km}^{2} = 1156 \: {km}^{2} \\ ⇢ {a}^{2} = 1156 \: {km}^{2} - 256 \: {km}^{2} \\ ⇢ {a}^{2} = 900 \: {km}^{2} \\ ⇢a \: = \sqrt{900 \: {km}^{2} } \\ ⇢a = \sqrt{30 \times 30 \: {km}^{2} } \\ ⇢a = 30 \: km$

$\sf\blue{Therefore,\:the\:length\:of\:the\:missing\:leg\:"a"\:is\:30\:km.}$

$\huge\bold{To\:verify :}$

$( {30 \: km})^{2} + ({16 \: km})^{2} =( {34 \: km})^{2} \\ ⇝900 \: {km}^{2} + 256 \: {km}^{2} = 1156 \: {km}^{2} \\⇝1156 \: {km}^{2} = 1156 \: {km}^{2} \\ ⇝L.H.S.=R. H. S$

Hence verified. ✔

$\circ \: \: { \underline{ \boxed{ \sf{ \color{green}{Happy\:learning.}}}}}∘$

7 0

2 years ago