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4 years ago

Please help me now!!!

2 answers:

6 0

Any equilateral triangle has 3 angles of 60° each as all the angles have to add to 360° and there must be 3 equal sides. As no side length is specified you can make infinitely many different equilateral triangles provided they all have different side lengths.

Korvikt [17]4 years ago

3 0

Answer:

<h2>D. Infinitely many</h2>

Step-by-step explanation:

In this case, the number of equilateral triangles we could make is restricted by the length of its sides. However, this problem doesn't restrict any side, it doesn't specifies a length for all sides, it only gives the angles, which have to be equal to 60° because it's a equilateral triangle.

This means that we could make infinitely many triangles with angles of 60°, it could be a triangles with sides of 1 unit long, 2 units, or 100 units long, all of them would have angles of 60°.

Therefore, the right answer is D.

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Complete th equivalent equation for -7x-60=x^2+10x.

algol13

Answer:

1. (x+ 5)(x+12)=0

2. -12 or -5

Hope this helps!!

6 0

3 years ago

DECIDE WHETHER THE CONGRUENCE STATEMENT IS TRUE. EXPLAIN YOUR REASONING.

kaheart [24]

Yes it is because ejehhjekeb

6 0

3 years ago

Of the 45 animals on the farm, there are only pigs and chickens. The animals have a total of 124 legs. How many pigs and how man

Bess [88]

Answer:

$on \: this \: farm \: there \: are \to \\ \underline{ \boxed{17 \: pigs }}\: \\ and \\ \underline{ \boxed{28 \:chickens}}$

Step-by-step explanation:

$\\ \underline{ \boxed{solution \: to \: question \: \boxed{ A}}} \\ \\ let \: the \: pigs \: be \to \: p \\ let \: the \: chicks \: be \to \: c \\let \: the \: total \: animals \: be \to \: 45 \\ hence \to \\ p + c = 45.......eq(1) \\ since \: chickens \: have \: two \: legs : \\ while \: pigs \: have \: four \: legs : \\ let \: animals \: with \: 2 \: legs \: be \to \: 2c \\ let \: animals \: with \: 4\: legs \: be \to \: 4p \\ the \: total \: number \: of \: animal \: legs = 124 \\ hence \to \\ 4p + 2c = 124..........eq(2) \\ therefore \: the \: system \: of \: equations \: \\ to \: describe \: the \: situation \: is \to \\ \\ p + c = 45 \\ 4p + 2c = 124 \\ \\ \\ \underline{ \boxed{solution \: to \: question \: \boxed{B }}} \\ \\ \\ we \: could \: apply \: any \: of \: the \: simultanious \\ principles \to \\ here : am \: appling \: the \: elimination \: method \to \\ p + c = 45 \\ 4p + 2c = 124 \\ to \: elimnate \: one \: varaible \: we \: multiply \: by \to \\ (4) \times p + c = 45 \\ (1) \times 4p + 2c = 124 \\ hence \to \\ - \frac{4p + 4c = 180}{4p + 2c = 124} \\next \to \\ 2c = 56 \\ c = \frac{56}{2} \\ \boxed{ c = 28} \\ lets \: work \: for \: p \to \\ p + c = 45 \\ p + 28 = 45 \\ p = 45 - 28 \\ \boxed{p = 17}$

6 0

3 years ago

6 165.1 159.2 163.5 174.8 173.2 177.6 174.3 164.1 171.4 find the median

Katen [24]

159.2, 163.5, 164.1, 171.4, 173.2, 174.3, 174.8, 177.6, 6165.1

mean= 835.9
median= 173.2
mode= N/A

5 0

4 years ago

While flying at an altitude of 1.5 km, a plane measures angles or depression to opposite ends of a large crater, shown in the im

Anna35 [415]

Check the picture below.

notice the alternate interior angles in the picture.

$\bf tan(68^o)=\cfrac{\stackrel{opposite}{1.5}}{\stackrel{adjacent}{x}}\implies x=\cfrac{1.5}{tan(68^o)}\implies x\approx 0.61 \\\\[-0.35em] ~\dotfill\\\\ tan(56^o)=\cfrac{\stackrel{opposite}{1.5}}{\stackrel{adjacent}{w}}\implies w=\cfrac{1.5}{tan(56^o)}\implies w\approx 1.01 \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ ~\hfill \stackrel{\textit{width of the crater}}{x+w\implies 1.62}~\hfill$

6 0

3 years ago