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4 years ago

What does it mean for two triangles to be similar

Mathematics

2 answers:

Zielflug [23.3K]4 years ago

3 0

When two triangles are similar they are congruent

lawyer [7]4 years ago

3 0

^ very very wrong do NOT ever assume they are congruent. Congruent triangles are always similar, but similar triangles are NOT ALWAYS congruent.

similar triangles are triangles that have proportioned sides and congruent angles.

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IF 24 IS ADDED TO A NUMBER, IT WILL BE 3 TIMES AS LARGE AS IT WAS ORIGINALLY. FIND THE NUMBER

Romashka [77]

$x-the\ number\\\\x+24=3x\ \ \ \ \ \ |subtract\ 24\ from\ both\ sides\\\\x=3x-24\ \ \ \ \ \ \ |subtract\ 3x\ from\ both\ sides\\\\-2x=-24\ \ \ \ \ \ \ \ |divide\ both\ sides\ by\ (-2)\\\\\boxed{x=12}\leftarrow answer$

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3 years ago

HELP ASAP GET 5 POINTS

tamaranim1 [39]

5 because 2 - -3 = 5.

8 0

3 years ago

Read 2 more answers

Cam has 32 dollars. How many roses can he afford to buy?

Vlad [161]

Answer:

8 roses

Step-by-step explanation:

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3 years ago

Evaluate the algebraic expression.<br> z^2+1+y when y=4, and z=3.

Luden [163]

Answer:

your question is

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3²+1+4

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5 0

2 years ago

A given field mouse population satisfies the differential equation dp dt = 0.5p − 410 where p is the number of mice and t is the

ohaa [14]

Answer:

a) $t = 2 *ln(\frac{82}{5}) =5.595$

b) $t = 2 *ln(-\frac{820}{p_0 -820})$

c) $p_0 = 820-\frac{820}{e^6}$

Step-by-step explanation:

For this case we have the following differential equation:

$\frac{dp}{dt}=\frac{1}{2} (p-820)$

And if we rewrite the expression we got:

$\frac{dp}{p-820}= \frac{1}{2} dt$

If we integrate both sides we have:

$ln|P-820|= \frac{1}{2}t +c$

Using exponential on both sides we got:

$P= 820 + P_o e^{1/2t}$

Part a

For this case we know that p(0) = 770 so we have this:

$770 = 820 + P_o e^0$

$P_o = -50$

So then our model would be given by:

$P(t) = -50e^{1/2t} +820$

And if we want to find at which time the population would be extinct we have:

$0=-50 e^{1/2 t} +820$

$\frac{820}{50} = e^{1/2 t}$

Using natural log on both sides we got:

$ln(\frac{82}{5}) = \frac{1}{2}t$

And solving for t we got:

$t = 2 *ln(\frac{82}{5}) =5.595$

Part b

For this case we know that p(0) = p0 so we have this:

$p_0 = 820 + P_o e^0$

$P_o = p_0 -820$

So then our model would be given by:

$P(t) = (p_o -820)e^{1/2t} +820$

And if we want to find at which time the population would be extinct we have:

$0=(p_o -820)e^{1/2 t} +820$

$-\frac{820}{p_0 -820} = e^{1/2 t}$

Using natural log on both sides we got:

$ln(-\frac{820}{p_0 -820}) = \frac{1}{2}t$

And solving for t we got:

$t = 2 *ln(-\frac{820}{p_0 -820})$

Part c

For this case we want to find the initial population if we know that the population become extinct in 1 year = 12 months. Using the equation founded on part b we got:

$12 = 2 *ln(\frac{820}{820-p_0})$

$6 = ln (\frac{820}{820-p_0})$

Using exponentials we got:

$e^6 = \frac{820}{820-p_0}$

$(820-p_0) e^6 = 820$

$820-p_0 = \frac{820}{e^6}$

$p_0 = 820-\frac{820}{e^6}$

8 0

3 years ago