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
tester [92]
3 years ago
13

6t + 4 ≤ 22 oh god help what is t I'm doing finals help help help

Mathematics
1 answer:
loris [4]3 years ago
6 0
T is less then or equal to 3
You get this by solving it like normal but keeping the less than or equal to
You might be interested in
WHO IS REALLY GOOD AT MATH AND CAN DO THIS FOR ME ILL MARK BRAINLIEST
Rashid [163]

Answer:

Step-by-step explanation:

4 0
3 years ago
Read 2 more answers
PLZZZZZZZZZZZZZZZZZ HELP!
Svet_ta [14]

answer: -5x+12=20

                   -12    -12

            -5x/-5= 8/-5

             x= -1.6

       Check: rewrite the orignal problem

          plug your variable

       

             

5 0
3 years ago
What is the y-intercept of the line with the equation 4x + 2y = 12?
makvit [3.9K]
2y = - 4x + 12
y = - 2x + 6

y intercept = 6

the answer is D
4 0
3 years ago
Royce Bernstein’s group medical insurance coverage costs $5,480 a year. His employer pays 65% of the cost. What is the amount of
iVinArrow [24]
84? I think u just divide if I’m not Erie try and divide because I’m doing it the way of average speed
5 0
3 years ago
Find n for which the nth iteration by the Bisection Method guarantees to approximate the root of f(x) = 2x^2 − 3x − 2 on [−2, 1]
Lady_Fox [76]

Answer:

n = 29 iterations would be enough to obtain a root of f(x)=2x^2-3x-2 that is at most 10^{-8} away from the correct solution.

Step-by-step explanation:

You can use this formula which relates the number of iterations, n, required by the bisection method to converge to within an absolute error tolerance of ε starting from the initial interval (a, b).

n\geq \frac{log(\frac{b-a}{\epsilon} )}{log(2)}

We know

a = -2, b = 1 and ε = 10^{-8} so

n\geq \frac{log(\frac{1+2}{10^{-8}} )}{log(2)}\\n \geq 29

Thus, n = 29 iterations would be enough to obtain a root of f(x)=2x^2-3x-2 that is at most 10^{-8} away from the correct solution.

<u>You can prove this result by doing the computation as follows:</u>

From the information given we know:

  • f(x)=2x^2-3x-2
  • \epsilon = 10^{-8}

This is the algorithm for the Bisection method:

  1. Find two numbers <em>a</em> and <em>b</em> at which <em>f</em> has different signs.
  2. Define c=\frac{a+b}{2}
  3. If b-c\leq \epsilon then accept c as the root and stop
  4. If f(a)f(c)\leq 0 then set <em>c </em>as the new<em> b</em>. Otherwise, set <em>c </em>as the new <em>a</em>. Return to step 1.

We know that f(-2)=2(-2)^2-3(-2)-2=12 and f(1)=2(1)^2-3(1)-2=-3 so we take a=-2 and b=1 then c=\frac{-2+1}{2} =-0.5

Because 1-(-0.5)\geq 10^{-8} we set c=-0.5 as the new <em>b.</em>

The bisection algorithm is detailed in the following table.

After the 29 steps we have that 6\cdot 10^{-9}\leq 10^{-8} hence the required root approximation is c = -0.50

8 0
3 years ago
Other questions:
  • Please help me please !!
    13·1 answer
  • Mallory bought 6 roses for her mother.two sixth of the roses are red 4/6 are yellow.Did Mallory buy fewer red roses or yellow ro
    7·2 answers
  • Please help me with the correct answer !!!
    14·1 answer
  • 36=5/9k a.4/5 b. 20 c.64 4/5 d. 180
    12·1 answer
  • What is 2 with an exponent of negative 2 equal to?
    9·2 answers
  • Figure B is the image of Figure A when reflected across line ℓ. Are Figure A and Figure B congruent
    6·1 answer
  • Simplify each expression by combining like terms. <br>7a - 6 + 8a ​
    13·1 answer
  • Someone answer please and thank you! No photos please those are blocked for me!
    15·1 answer
  • Given a circle with a diameter of 4, what is the circumstances
    12·2 answers
  • PLEASE ANSWER ASAP!!!! (and actually answer the question, no begging for "brainliest")
    5·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!