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Dmitrij [34]
3 years ago
13

Where is the 6 located in the number 4.6875

Mathematics
2 answers:
Zepler [3.9K]3 years ago
7 0

Answer:

In the tenths spot

Step-by-step explanation:

Nadya [2.5K]3 years ago
3 0

Answer:

In the tenths place

Step-by-step explanation:

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Yosef typed a 36 word paragraph in 2/3 minute. What is his typing speed, in words per minute?
Yakvenalex [24]

Answer:

the answer is:

Yousef typing speed, in words per minute is

equal to

54 words

minute

8 0
2 years ago
Which exponential function has an initial value of 2?<br><br> f(x) = 2(3x)<br><br> f(x) = 3(2x)
Sidana [21]
<span>f(x) = 2(3x)
Exponential functions represent the initial value outside of the parentheses so if 2 is the initial value it has to be on the outside of the parentheses. 
Exponential growth formula.
</span>y=a(1+r)^{x}
<span>a represents the initial value.</span>
7 0
3 years ago
Read 2 more answers
Solve 7−3n ≤ n+3<br> ................
vredina [299]

Answer:

1 ≤ n

Step-by-step explanation:

7−3n ≤ n+3

  +3n   +3n

7 ≤ 4n+3

-3       -3

4 ≤ 4n

divide by 4 on each side

1 ≤ n

4 0
2 years ago
How many times does 7 go into 48
givi [52]
6 times because 6•7=42 so it must be 6 because 7•7=49
Good luck!!
4 0
3 years ago
Which equation has the solutions x=1+or-square root of 5?
stiv31 [10]

We will proceed to solve each case to determine the solution of the problem.

<u>case a)</u> x^{2}+2x+4=0

Group terms that contain the same variable, and move the constant to the opposite side of the equation

x^{2}+2x=-4

Complete the square. Remember to balance the equation by adding the same constants to each side.

x^{2}+2x+1=-4+1

x^{2}+2x+1=-3

Rewrite as perfect squares

(x+1)^{2}=-3

(x+1)=(+/-)\sqrt{-3}\\(x+1)=(+/-)\sqrt{3}i\\x=-1(+/-)\sqrt{3}i

therefore

case a) is not the solution of the problem

<u>case b)</u> x^{2}-2x+4=0

Group terms that contain the same variable, and move the constant to the opposite side of the equation

x^{2}-2x=-4

Complete the square. Remember to balance the equation by adding the same constants to each side.

x^{2}-2x+1=-4+1

x^{2}-2x+1=-3

Rewrite as perfect squares

(x-1)^{2}=-3

(x-1)=(+/-)\sqrt{-3}\\(x-1)=(+/-)\sqrt{3}i\\x=1(+/-)\sqrt{3}i

therefore

case b) is not the solution of the problem

<u>case c)</u> x^{2}+2x-4=0

Group terms that contain the same variable, and move the constant to the opposite side of the equation

x^{2}+2x=4

Complete the square. Remember to balance the equation by adding the same constants to each side.

x^{2}+2x+1=4+1

x^{2}+2x+1=5

Rewrite as perfect squares

(x+1)^{2}=5

(x+1)=(+/-)\sqrt{5}\\x=-1(+/-)\sqrt{5}

therefore

case c) is not the solution of the problem

<u>case d)</u> x^{2}-2x-4=0

Group terms that contain the same variable, and move the constant to the opposite side of the equation

x^{2}-2x=4

Complete the square. Remember to balance the equation by adding the same constants to each side.

x^{2}-2x+1=4+1

x^{2}-2x+1=5

Rewrite as perfect squares

(x-1)^{2}=5

(x-1)=(+/-)\sqrt{5}\\x=1(+/-)\sqrt{5}

therefore

case d) is the solution of the problem

therefore

<u>the answer is</u>

x^{2}-2x-4=0

5 0
2 years ago
Read 2 more answers
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