0.067 as a fraction

Mathematics

2 answers:

yarga [219]4 years ago

6 0

Answer:

67/1000

Step-by-step explanation:

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tangare [24]4 years ago

4 0

Answer: 67/1000

Step-by-step explanation:

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Help with this question Asap!! I need all the help I can get!

tankabanditka [31]

Answer:

$\large{YES}\ \dfrac{BC}{YZ}=\dfrac{AC}{XY}=\dfrac{AB}{XZ}=\dfrac{2}{1}\\\\\text{and}\ \angle B\cong\angle Z,\ \angle C\cong\angle Y,\ \angle A\cong\angle X$

Step-by-step explanation:

$BC\to ZY\\AB\to XZ\\AC\to XY\\\angle A\to\angle X\\\angle B\to\angle Z\\\angle C\to\angle Y$

$\text{We have:}\\\\\triangle ABC:\ AB=20,\ BC=12,\ AC=18\\\triangle XZY:\ XZ=10,\ YZ=6,\ XY=9\\\\\text{check the ratio:}\\\\\dfrac{AB}{XZ}=\dfrac{20}{10}=2\\\\\dfrac{BC}{YZ}=\dfrac{12}{6}=2\\\\\dfrac{AC}{XY}=\dfrac{18}{9}=2\\\\\text{CORRECT :)}\\\\\angle B\cong\angle Z,\ \angle C\cong\angle Y\\\\\text{We know that: The sum of the acute angles in a right triangle is}\ 90^o.\\\\m\angle B+m\angle A=90^o\ \text{and}\ m\angle Z+m\angle X=90^o$

$\text{We know}\ m\angle B=m\angle Z\to \ m\angle A=m\angle X.\\\text{therefore}\ \angle A\cong\angle X$

5 0

4 years ago

Solve for a and b in the following circles:<br> B)<br> a а<br> b<br> 82°

Crank

Answer:

A.po ang sagot

Step-by-step explanation:

ty po hope it help wag nyo na lang po kopyhin kung mali yan po kasi answer ko

8 0

3 years ago

Read 2 more answers

How do u do b??????????

horrorfan [7]

the assumption being that the first machine is the one on the left-hand-side and the second is the one on the right-hand-side.

the input goes to the 1st machine and the output of that goes to the 2nd machine.

if she uses and input of 6 on the 2nd one, the result will be 6² - 6 = 30, if we feed that to the 1st one the result will be √( 30 - 5) = √25 = 5, so, simply having the machines swap places will work to get a final output of 5.

clearly we can never get an output of -5 from a square root, however we can from the quadratic one, the 2nd machine/equation.

let's check something, we need a -5 on the 2nd, so

$\bf \underset{final~out put}{\stackrel{y}{-5}}=x^2-6\implies 1=x^2\implies \sqrt{1}=x\implies 1=x$

so if we use a "1" as the output on the first machine, we should be able to find out what input we need, let's do that.

$\bf \underset{first~out put}{\stackrel{y}{1}}=\sqrt{x-5}\implies 1^2=x-5\implies 1=x-5\implies 6=x$

so if we use an input of 6 on the first machine, we should be able to get a -5 as final output from the 2nd machine.

$\bf \stackrel{first~machine}{y=\sqrt{\boxed{6}-5}}\implies y=\sqrt{1}\implies y=1 \\\\\\ \stackrel{second~machine}{y = \boxed{1}^2-6}\implies y = 1-6\implies y = -5$