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Flauer [41]
3 years ago
12

4x^3(5x) = what, will mark brainliest

Mathematics
2 answers:
Rzqust [24]3 years ago
6 0

Answer:

25

Step-by-step explanation:

Tomtit [17]3 years ago
4 0
Answer: 25x^4
Hope you have luck with this answer
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Factor completely, x^2+25.<br><br> A).(X+5)(X+5).<br> B).(X+5)(X-5).<br> C).(X-5)(X-5).<br> D.Prime.
Tatiana [17]
The answer you are looking for is B. If you use the FOIL method (first,outer,inner,last) of multiplication, you will find that (x+5) times (x-5) is actually x^2+25. Hope this helps! Comment if you have questions!!! :)
4 0
3 years ago
Read 2 more answers
13. Find the measure of ZABC. Show your work.<br>​
ale4655 [162]

Answer:

47°

Step-by-step explanation:

In the picture attached, the problem is shown.

When two secants intersect at a point, the angle made is one-half of the difference between the greater arc and the minor arc formed. In this case:

∠ABC = 1/2*(arc DE - arc AC)

Replacing with data from the picture:

∠ABC = 1/2*(142° - 48°)

∠ABC = 47°

6 0
3 years ago
Which one of the options is not a factor
ki77a [65]

Answer:

its X+1 D)X-1

Step-by-step explanation:

4 0
3 years ago
How are the power rules for quotients and products similar? How are they different? Refer specifically to the way that the rule
Lady_Fox [76]
Power rules for both quotient and product sums are useful to simplify large exponential form (of the same base)

The difference is in the rule. For quotient sum, the powers are subtracted, while for product sum, the powers are added up.

An example for quotient sum
\frac{7^{13} }{7^{8} } = \frac{7*7*7*7*7*7*7*7*7*7*7*7*7}{7*7*7*7*7*7*7*7*7*7}
Using the principle of simplifying fractions, we can cancel out ten 7s from both numerator and denominator, leaving us with only three 7s on the numerator which gives us 7^{3}. This working out could be simplified by doing 7^{13-10}= 7^{3}

An example for product sum
9^{5}* 9^{6} =(9*9*9*9*9)*(9*9*9*9*9*9)= 9^{11}. There is a total of eleven 9s if we were to work out the product sum the long way. This could be simplified by doing 9^{5+6} = 9^{11}
6 0
4 years ago
If x= 1/2-√3 , prove that x³-2x²-7x+5=3
prohojiy [21]
x=\frac{1}{2}-\sqrt3\\\\x^3-2x^2-7x+5=3\\\\(\frac{1}{2}-\sqrt3)^3-2(\frac{1}{2}-\sqrt3)^2-7(\frac{1}{2}-\sqrt3)+5\\\\=(\frac{1}{2})^3-3\cdot(\frac{1}{2})^2\cdto\sqrt3+3\cdot\frac{1}{2}\cdot(\sqrt3)^2-(\sqrt3)^3-...\\\\...-2[(\frac{1}{2})^2-2\cdot\frac{1}{2}\cdot\sqrt3+(\sqrt3)^2]-\frac{7}{2}+7\sqrt3+5

=\frac{1}{8}-\frac{3\sqrt3}{4}+\frac{9}{2}-3\sqrt3-2(\frac{1}{4}-\sqrt3+3)-\frac{7}{2}+7\sqrt3+5\\\\=\frac{1}{8}+\frac{9}{2}-\frac{1}{2}-6-\frac{7}{2}+5-\frac{3}{4}\sqrt3-3\sqrt3+2\sqrt3+7\sqrt3\\\\=\frac{1}{8}+\frac{1}{2}-1+5\frac{1}{4}\sqrt3=\frac{1}{8}-\frac{1}{2}+5\frac{1}{4}\sqrt3=\frac{1}{8}-\frac{4}{8}+5\frac{1}{4}\sqrt3\\\\=-\frac{3}{8}+5\frac{1}{4}\sqrt3\neq3
6 0
3 years ago
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