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2 years ago

5x 10^4 + 2.5 x 10^2 =

Mathematics

2 answers:

cupoosta [38]2 years ago

8 0

Step-by-step explanation:

step1. 5 multiply by 10 power 4 is equal to 50000

step2. 2.5 multiply by 10 power 2 is equal to 250

final step Add up step1 and 2 to have 50250 i.e 50000+250= 50250

professor190 [17]2 years ago

6 0

Answer: 2000000

Step-by-step explanation:

10^4 times 5 times 10^2 =10^6 times 5

5 times 10^6= 5000000

5000000/2.5

=2000000

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Simply the problem.

Marat540 [252]

Answer:

Step-by-step explanation:

$\sqrt{3}\cdot\sqrt{21}= \\\\\sqrt{3\cdot 21}= \\\\\sqrt{63}= \\\\3\sqrt{7}$

Therefore, the correct answer is choice C. Hope this helps!

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3 years ago

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1 and 5/7 x 1/6 simplify

konstantin123 [22]

Answer:

0.286

Step-by-step explanation:

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2 years ago

In 687,332,195, how is the value of the three in the ten thousands place related to the value of the 3 in the hundred thousands

olganol [36]

<span>In this case, the value of the first 3 (the ten-thousands) has a value of 30,000. The 3 next to it, in the hundred-thousands place, has a value of 300,000. To compare the two, the 3 on the right has a value one-tenth as much as that on the left.</span>

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3 years ago

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Find the equivalent expression 9 ( 2a + 1

aivan3 [116]

Answer:

Equivalent expression of $9(2a+1)$ is, $18a+9$

Step-by-step explanation:

The distributive property says that:

$a \cdot (b+c) = a\cdot b + a\cdot c$

Given the expression: $9(2a+1)$

Apply the distributive property:

$9 \cdot (2a) +9 \cdot 1$

Simplify:

$18a+9$

Therefore, the equivalent expression of $9(2a+1)$ is, $18a+9$

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3 years ago

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Find the six trigonometric function values for angle ∅ where its adjacent side is -9 and its hypotenuse is 41. (Theta is located

arsen [322]

Check the picture below.

$\bf \textit{using the pythagorean theorem}
\\\\
c^2=a^2+b^2\implies \sqrt{c^2-a^2}=b
\qquad 
\begin{cases}
c=hypotenuse\\
a=adjacent\\
b=opposite\\
\end{cases}
\\\\\\
\sqrt{41^2-(-9)^2}=b\implies \sqrt{1681-81}=b\\\\\\ \sqrt{1600}=b\implies 40=b\\\\
-------------------------------$

$\bf sin(\theta )=\cfrac{\stackrel{opposite}{40}}{\stackrel{hypotenuse}{41}}\qquad~~ cos(\theta )=\cfrac{\stackrel{adjacent}{-9}}{\stackrel{hypotenuse}{41}}\qquad~~ tan(\theta )=\cfrac{\stackrel{opposite}{40}}{\stackrel{adjacent}{-9}}
\\\\\\
csc(\theta )=\cfrac{\stackrel{hypotenuse}{41}}{\stackrel{opposite}{40}}\qquad ~~sec(\theta )=\cfrac{\stackrel{hypotenuse}{41}}{\stackrel{adjacent}{-9}}\qquad ~~cot(\theta )=\cfrac{\stackrel{adjacent}{-9}}{\stackrel{opposite}{40}}$

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2 years ago