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Ostrovityanka [42]
2 years ago
12

9-x-x=-1 Solve for x

Mathematics
2 answers:
aksik [14]2 years ago
5 0

Answer:

x = 5

Step-by-step explanation:

9 - -1 = ?

9 + 1 = 10

10 ÷ 2 = 5

astra-53 [7]2 years ago
4 0

x = 2

Hope this helps!

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Guys hi am i correct ??
Sloan [31]
Yes, you are correct.

Hope my answer helped u :)
4 0
2 years ago
What is the product of 5x10 to the power of 2
ratelena [41]

50^2 = 50 to the power of 2 = 2.5 × 10^3

Explanation:

(50) to the 2nd power or simply '50 to the 2nd' is obtained by multiplying 2 times the base 50 by itself. So,

50^2 = 50 × 50 = 2.5 × 10^3

Note: We say that 50 is the base, 2 is the exponent, and the whole thing or the result is a power of 50.

5 0
3 years ago
Read 2 more answers
Figure A is a scale image of figure B. Figure A maps to figure B with a scale factor of 0.80. What is the value of x?
Ivan

Answer:

4

Step-by-step explanation:

x=5*0.8=4

5 0
3 years ago
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Corey gets paid $6 per hour to work at a snack stand. This weekend , he worked on both Saturday and Sunday. he made total of $48
Dahasolnce [82]

Answer:

5 hours

Step-by-step explanation:

1. Find out how many dollars he earned on Saturday

3 x 6 = 18

He earned 18 dollars on Saturday.

2. Subtract 18 from 48 since that is from Saturday, not Sunday

48 - 18 = 30

3. Divide 30 by 6 to see how many hours he worked on Sunday

30/6 = 5

6 0
3 years ago
Help ASAP show work please thanksss!!!!
Llana [10]

Answer:

\displaystyle log_\frac{1}{2}(64)=-6

Step-by-step explanation:

<u>Properties of Logarithms</u>

We'll recall below the basic properties of logarithms:

log_b(1) = 0

Logarithm of the base:

log_b(b) = 1

Product rule:

log_b(xy) = log_b(x) + log_b(y)

Division rule:

\displaystyle log_b(\frac{x}{y}) = log_b(x) - log_b(y)

Power rule:

log_b(x^n) = n\cdot log_b(x)

Change of base:

\displaystyle log_b(x) = \frac{ log_a(x)}{log_a(b)}

Simplifying logarithms often requires the application of one or more of the above properties.

Simplify

\displaystyle log_\frac{1}{2}(64)

Factoring 64=2^6.

\displaystyle log_\frac{1}{2}(64)=\displaystyle log_\frac{1}{2}(2^6)

Applying the power rule:

\displaystyle log_\frac{1}{2}(64)=6\cdot log_\frac{1}{2}(2)

Since

\displaystyle 2=(1/2)^{-1}

\displaystyle log_\frac{1}{2}(64)=6\cdot log_\frac{1}{2}((1/2)^{-1})

Applying the power rule:

\displaystyle log_\frac{1}{2}(64)=-6\cdot log_\frac{1}{2}(\frac{1}{2})

Applying the logarithm of the base:

\mathbf{\displaystyle log_\frac{1}{2}(64)=-6}

5 0
2 years ago
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