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Ivenika [448]
3 years ago
8

Describing each step.

Mathematics
1 answer:
Firlakuza [10]3 years ago
8 0

Answer:

1-2

Step-by-step explanation:

It's 1-2 beacause from 1-2 the oriantation did not change, the size did not change, the shape did not change, and it was not a mirror image

You might be interested in
Earl is ordering supplies. Yellow paper costs $4.00 per ream while white paper costs $7.50 per ream. He would like to order 100
Eddi Din [679]

Answer:

Earl should order 76 yellow paper reams and 24 white paper reams.

Step-by-step explanation:

This problem can be computed as system of equations.

I will say that x is the number of yellow paper reams and that y is the number of white paper reams.

Earl will order 100 reams in total, so

x + y = 100

He has a budget of $484, so this is what he is going to spend. Each yellow paper ream(x) costs $4.00 and each white paper ream costs $7.50, so we know that

4x + 7.5y = 484

To know how many reams of each color should he order, we need to solve the following system of equations

1) x + y = 100

2) 4x + 7.5y = 484

I am going to write x as a function of y in equation 1), and then replace it in equation 2).

x = 100-y

Now replacing in equation 2)

4(100-y) + 7.5y = 484

400 - 4y + 7.5y = 484

3.5y = 84

y = 24.

Returning to equation 1)

x = 100-y = 100-24 = 76

So, Earl should order 76 yellow paper reams and 24 white paper reams.

5 0
3 years ago
Find the area.<br> 5 cm<br> 5 cm<br> 16 cm---<br> 7 cm<br> 14 cm
zalisa [80]

Answer:

if it is area of triangle it would be 49

Step-by-step explanation:

3 0
3 years ago
A skier has decided that on each trip down a slope, she will do 2 more jumps than before. On her first trip she did 6 jumps. Der
elixir [45]

The <em>correct answers</em> are:

\Sigma_{n=4}^{12} 6+2(n-1) \\ \\=180

Explanation:

Since she is adding two more jumps every time she goes down hill, this is an arithmetic sequence. The general form of an arithmetic sequence is

a_n=a_1+d(n-1), \\ \text{where } a_n \text{represents the nth term, } a_1 \text{represents the first term, and n represents the term number}

Since we want the number of jumps on her 4th through 12th trips, we will set n in the summation from 4 to 12. n=4 goes at the bottom of Σ and 12 goes at the top, to represent the values we are interested in.

Beside this, we write our general form. The first term is 6 and d, the common difference, is 2. This gives us 6+2(n-1) beside the summation:

\Sigma_{n=4}^{12} 6+2(n-1)

To evaluate this, we substitute the values 4, 5, 6, 7, 8, 9, 10, 11 and 12 in for n, adding all of the values together:

6+2(4-1)+6+2(5-1)+6+2(6-1)+6+2(7-1)+6+2(8-1)+6+2(9-1)+6+2(10-1)+6+2(11-1)+6+2(12-1)

=6+6+6+8+6+10+6+12+6+14+6+16+6+18+6+20+6+22

=180

8 0
3 years ago
Read 2 more answers
Ellis is painting wooden fenceposts before putting them in his yard. They are each 7 feet tall and have a diameter of 1 foot. Th
Zina [86]

what class is this for send to me in message

7 0
3 years ago
Two students are registered for the same class and attend independently of each other, student a 90% of the time and student b 7
IRISSAK [1]

The probability of student A is 15/44

Given student A is 70% of time and student B is 60% of time

We need to find the probability of student A

Using formula of conditional probability is P(X/Y) = P(X and Y) / P(Y)

Here X = A and Y = A or B

Therefore,

P(A/A or B) = P(A and (A OR B) / P (A or B)

Now, There is none complement for at least 1

We know that Student A attends 70 % of time

So , his absence is 30% of the time .

Hence the probability of absence is 0.3

Now Considering B in the similar way

We get,

Probability of the absence is 0.4

They are both absent (0.3)(0.4)= 0.12

Here we can say that 12 % of time both are absent

So one or another present on that time is 88%

The probability of present of the time is 0.88

Now calculating the probability,

P(A/A or B) = P(A and (A OR B) / P (A or B)

= 0.3/0.88

= 30/88

= 15/44

Hence the Probability Of A is 15/44

Learn more about Probability here brainly.com/question/24756209

#SPJ10

4 0
2 years ago
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