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zhenek [66]
3 years ago
13

Sarah had a sheet of paper that was 3 inches long and 7 inches wide. What is the perimeter of the paper

Mathematics
2 answers:
notsponge [240]3 years ago
4 0

Answer:

20 square inches

Step-by-step explanation:

Illusion [34]3 years ago
3 0

Answer: 20

Step-by-step explanation:

(3+7)*2 = 20

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Which ordered pair makes both inequalities true? y < –x + 1 y > x On a coordinate plane, 2 straight lines are shown. The f
goldenfox [79]

Answer:

Option B.

Step-by-step explanation:

The given inequalities are

y

y>x

We need to find the ordered pair which makes both inequalities true.

Check the above inequalities for each given ordered pair.

For (-3,5),

(5) (False)

For (-2,2),

(2) (True)

y>x\Rightarrow 2>-2 (True)

So, both inequalities are true for (-2,2). Option B is correct.

For (-1,-3),

y>x\Rightarrow -3>-1 (False)

For (0,-1),

y>x\Rightarrow -1>0 (False)

Both inequalities are not true for (-3,5), (-1,-3) and (0,-1).

Therefore, the correct option is B.

3 0
2 years ago
Read 2 more answers
Pls Help! John wants to prove ABC = to DEF. He Knoes AB = DE and AC = DF. With Screenshot
matrenka [14]

Answer:

\angle A \cong \angle D

Step-by-step explanation:

The Side-Angle-Side method cana only be used when information given shows that an included angle which is between two sides of a ∆, as well as the two sides of the ∆ are congruent to the included side and two sides of the other ∆.

Thus, since John already knows that \overline{AB} \cong \overline{DE} and \overline{AC} \cong \overline{DF}, therefore, an additional information showing that the angle between \overline{AB} and \overline{DE} in ∆ABC is congruent to the angle between \overline{AC} and \overline{DF} in ∆DEF.

For John to prove that ∆ABC is congruent to ∆DEF using the Side-Angle-Side method, the additional information needed would be \angle A \cong \angle D.

See attachment for the diagram that has been drawn with the necessary information needed for John to prove that ∆ABC is congruent to ∆DEF.

7 0
2 years ago
For a saturday matinee, adult tickets cost $5.50 and kids under 12 pay only $3.00. if 90 tickets are sold for a total of $395, h
juin [17]
Kids - 40 ticketsadults - 50 tickets
8 0
3 years ago
Formulate the situation as a system of two linear equations in two variables. Be sure to state clearly the meaning of your x- an
slega [8]

Answer:

1) There were 33 $4,000 investors and 27 $8,000 investors.

2) The solution in x = 4, y = 9.

3) There were 24 nickels and 56 dimes.

Step-by-step explanation:

1) A lawyer has found 60 investors for a limited partnership to purchase an inner-city apartment building, with each contributing either $4,000 or $8,000. If the partnership raised $348,000, then how many investors contributed $4,000 and how many contributed $8,000?

I am going to say that:

x is the number of investors that contributed 4,000.

y is the number of investors that contributed 8,000.

Building the system:

There are 60 investors. So:

x + y = 60

In all, the partnership raised $348,000. So:

4000x + 8000y = 348000

I am going to simplify by 4000. So:

x + 2y = 87

Solving the system:

The elimination method is a method in which we can transform the system such that one variable can be canceled by addition. So:

1)x + y = 60

2)x + 2y = 87

I am going to multiply 1) by -1. So we have

1)-x - y = -60

2)x + 2y = 87

By addition, the x are going to cancel each other

-x + x - y + 2y = -60 + 87

y = 27

For x:

x + y = 60

x = 60-y = 60-27 = 33

There were 33 $4,000 investors and 27 $8,000 investors.

2) Solve the system by row-reducing the corresponding augmented matrix.

2x + y = 17

x + y = 13

This system has the following augmented matrix:

\left[\begin{array}{ccc}2&1&17\\1&1&13\end{array}\right]

To help the row reducing, i am going to swap the first with the second line:

L1  L2

So we have:

\left[\begin{array}{ccc}1&1&13\\2&1&17\end{array}\right]

Now, reducing the first column.

L2 = L2 - 2L1

So we have:

\left[\begin{array}{ccc}1&1&13\\0&-1&-9\end{array}\right]

Now we do:

L2 = -L2

And the matrix is:

\left[\begin{array}{ccc}1&1&13\\0&1&9\end{array}\right]

Now to reduce the second column, we do:

L1 = L1 - L2

\left[\begin{array}{ccc}1&0&4\\0&1&9\end{array}\right].

So the solution is:

x = 4, y = 9.

3) A jar contains 80 nickels and dimes worth $6.80. How many of each kind of coin are in the jar?

I am going to say that x is the number of nickels and y is the number of dimes.

Each nickel is worth 5 cents and each dime is worth 10 cents.

Building the system:

There are 80 coins in all:

x + y = 80

They are worth $6.80. So:

0.05x + 0.10y = 6.80

Solving the system:

1)x + y = 80

2)0.05x + 0.10y = 6.80

I am going to divide 1) by -10, so we can cancel y. So:

1)-0.10x - 0.10y = -8

2)0.05x + 0.10y = 6.80

Adding:

-0.10x + 0.05x - 0.10y + 0.10y = -8 + 6.80

-0.05x = -1.2 *(-100)

5x = 120

x = \frac{120}{5}

x = 24

Also

x + y = 80

y = 80-x = 80-24 = 56

There were 24 nickels and 56 dimes.

8 0
3 years ago
URGENT!!! PLEASE HELP!! two cars leave at the same time from two cities that are 374 miles apart and travel towards each other.
Brilliant_brown [7]

Answer:

3 hours and 40 minutes I think

Step-by-step explanation:

8 0
3 years ago
Read 2 more answers
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