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Andrej [43]
3 years ago
14

Which congruency statements are true?

Mathematics
2 answers:
MArishka [77]3 years ago
7 0

Answer:

QH≅MF, MN≅QR, M≅Q,

djyliett [7]3 years ago
4 0

Answer: 1st answer, 2nd answer, and 4th answer

Step-by-step explanation:

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What is −2/5 + 4/5
igor_vitrenko [27]

Answer:

5/2 or 0.4

Step-by-step explanation:

6 0
3 years ago
Read 2 more answers
What is the axis of symmetry of h(x) = –2x2 + 12x – 3?
murzikaleks [220]
The axis of symmetry is the x value of the vertex
we have a handy-dandy way of finding that from standard form, ax^2+bx+c=y

for
ax^2+bx+c=y
the x value of the vertex is -b/2a

y=-2x^2+12x-3
-b/2a=-12/(2*-2)=-12/-4=3

x=3 is the axis of symmetry
5 0
4 years ago
Read 2 more answers
A bus can hold 42 students
dsp73

Answer:

10, there will be 20 left open

Step-by-step explanation:

if each bus holds 42 students and there are 400 going then you need to divide 400 by 42, and then round up, to find how many seats will be left open, you take the answer and then multiply it by 42 and then subtract 400.

sorry for not putting my work last time.

8 0
4 years ago
What is the least natural number which when divided by 3,5,6,8,10 and 12 leaves in each case remainder 2 but when divided by 13
bazaltina [42]

Answer:

962

Step-by-step explanation:

MMM for step by step its kinda just plug and check

962/3 = 320 R2

962/6 = 160 R2

962/8 = 120 R2

962/10 = 96 R2

962/12 = 80 R2

and finally 962/13 = 74

4 0
2 years ago
PLEASE HELP!!
erik [133]
2.=c.\\\\2\sin4x\cos4x=2\sin(2\cdot4x)=2\sin8x\\\\Used:\\\sin2\alpha=2\sin\alpha\cos\alpha

1.=b.\\\\\csc x-\sin x=\dfrac{1}{\sin x}-\dfrac{\sin^2x}{\sin x}=\dfrac{1-\sin^2x}{\sin x}=\dfrac{\cos^2x}{\sin x}\\\\=\dfrac{\cos x\cos x}{\sin x}=\cos x\cdot\dfrac{\cos x}{\sin x}=\cos x\cot x\\\\Used:\\\csc x=\dfrac{1}{\sin x}\\\\\sin^2x+\cos^2x=1\to\cos^2x=1-\sin^2x\\\\\cot x=\dfrac{\cos x}{\sin x}

3.=a.\\\\\dfrac{\sin x-1}{\sin x+1}=\dfrac{\sin x-1}{\sin x+1}\cdot\dfrac{\sin x+1}{\sin x+1}=\dfrac{\sin^2x-1^2}{(\sin x+1)^2}=\dfrac{\sin^2x-1}{(\sin x+1)^2}\\\\=\dfrac{-(1-\sin^2x)}{(\sin x+1)^2}=\dfrac{-\cos^2x}{(\sin x+1)^2}\\\\Used:\\(a-b)(a+b)=a^2-b^2\\\\\sin^2x+\cos^2x=1\to \cos^2x=1-\sin^2x
8 0
3 years ago
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