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Elenna [48]
2 years ago
8

Which congruence theorem can be used to prove △MLQ ≅ △NPQ? AAS SSS ASA SAS

Mathematics
2 answers:
zvonat [6]2 years ago
8 0

Answer:

the answer is SSS my child

storchak [24]2 years ago
4 0
Answer: SSS

Proof:
In ΔMLQ and ΔNPQ,
MQ = NQ (given) S
Since Q is the midpoint of LP, by definition, LQ = QP (S)
LM = PN (given) S

∴ ΔMLQ ≡ ΔNPQ (SSS)
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Ella’s test grades in her history class are 89, 94, 82, 84, and 98. What score must Ella make on her next test to have a mean te
aksik [14]
Mean is same as average

to find average
add up terms
divided that sum by the number of terms you have
example

mean of x,y,z
x+y+z
3 terms
(x+y+z)/3=mean



so
represent next test score as x
89+94+82+84+98+x
count how many terms ther are (6 terms)

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(89+94+82+84+98+x)/6=90
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8 0
2 years ago
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The graph of a sinusoidal function has a maximum point at (0,5)(0,5)left parenthesis, 0, comma, 5, right parenthesis and then ha
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Answer:

sinusoidal functions have the standard form of Asin(Bx - C) + D

Where A is the amplitude, 2pi/B gives us our period, C gives us our horizontal shifts in the opposite direction of the sign (because it's inside the parenthesis), and D gives us our vertical shifts in the same direction as the sign of D (positive or negative).

The easiest part is assigning the amplitude which is 3 as stated in the problem. This means that all of the max/mins of the graph will be multiplied by a factor of 3. However, this could be a positive or negative 3 depending on other info in the problem. We'll come back to this in a second.

We know 2pi/B = 8 in our problem. Solving for B gives us B = pi/4.

Now we need to find the corresponding maximum and minimums for a sin/cos function with a period of 8 by taking x values 0, pi/2, pi, 3pi/2, and 2pi (these are all values where normal sin/cos functions are either at their max/min or zero) and then divide each by pi/4. This will give us our new max/mins for a function with period 8. These values are 0, 2, 4, 6, and 8.

Since 2 is a minimum, then we know that there are no horizontal shifts. So C = o

Now we need to figure out if this is a sin or cos graph. A normal cos graph has a value of 0 at x = pi/2 which corresponds x = 2 in our problem. Your problem says that x = 2 will give us a minimum value so this tells us that our function must be a sin graph, not a cos graph. However, a sin graph has a maximum at 2 (which is pi/2 in a normal sin graph) while your problem calls for a minimum. This means that the amplitude we found earlier of 3 must actually be a -3 (I told you we'd get back to this!). The negative sign flips all of the maximums to minimums and vice versa of the sinusoidal graph.

Ok, let's put together what we know so far. A = -3, B = pi/4, and C = 0. So f(x) = -3sin((pi/4)x) + d

......But what about D?

Well, your problem says that one minimum of the graph is (2,1). In a graph without vertical shifts, we would expect the minimum to be at (2,-3). The difference between -3 and 1 is 4. This means in order to get from -3 to 1 we have to shift upwards of 4 units. In other words, D = 4.

Answer: -3sin((pi/4)x) + 4

Step-by-step explanation:

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Step-by-step explanation:

The desired equation equates 1/15 of the amount Lindsay earned with 1/10 the amount without the bonus:

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_____

<em>Comment on the form of the answer</em>

If you don't like answers in this form, please do not post questions in this form. Ordinary math symbols are much appreciated.

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Hiii

all you do is subtract...
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