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Rom4ik [11]
2 years ago
9

Pls answer quick ill give more points if right.

Mathematics
2 answers:
slega [8]2 years ago
6 0
C I believe it makes the most sense.
Vikentia [17]2 years ago
3 0

Answer:

I WOULD SAY C! IT MAKES THE MOST SENES

Step-by-step explanation:

   Defining the problem.

   Generating alternatives.

   Evaluating and selecting alternatives.

   Implementing solutions.

C says "Write the facts, organize your thoughts and define your strategy."

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△QRS is rotated about point P to △Q′R′S′ . Which statements are true about the pre-image and image?
yuradex [85]

Note that rotation will not affect the shape and size of an object.

Rotation with respect to a point preserves the corresponding sides and the corresponding angles of the original image.

Hence, the statements

The corresponding angle measurements in each triangle between the pre-image and the image are preserved and

The corresponding lengths, from the point of rotation, between the pre-image and the image are preserved

are true.

3 0
3 years ago
14(a + 2) = 168 find a
valkas [14]

\bold\red{Answer:}

14a+28=168

14a=140

a=10

hope it helps

8 0
2 years ago
Plz help. I need number 6 and 7
Cerrena [4.2K]
80 possible 2 digit passwords
5040 possible 4 digit PIN's with no repeats

can't see 7 to answer pls tell
8 0
3 years ago
Find the distance between each pair of points.round your awnser to the nearest tenths,if nessesary (-3,-4),(0,0) problem 2 (-8,0
Arte-miy333 [17]

Answer:

1st one: -3,-4 away

2nd one: Lets just say I got kinda confused on that one so sorry for any inconvenience on my behalf from this question.

Step-by-step explanation:

I will call (-3,-4) Point A and I'll call (0,0) Point B. Point A is at (-3,-4) and Point B is obviously at the origin. That makes it easy, because you just subtract Point A by Point B to find how far away they are.

3 0
3 years ago
A small metal bar, whose initial temperature was 10° C, is dropped into a large container of boiling water. How long will it tak
IgorC [24]

Answer

According to newton law of cooling

\dfrac{d T}{dt} = k (T - T_a)

and

T = Ce^{kt} + T_a

now At Ta = 100

T₀ = 10 °C

T₁ = 12° C

T(0) = Ce^{0} + T_a

10 = C + 100

C = -90

now,

12 = -90e^k + 100

e^k = \dfrac{88}{90}

k = ln{\dfrac{88}{90}}

at time equal to t

T(t) = 70

T(t) = -90 (\dfrac{88}{90})^t + 100

70 = -90(\dfrac{88}{90})^t + 100

(\dfrac{88}{90})^t = \dfrac{1}{3}

t\times ln(\dfrac{88}{90}) = ln(\dfrac{1}{3})

t = 48.88 s

hence, after 48.88 s the temperature of the body will be 70°C

b) time taken to reach 98°C

T(t) = -90 (\dfrac{88}{90})^t + 100

98 = -90(\dfrac{88}{90})^t + 100

(\dfrac{88}{90})^t = \dfrac{1}{45}

t\times ln(\dfrac{88}{90}) = ln(\dfrac{1}{45})

t = 390 s

hence, after 390 s the temperature of the body will be 98°C

5 0
2 years ago
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