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3 years ago

Indicate the method you would use

Mathematics

1 answer:

Alexeev081 [22]3 years ago

7 0

Answer:

<em> SSS</em><em> can be used to prove that the given triangles are congruent.</em>

Step-by-step explanation:

In the given two triangles ΔADC and ΔABC,

AD = AB = 7 units,
CD = BC = 8 units,
AC is common to both triangles,

Hence, ΔADC ≅ ΔABC by Side-Side-Side (SSS) congruence.

SSS congruence-

If three sides of one triangle are congruent to three sides of another triangle, then the triangles are congruent.

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Two mechanics worked on a car. The first mechanic worked for

Dima020 [189]

X - the rate charged by the first mechanic
y - the rate charged by the second mechanic

The sum of the two rates was $155.
$x+y=155 \\
y=155-x$

The first mechanic worked for 10 hours, the second mechanic worked for 5 hours, and together they charged $1000.
$10x+5y=1000 \ \ |\div 5 \\
2x+y=200 \\
y=200-2x$

Set 155-x and 200-2x equal to each other:
$155-x=200-2x \\
-x+2x=200-155 \\
x=45 \\ \\
y=155-x=155-45=110$

The rate charged by the first mechanic was $45, and the rate charged by the second mechanic was $110.

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3 years ago

<img src="https://tex.z-dn.net/?f=%5Cfrac%7Bcos%2030%7D%7B1%2Bsin30%7D%3Dtg30" id="TexFormula1" title="\frac{cos 30}{1+sin30}=tg

Kay [80]

$\cos 30=\frac{\sqrt{3}}{2}$

$\sin 30 = \frac{1}{2}$

$\tan 30=\frac{1}{\sqrt{3}}$

Now we insert that into the equation:

$\dfrac{\frac{\sqrt{3}}{2}}{1+\frac{1}{2}} = \frac{1}{\sqrt{3}}$

$\dfrac{\frac{\sqrt{3}}{2}}{\frac{3}{2}} = \frac{1}{\sqrt{3}}$

$\frac{\sqrt{3}}{3} = \frac{1}{\sqrt{3}}$

$\frac{\sqrt{3}}{3} = \frac{1}{\sqrt{3}}\cdot \frac{\sqrt{3}}{\sqrt{3}}\\$

$\frac{\sqrt{3}}{3} = \frac{\sqrt{3}}{\sqrt{3}\cdot \sqrt{3}}\\$

$\frac{\sqrt{3}}{3} = \frac{\sqrt{3}}{3}$

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4 years ago

Find the perimeter of the polygon with the vertices A (-1,5), B (-1,-4), C (-6,-4), and D (-6,5)

ch4aika [34]

Schools almost over, good luck!!

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3 years ago

How many terms are in the arithmetic sequence 7, 0, −7, . . . , −175?

Sav [38]

So hmmm 7, 0, -7.... what the dickens is going on? hmmm is really dropping each time by 7, so, 7-7, 0, and 0-7, -7 and so on.

so, the "common difference" is then -7, and our first term is 7, now, who's -175? let's check.

$\bf a_n=a_1+(n-1)d\qquad 
\begin{cases}
n=n^{th}\ term\\
a_1=\textit{first term's value}\\
d=\textit{common difference}\\
----------\\
d=-7\\
a_1=7\\
a_n=-175
\end{cases}
\\\\\\
-175=7+(n-1)(-7)\implies -175=7+7-7n
\\\\\\
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3 years ago

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max2010maxim [7]

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