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VMariaS [17]
3 years ago
7

Can someone help me with this math question 2^x=2^3

Mathematics
1 answer:
Soloha48 [4]3 years ago
7 0
2ˣ = 2³

Since 2ˣ is EQUAL to  2³  and since 2 is equal to 2, then the solution is a must that the 2 exponents are equal.
Then x = 3
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I’m honestly done with math
rewona [7]

Answer:

15.

Step-by-step explanation:

8x - 10 = 110    (vertical angles are equal in measure).

8x - 10+ 10 = 110 + 10

8x = 120

x = 120 /8

x = 15.

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What is the equation of the graphed line written in standard form?
Dmitry_Shevchenko [17]

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A

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enyata [817]

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3 0
3 years ago
Write a fraction and a percent for the shaded portion of each one.
Cerrena [4.2K]
I dunno if im correct, but for A. I’d say 1/4 for B. I’d say 1 1/2 for C. 2/3
4 0
3 years ago
Read 2 more answers
How many complex zeros does the polynomial function have?<br> f(x)=−3x^6−2x^4+5x+6
REY [17]

one way would be to factor

I can't factor it so we will have to use Descartes' Rule of Signs which is helpful for finding how many real roots you have


it goes like this:

for a polynomial with real coefients, consider f(x)=-3x^6-2x^4+5x+6.

after arranging the terms in decending order in terms of degree, count how many times the signs of the coeffients change direction and minus 2 from that number until you get to 1 or 0. that will be the number of even roots the function can have

We have (-, -, +, +). the signs changed 1 times, so it has 1 real positive root


to get the negative roots, we evaluate f(-x) and see how many times the root changes

f(-x)=-3x^6-2x^4-5x+6

signs are (-, -, -, +). there was 1 change in sign

so the function has 1 real negative root



a total of 2 real roots


a function of degree n can have at most, n roots


our function is degree 6 so it has 6 roots

if 2 are real, then the others must be complex

6-2=4 so there are 4 complex roots


you can also show that there are only 2 real roots by using a graphing utility to see that there are only 2 real roots

7 0
3 years ago
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