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Natasha2012 [34]
3 years ago
6

Could someone help me with these 2 questions

Mathematics
1 answer:
Lunna [17]3 years ago
4 0
∠P~=∠S
∠Q ~= ∠T
∠R~= ∠M
P/S=Q/T=RM
hope this helps!
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HELP!! Not hard, just confusing.. ΔABC is reflected across the x-axis and then translated 4 units up to create ΔA′B′C′. What are
allsm [11]
The coordinates of the vertices of ΔABC are:A( x1, y1), B( x2, y2) and C( x3, y 3 ). After it is reflected across the x-axis, coordinates are ( x1, -y1), (x2, -y2), (x3, -y3). Finally, the coordinates of the vertices of ΔA´B´C´ after translation are: A´( x1, 4-y1), B´( x2, 4- y2), C´( x3, 4-y3 ) 
8 0
3 years ago
What is the answer a, b, c, or d
podryga [215]

Answer: Choice B

(-2, 5)

==================================================

Explanation:

The original system is

\begin{cases}-4x+3y = 23\\ x-y = -7\end{cases}

Multiply both sides of the second equation by 3. Doing so leads to this updated system of equations

\begin{cases}-4x+3y = 23\\ 3x-3y = -21\end{cases}

Now add straight down

The x terms add to -4x+3x = -1x = -x

The y terms add to 3y+(-3y) = 0y = 0

The terms on the right hand sides add to 23+(-21) = 2

We end up with the equation  -x = 2 which solves to x = -2

Now use this to find y. You can pick any equation with x,y in it

----------------

-4x+3y = 23

-4(-2)+3y = 23

8+3y = 23

3y = 23-8

3y = 15

y = 15/3

y = 5

Or

x-y = -7

-2-y = -7

-y = -7+2

y = -5

y = 5

Either way, we get the same y value.

So that's why the solution is (x,y) = (-2, 5)

6 0
3 years ago
Making an
Tems11 [23]
1. the combinations are 5 and 3 twice and 3 and 4 twice
5+3+5+3+3+4+3+4= 30 L

2. Trya can ride 12 , 18, 24,30 kilometres are the possible distance

3. lexi can put 4 ribbons and 2 photos on her bulletin that is 40 cm meters.
ribbons 4×4=16
bulletin 12×2=24
altogether 40 cm with nothing overlapping

4. don't know dollars and centd sorry I know pounds as I am British
7 0
3 years ago
A student simplifies the given expression. Which statement BEST describes the error in the student's work? 12 + 24 ÷ 6(4) 36 ÷ 2
Marrrta [24]
A. The student added before multiplying.
If they had multiplied 6 and 4 first, the next step would have been too divide. But since they added first, they got 36/24 or 3/2 when the answer should be 13.
3 0
3 years ago
a slitter assembly contains 48 blades five blades are selected at random and evaluated each day for sharpness if any dull blade
son4ous [18]

Answer:

P(at least 1 dull blade)=0.7068

Step-by-step explanation:

I hope this helps.

This is what it's called dependent event probability, with the added condition that at least 1 out of 5 blades picked is dull, because from your selection of 5, you only need one defective to decide on replacing all.

So if you look at this from another perspective, you have only one event that makes it so you don't change the blades: that 5 out 5 blades picked are sharp. You also know that the probability of changing the blades plus the probability of not changing them is equal to 100%, because that involves all the events possible.

P(at least 1 dull blade out of 5)+Probability(no dull blades out of 5)=1

P(at least 1 dull blade)=1-P(no dull blades)

But the event of picking one blade is dependent of the previous picking, meaning there is no chance of picking the same blade twice.

So you have 38/48 on getting a sharp one on your first pick, then 37/47 (since you remove 1 sharp from the possibilities, and 1 from the whole lot), and so on.

Also since are consecutive events, you need to multiply the events.

The probability that the assembly is replaced the first day is:

P(at least 1 dull blade)=1-P(no dull blades)

P(at least 1 dull blade)=1-(\frac{38}{48}* \frac{37}{47} *\frac{36}{46}*\frac{35}{45}*\frac{34}{44})

P(at least 1 dull blade)=1-0.2931

P(at least 1 dull blade)=0.7068

5 0
3 years ago
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