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Wittaler [7]
3 years ago
7

Which of these expressions is equivalent to log(25 • 18)?

Mathematics
1 answer:
riadik2000 [5.3K]3 years ago
5 0

Answer: B. log(25) + log(18)

Step-by-step explanation:

Use the rule for the sum of logs: the log of a product is the sum of the logs, which means log(x) + log(y) = log(xy) if they have the same base.

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3. What is the area of the triangle below? Round your
Jobisdone [24]

Answer is A because when you work it out the area is 110.500 and rounding it makes it -

8 0
3 years ago
At short company has initial costs of $800 per month and it costs them $1.50 per shirt. Let y represent the cost to run the comp
adelina 88 [10]

Answer:

A. y = $800 + $1.50x

B. The cost to run the company each month = y

C. The initial cost ($800) plus the cost of each t-shirt ($1.50) and the number of t-shirts (x)

Step-by-step explanation:

I hope this helps :)

4 0
2 years ago
Do this pls 35 points
love history [14]

Answer:

x = 68

Step-by-step explanation:

If you look at the triangle properly, it is a isosceles triangle where two legs are equal to each other.

The answer for x will be:

56 + 56 + x =180

112 + x = 180

x = 68

7 0
3 years ago
P(x)= 3x^3-5x^2-14x-4
nexus9112 [7]
   
\displaystyle\\
P(x)=3x^3-5x^2-14x-4\\\\
D_{-4}=\{-4;~-2;~\underline{\bf -1};~1;~2;~4\}\\\\
\text{We observe that } \frac{-1}{3} \text{ is a solution of the equation:}\\
3x^3-5x^2-14x-4=0\\\\


\displaystyle\\
\text{Verification}\\\\
3x^3-5x^2-14x-4=\\\\
=3\times\Big(-\frac{1}{3}\Big)^3-5\times\Big(-\frac{1}{3}\Big)^2-14\times\Big(-\frac{1}{3}\Big)-4=\\\\
=-\frac{1}{9}-\frac{5}{9}+\frac{14}{3}-4=\\\\
=-\frac{6}{9}+\frac{14}{3}-4=\\\\
=-\frac{6}{9}+\frac{42}{9}- \frac{4\times 9}{9}=\\\\
 =-\frac{6}{9}+\frac{42}{9}- \frac{36}{9}= \frac{42-6-36}{9}=\frac{42-42}{9}=\frac{0}{9}=0\\\\
\Longrightarrow~~~P(x)~\vdots~\Big(x+ \frac{1}{3}\Big)\\\\
\Longrightarrow~~~P(x)~\vdots~(3x+1)


\displaystyle\\
3x^3-5x^2-14x-4=0\\
~~~~~-5x^2 = x^2 - 6x^2\\
~~~~~-14x =-2x-12x \\
3x^3+x^2 - 6x^2-2x-12x-4=0\\
x^2(3x+1)-2x(3x+1) -4(3x+1)=0\\
(3x+1)(x^2-2x -4)=0\\\\
\text{Solve: } x^2-2x -4=0\\\\
x_{12}= \frac{-b\pm  \sqrt{b^2-4ac}}{2a}=\\\\=\frac{2\pm  \sqrt{4+16}}{2}=\frac{2\pm  \sqrt{20}}{2}=\frac{2\pm  2\sqrt{5}}{2}=1\pm\sqrt{5}\\\\
x_1 =1+\sqrt{5}\\
x_2 =1-\sqrt{5}\\
\Longrightarrow P(x)= 3x^3-5x^2-14x-4 =\boxed{(3x+1)(x-1-\sqrt{5})(x-1+\sqrt{5})}



7 0
3 years ago
Pls Help its only one question its about Triangles Thank you!!
Alex_Xolod [135]
Does it tell you what (x) stands for
4 0
3 years ago
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