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poizon [28]
3 years ago
15

In fractions what is between 16.4 and16.5

Mathematics
2 answers:
earnstyle [38]3 years ago
8 0
16.4/5 I not sure just trying
stepladder [879]3 years ago
3 0
16.45 is in between 16.4 and 16.5

16.45 in fractions would be 16 \frac{45}{100}=16 \frac{9}{20}= \boxed{\frac{329}{20}}
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Find the midpoint of the line segment with end coordinates of: (-2, -4) and (2, -10)
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(0,-7)

Step-by-step explanation:

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7 0
2 years ago
Angle A is supplementary to angle B. If A. 40<br> B. 85<br> C. 5<br> D. 10
NeX [460]
So if both were 480 and a 180 I need the rest of the question to answer
4 0
3 years ago
I will give a brainlist if someone could answer this problem with a explanation:)
egoroff_w [7]

Answer:

Step-by-step explanation:

The two base angles in each triangle are equal. (The triangle is isoceles and the property used is angles opposite equal sides are equal).

Now the tricky part. The peak angles are also equal. That's because both peak angles are made equal by Peak angle + 2*base angle = 180

Peak angle = 180 - 2*base angle

Therefore the triangles are congruent by SAS.  

I suppose you could get them equal by taking one of the base angles (all 4 are equal to each other) and the peak angles and claim equality by ASA, but it seems a little bit tortuous to me.

5 0
3 years ago
Given sin x = -4/5 and x is in quadrant 3, what is the value of tan x/2
love history [14]

bearing in mind that, on the III Quadrant, sine as well as cosine are both negative, and that hypotenuse is never negative, so, if the sine is -4/5, the negative number must be the numerator, so sin(x) = (-4)/5.


\bf sin(x)=\cfrac{\stackrel{opposite}{-4}}{\stackrel{hypotenuse}{5}}\impliedby \textit{let's find the \underline{adjacent}} \\\\\\ \textit{using the pythagorean theorem} \\\\ c^2=a^2+b^2\implies \sqrt{c^2-b^2}=a \qquad \begin{cases} c=hypotenuse\\ a=adjacent\\ b=opposite\\ \end{cases} \\\\\\ \pm\sqrt{5^2-(-4)^2}=a\implies \pm\sqrt{9}=a\implies \pm 3=a \\\\\\ \stackrel{III~Quadrant}{-3=a}~\hfill cos(x)=\cfrac{\stackrel{adjacent}{-3}}{\stackrel{hypotenuse}{5}} \\\\[-0.35em] ~\dotfill

\bf tan\left(\cfrac{\theta}{2}\right)= \begin{cases} \pm \sqrt{\cfrac{1-cos(\theta)}{1+cos(\theta)}} \\\\ \cfrac{sin(\theta)}{1+cos(\theta)}\qquad \leftarrow \textit{let's use this one} \\\\ \cfrac{1-cos(\theta)}{sin(\theta)} \end{cases} \\\\[-0.35em] ~\dotfill

\bf tan\left( \cfrac{x}{2} \right)=\cfrac{~~\frac{-4}{5}~~}{1-\frac{3}{5}}\implies tan\left( \cfrac{x}{2} \right)=\cfrac{~~\frac{-4}{5}~~}{\frac{2}{5}}\implies tan\left( \cfrac{x}{2} \right)=\cfrac{-4}{5}\cdot \cfrac{5}{2} \\\\\\ tan\left( \cfrac{x}{2} \right)=\cfrac{-4}{2}\cdot \cfrac{5}{5}\implies tan\left( \cfrac{x}{2} \right)=-2

4 0
4 years ago
Read 2 more answers
Help please Um idk what’s going on
Arada [10]
The answer would be A.) since f(-2)=x+3 leads to f(-2)= -1/2x + -3/2 which is -2 since -4/2 is -2, so then we have f=-1
3 0
4 years ago
Read 2 more answers
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