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Ronch [10]
2 years ago
11

Help me out please!!​

Mathematics
2 answers:
deff fn [24]2 years ago
6 0
It is like 2•2•2•2 X 2•2•2
= 16 X 8
= 128
Alexeev081 [22]2 years ago
4 0

Answer:

Form: 2^7, 2^7 = 128

Step-by-step explanation:

You cannot multiply exponent, you must add the exponents.

4 + 3 = 7.

Keep the base the same.

Then find what 2^7 is.

Multiply:

2 x 2 x 2 x 2 x 2 x 2 x 2.

2 x 2 =

4 x 2 =

8 x 2 =

16 x 2 =

32 x 2 =

64 x 2 =

128.

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Write an inequality to represent the phrase: The sum of a number an 6 is no less than 31
Leviafan [203]
The inequality that represents this phrase is n+6 > 31.
5 0
3 years ago
Determine whether the given measures can be the lengths of the sides of a triangle write yes or no
ohaa [14]

NO.,the given measures can not be the lengths of the sides of a triangle

Step-by-step explanation

The sum of the lengths of any two sides of a triangle must be greater than the length of the third side.

so, Find the range for the measure of the third side of a triangle given the measures of two sides.

here given measures are 2,2,6

2+2 = 4 which is less than the third side 6

        = 4 < 6    

This not at all a triangle.

Hence, the given measures can not be the lengths of the sides of a triangle

3 0
3 years ago
What is the value of k?<br> k=
noname [10]

Answer:

no equation given ,pls mention it in the comments

8 0
3 years ago
0.88 kilogram what is the value of the digit in the tents place
Orlov [11]
The value of the tenths place is 8
7 0
3 years ago
2v^2-12 =-12v <br> Would really appreciate it loves❤️
Black_prince [1.1K]

For this case we have the following equation:

2v ^ 2-12 = -12v

Rewriting we have:

2v ^ 2 + 12v-12 = 0

Dividing by 2 to both sides of the equation:

v ^ 2 + 6v-6 = 0

We apply the quadratic formula:

x = \frac {-b \pm \sqrt {b ^ 2-4 (a) (c)}} {2a}

We have to:

a = 1\\b = 6\\c = -6

Substituting:

x = \frac {-6 \pm \sqrt {6 ^ 2-4 (1) (- 6)}} {2 (1)}\\x = \frac {-6 \pm \sqrt {36 + 24}} {2}\\x = \frac {-6 \pm \sqrt {60}} {2}\\x = \frac {-6 \pm \sqrt {4 * 15}} {2}\\x = \frac {-6 \pm2 \sqrt {15}} {2}

Thus, we have two roots:

x_ {1} = - 3+ \sqrt {15}\\x_ {2} = - 3- \sqrt {15}

ANswer:

x_ {1} = - 3+ \sqrt {15}\\x_ {2} = - 3- \sqrt {15}

7 0
3 years ago
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