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Licemer1 [7]
3 years ago
12

Solve (2x-1)/(y)=(w+2)/(2z) for w

Mathematics
1 answer:
zubka84 [21]3 years ago
4 0

Answer:

w= 4xz/y-2

Step-by-step explanation:

2x-1/y=w+2/2z

2x-1x2z=w+2xy

4xz-1=wy+2

4xz-1-2=wy

4xz/y-2=w

w=4xz/y-2

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How many strings are there of four lowercase letters that have the letter x in them?
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4 years ago
I WILL MARK BRAINLIEST!!
vagabundo [1.1K]

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Step-by-step explanation:

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PLEASE HELP ASAP!!! I NEED CORRECT ANSWERS ONLY PLEASE!!!
Mila [183]

Answer:

m\angle H = 44.4\°.

Step-by-step explanation:

Given:

In Right Angle Triangle GIH

∠ I = 90°

GI = 7    ....Side opposite to angle H

GH = 10  .... Hypotenuse

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m∠H = ?

Solution:

In Right Angle Triangle ABC ,Sine Identity,

sin \ H = \frac{Oppsite\  side\  to\  \angle H}{Hypotenuse}

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sin\ H = \frac{7}{10} = 0.7

Now taking sin^{-1} we get;

\angle H = sin^{-1}\ 0.7 = 44.427

rounding to nearest tenth we get.

m\angle H = 44.4\°.

Hence m\angle H = 44.4\°.

6 0
3 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
3 years ago
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