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3 years ago

4) Express 1697 in Roman figures:

Mathematics

2 answers:

evablogger [386]3 years ago

6 0

Answer:

$\huge{\textbf{\textsf{{\color{skyblue}{An}}{\blue{sw}}{\red{er}}{\color{red}{:}}}}}$

1697 ➳ MDCXCVII

Digiron [165]3 years ago

4 0

Answer:

MDCXCVII

Step-by-step explanation:

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Triangle $ABC$ has a right angle at $B$. Legs $\overline{AB}$ and $\overline{CB}$ are extended past point $B$ to points $D$ and

Lisa [10]

Answer:

Given :

ABC is a right triangle in which ∠ABC = 90°,

Also, Legs AB and CB are extended past point B to points D and E,

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$\angle EAC = \angle ACD = 90^{\circ}$

To prove :

$EB\times BD=AB\times BC$

Proof :

In triangles AEC and EBA,

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∠CEA = ∠AEB ( common angles )

By AA similarity postulate,

$\triangle AEC \sim \triangle EBA$ ,

Similarly,

$\triangle AEC \sim \triangle ABC$

$\implies\triangle EBA\sim \triangle ABC-----(1)$

Now, In triangles ADC and CBD,

∠ACD = ∠CBD ( right angles )

∠ADC= ∠BDC ( common angles )

By AA similarity postulate,

$\triangle ADC \sim \triangle CBD$ ,

Similarly,

$\triangle ADC \sim \triangle ABC$

$\implies \triangle CBD\sim \triangle ABC-----(2)$

From equations (1) and (2),

$\triangle EBA\sim \triangle CBD$

The corresponding sides of similar triangles are in same proportion,

$\frac{EB}{BC}=\frac{AB}{BD}$

$\implies EB\times BD=AB\times BC$

Hence, proved....

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3 years ago

What percent of 25$ is 5$ no links

liraira [26]

Answer:

20%

Step-by-step explanation:

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What is significant digit rules and how is it used?

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Answer:

Non-zero digits are always significant. Any zeros between two significant digits are significant. A final zero or trailing zeros in the decimal portion ONLY are significant.

Step-by-step explanation:

Non-zero digits are always significant.
Any zeros between two significant digits are significant.
A final zero or trailing zeros in the decimal portion ONLY are significant.

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3 years ago

A square piece of paper has an area of x2 square units. A rectangular strip with a width of 2 units and a length of x units is c

vlabodo [156]

Let

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we know that

the area of the original square paper is equal to

$A1=x*x=x^{2} \ units^{2}$

the area of the remaining piece of paper is equal to

$A2=x^{2} -2x\\ A2=120\ units^{2} \\ so\\ x^{2} -2x=120\\ x^{2} -2x-120=0$

therefore

the answer is the option

x2 − 2x − 120 = 0

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4) Express 1697 in Roman figures:​

4) Express 1697 in Roman figures: