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pantera1 [17]
3 years ago
7

Alazari rode her bike to school

Mathematics
2 answers:
nordsb [41]3 years ago
5 0
This answer applies to the question only if the 3 locations are collinear, if they are not I don’t believe that there is a possible solution for the problem

irga5000 [103]3 years ago
3 0
260meters because 70+70=140 and 60+60=120 which is 140+120=260
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Consider the triangles shown. which statement is true ​
Schach [20]
Answer: A

Step by step explanation:
6 0
4 years ago
Anyone who could do these 8 listed questions I will give a generous amount of points. Please make sure to list all the angles th
k0ka [10]

Answer:

1) 0, 180

2) 90

3) 3pi/2

4) pi/2, -3pi/2

5) 90, 270

6) 0

7) pi

8) -2pi, 0, 2pi

Step-by-step explanation:

1) sinx = 0

x = 0, 180, 360

2) sinx = 1

x = 90

3) sinx = -1

x = 270 or 3pi/2

4) sinx = 1

x = pi/2, pi/2 - 2pi = -3pi/2

5) cosx = 0

x = 90, 360

6) cosx = 1

x = 0, 360

7) cosx = -1

x = pi

8) cosx = 1

-2pi, 0 , 2pi

5 0
3 years ago
State the operations used to solve the following equation 2x+8=9
jeka94

Answer:

Subtraction and Division

Step-by-step explanation:

2x+8=9

2x+8-8=9-8

2x=1

x=1/2

3 0
3 years ago
I need help.. i really want to go sleep.. thank you so much...
Effectus [21]

Answer:

1) True 2) False

Step-by-step explanation:

1) Given  \sum\limits_{k=0}^8\frac{1}{k+3}=\sum\limits_{i=3}^{11}\frac{1}{i}

To verify that the above equality is true or false:

Now find \sum\limits_{k=0}^8\frac{1}{k+3}

Expanding the summation we get

\sum\limits_{k=0}^8\frac{1}{k+3}=\frac{1}{0+3}+\frac{1}{1+3}+\frac{1}{2+3}+\frac{1}{3+3}+\frac{1}{4+3}+\frac{1}{5+3}+\frac{1}{6+3}+\frac{1}{7+3}+\frac{1}{8+3} \sum\limits_{k=0}^8\frac{1}{k+3}=\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+\frac{1}{7}+\frac{1}{8}+\frac{1}{9}+\frac{1}{10}+\frac{1}{11}

Now find \sum\limits_{i=3}^{11}\frac{1}{i}

Expanding the summation we get

\sum\limits_{i=3}^{11}\frac{1}{i}=\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+\frac{1}{7}+\frac{1}{8}+\frac{1}{9}+\frac{1}{10}+\frac{1}{11}

 Comparing the two series  we get,

\sum\limits_{k=0}^8\frac{1}{k+3}=\sum\limits_{i=3}^{11}\frac{1}{i} so the given equality is true.

2) Given \sum\limits_{k=0}^4\frac{3k+3}{k+6}=\sum\limits_{i=1}^3\frac{3i}{i+5}

Verify the above equality is true or false

Now find \sum\limits_{k=0}^4\frac{3k+3}{k+6}

Expanding the summation we get

\sum\limits_{k=0}^4\frac{3k+3}{k+6}=\frac{3(0)+3}{0+6}+\frac{3(1)+3}{1+6}+\frac{3(2)+3}{2+6}+\frac{3(3)+4}{3+6}+\frac{3(4)+3}{4+6}

\sum\limits_{k=0}^4\frac{3k+3}{k+6}=\frac{3}{6}+\frac{6}{7}+\frac{9}{8}+\frac{12}{8}+\frac{15}{10}

now find \sum\limits_{i=1}^3\frac{3i}{i+5}

Expanding the summation we get

\sum\limits_{i=1}^3\frac{3i}{i+5}=\frac{3(0)}{0+5}+\frac{3(1)}{1+5}+\frac{3(2)}{2+5}+\frac{3(3)}{3+5}

\sum\limits_{i=1}^3\frac{3i}{i+5}=\frac{3}{6}+\frac{6}{7}+\frac{9}{8}

Comparing the series we get that the given equality is false.

ie, \sum\limits_{k=0}^4\frac{3k+3}{k+6}\neq\sum\limits_{i=1}^3\frac{3i}{i+5}

6 0
4 years ago
Solve for x. <br><br> x^3=8/125<br><br> Enter your answer in the box as a fraction in simplest form.
stepan [7]
<span>x^3=8/125
x^3 = 2^3  /  5^3
x = 2/5

hope it helps</span>
6 0
3 years ago
Read 2 more answers
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