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2 years ago

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.

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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:

Step-by-step explanation:

x=5*0.8=4

5 0

3 years ago

Read 2 more answers

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