1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
KonstantinChe [14]
3 years ago
7

5/2b+(-25/2)=-10 what is the value of b?

Mathematics
1 answer:
gulaghasi [49]3 years ago
8 0
The value of b in the equation 5/2b+(-25/2) is 1. 

You might be interested in
Why resultant magnitude by using Pythagorean theorem is different than using ( x,y )components addition vectors ???
Andreyy89

9514 1404 393

Explanation:

Your different answers came about as a result of an error you made in the Pythagorean theorem calculation.

  √(100 +100) = √200 = 14.142 . . . . same as vector calculation

4 0
3 years ago
Add 2/5 plus 27/9 -0
Leviafan [203]
2/5 plus 27/9 minus 0 equals 0
4 0
3 years ago
Hi please help :)))))))
lesya [120]

Answer:

there is no difference

Step-by-step explanation:

3 0
3 years ago
Read 2 more answers
Round 5.9496 to the nearest thousandth
Evgesh-ka [11]

Answer:

5.95

Step-by-step explanation:

that is because 96 is rounded too 100

so its 5.9500 = 5.95

4 0
3 years ago
Read 2 more answers
How to solve logarithmic equations as such
Serga [27]

\bf \textit{exponential form of a logarithm} \\\\ \log_a b=y \implies a^y= b\qquad\qquad a^y= b\implies \log_a b=y \\\\\\ \begin{array}{llll} \textit{Logarithm of exponentials} \\\\ \log_a\left( x^b \right)\implies b\cdot \log_a(x) \end{array} ~\hspace{7em} \begin{array}{llll} \textit{Logarithm Cancellation Rules} \\\\ log_a a^x = x\qquad \qquad \stackrel{\textit{we'll use this one}}{a^{log_a x}=x} \end{array} \\\\[-0.35em] \rule{34em}{0.25pt}

\bf \log_2(x-1)=\log_8(x^3-2x^2-2x+5) \\\\\\ \log_2(x-1)=\log_{2^3}(x^3-2x^2-2x+5) \\\\\\ \log_{2^3}(x^3-2x^2-2x+5)=\log_2(x-1) \\\\\\ \stackrel{\textit{writing this in exponential notation}}{(2^3)^{\log_2(x-1)}=x^3-2x^2-2x+5}\implies (2)^{3\log_2(x-1)}=x^3-2x^2-2x+5

\bf (2)^{\log_2[(x-1)^3]}=x^3-2x^2-2x+5\implies \stackrel{\textit{using the cancellation rule}}{(x-1)^3=x^3-2x^2-2x+5} \\\\\\ \stackrel{\textit{expanding the left-side}}{x^3-3x^2+3x-1}=x^3-2x^2-2x+5\implies 0=x^2-5x+6 \\\\\\ 0=(x-3)(x-2)\implies x= \begin{cases} 3\\ 2 \end{cases}

5 0
3 years ago
Other questions:
  • Given f(x) = x − 7 and g(x) = x2 .<br><br> Find g(f(4)).<br><br> g(f(4))
    6·2 answers
  • Find the prime factorization of the following number. 110
    12·1 answer
  • Simplify the expression. HELP, I WANT TO MAKE SURE I DO IT RIGHT
    8·1 answer
  • Mason is taking a multiple choice test with a total of 100 points available. Each question is worth exactly 5 points . What woul
    12·1 answer
  • Order the numbers from least to greatest: -3, 5, -1, 0, 7
    15·2 answers
  • Meg is a veterinarian. In a given week, 50% of
    5·1 answer
  • In the figure below, ADBE is the result of a dilation of AABC with point B as the
    12·1 answer
  • How
    12·1 answer
  • I need help on this please​
    12·2 answers
  • Consider the graph of the following quadratic equation
    14·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!