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Ksenya-84 [330]
2 years ago
15

In golf, shooting par is scored as a 0. A score of 2 below par on a hole in golf is called an eagle. Jacob’s friend Michelle sco

red an eagle on each of her first five holes.
Explain how you would use addition to find the product of –2 and 5 using the integer tiles and the number line.

Mathematics
2 answers:
pav-90 [236]2 years ago
6 0

Answer:

what's the integer tiles you would add 5 groups of negative 2 tiles or removed two groups of five positive tiles on the number line you would Bounce by 5 to the left

Step-by-step explanation:

hope this helps :)

riadik2000 [5.3K]2 years ago
4 0

Answer:

with integers tiles you would add 5 groups of negative 2 tiles  or remove 2 groups of 5 positive tiles. On the number line, you would bounce by 5 to the left times

You might be interested in
How many 5-digit numbers are there that are divisible by either 45 or 60 but are not divisible by 90?
o-na [289]

Answer:

2500 numbers

Step-by-step explanation:

Problem;

 Solving for the number of 5-digit numbers divisible either by 45 or 60 but not divisible by 90;

 Let us approach this in a step wise manner;

  5 -digits numbers ranges from 10,000 to 99,999

To solve this problem, let us find the number of digits that are divisible by 45;  

  the first number divisible by 45 within this range is 10035

   the last number divisible by 45 in this range is 99990

    increment is 45

The total number in this range is:

           \frac{last number - first number}{increment}  + 1  = \frac{99990 - 10035}{45}  + 1

                                                      =  2000 numbers

To solve this problem, let us find the number of digits that are divisible by 60;  

  the first number divisible by 60 within this range is 10020

   the last number divisible by  in this range is 99960

    increment is 60

\frac{99990 - 10080 }{90}

The total number in this range is:

               \frac{last number - first number}{increment}  + 1  = \frac{99960 - 10020}{60}  + 1  

                                                         =  1500 numbers

To solve this problem, let us find the number of digits that are divisible by 90;  

  the first number divisible by 90 within this range is 10080

   the last number divisible by  in this range is 99990

    increment is 90

The total number in this range is:

         \frac{last number - first number}{increment} + 1  = \frac{99990 - 10080}{90} + 1

                                                  = 1000 numbers

Now solution:

  Number of 5 digits  = number of 45  + number of 60 - number of 90

                                       = 2000 + 1500  - 1000

                                        = 2500 numbers

3 0
2 years ago
Plot the function y(x)=e–0.5x sin(2x) for 100 values of x between 0 and 10. Use a 2- point-wide solid blue line for this functio
Flauer [41]

Answer:

The plot is attached.

Step-by-step explanation:

Plot the function y(x)=e^–0.5x sin(2x) for 100 values of x between 0 and 10. Use a 2- point-wide solid blue line for this function.

The step value for x is (10-0)/100=0.1.

Plot the function y(x)=e–0.5x cos(2x) on the same axes. Use a 3-point-wide dashed red line for this function.

The step value for x is the same as the previous function.

The plot is attached.

3 0
2 years ago
Read 2 more answers
Titus is asked to prove hexagon FEDCBA is congruent to hexagon
Goshia [24]

Answer:

<u><em>(x, y) -------> ( - x , y - 10 )</em></u>

Step-by-step explanation:

Titus is not correct.

There are two transformations will correctly prove FEDCBA ≅ F'E'D'C'B'A'

First transformation is Reflection over y-axis , then translation 10 units down.

The final formula is <em>( - x , y - 10 )</em>

5 0
2 years ago
Jar A contains four liters of a solution that is 45% acid. Jar B contains five liters of a solution that is 48% acid. Jar C cont
castortr0y [4]

Answer:

k= 80%

Step-by-step explanation:

Jar A contains 4*0.45 L acid, and 4 L of a solution  of acid.

Jar B contains 5*0.48 L acid., and 5 L of a solution of acid.

Jar C contains 1*k/100 = k/100 acid, and 1 L of a solution.

50% = 0.5

For jar A.

(2/3)*k/100 L acid  is added to jar A.

Now jar A contains   4*0.45 L + (2/3)*k/100 L acid, and it has (4+2/3)L of a solution.

L solute/L solution = 0.5

[4*0.45 L + (2/3)*k/100 L]/(4+2/3)L = 0.5

[1.8 + (2k/300)]/[(12+2)/3] = 0.5

[1.8 + (2k/300)]/[14/3] = 0.5

[1.8 + (2k/300)]= 0.5*(14/3)

(2k/300) = 0.5*(14/3) - 1.8

2k = (0.5*(14/3) - 1.8)*300

k = (0.5*(14/3) - 1.8)*300/2 =80

k= 80%

We also can find k using jar B.

(1/3)k/100 L acid is added  to jar B.

Now jar B contains 5*0.48 L+ (1/3)k/100 L acid, and it has (5+1/3) L of a solution.

L solute/L solution = 0.5

[5*0.48 L+ (1/3)k/100 L ]/(5+1/3)L= 0.5

[5*0.48 + (1/3)k/100 ]/(5+1/3)= 0.5

This equation also gives k=80%

Check.

We can check at least for jar A.

Jar A has 4L solution and 4*0.45=1.8 L acid.

2/3 L of the solution from jar C was added, and now we have 4 2/3 L of solution.

(2/3)* 80%= (2/3)*0.8 acid was added from jar C.

Now we have [1.8 +(2/3)*0.8] L acid in jar A.

L solute/L solution =  [1.8 +(2/3)*0.8] L /(4 2/3) L = 0.5 or 50%  as it is given that jar A has 50% at the end.

7 0
2 years ago
Smith High School offers a baseball camp that is $75 for 4 days of camp and a basketball camp that is $100 for 5 days of camp. W
castortr0y [4]
A) baseball camp by $1.25 a day

5 0
2 years ago
Read 2 more answers
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