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
julia-pushkina [17]
3 years ago
13

What is %15 of $35.15 What is %15 of $27.00

Mathematics
2 answers:
expeople1 [14]3 years ago
8 0
Just multiply each times 0.15 :)

$35.15 = $5.2725
&27.00 = $4.05
goldfiish [28.3K]3 years ago
6 0

Answer:

5.27 , 4.05

Step-by-step explanation:

35.15 times .15= 5.27

27.00 times .15= 4.05

To find the answer to a percent, take the percentage and place a decimal in front of the number than multiply it my the total amount. like so.

You might be interested in
9. Calcule el valor de la fuerza que experimenta un ascensor cuando levanta 150 kg en los siguientes casos: a) Cuando asciende c
GenaCL600 [577]

Answer:

a) La fuerza neta que experimenta un ascensor cuando asciende con una aceleración de 4 metros por segundo al cuadrado es 600 newtons hacia arriba.

b) La fuerza neta que experimenta un ascensor cuando desciende con una aceleración de 6 metros por segundo al cuadrado es 900 newtons hacia abajo.

c) Por la fuerza de tensión sobre el cable del ascensor.

Step-by-step explanation:

a) La fuerza neta experimentada por el ascensor (F), en newtons, es:

F = m\cdot a (1)

Donde:

m - Masa, en kilogramos.

a - Aceleración, en metros por segundo cuadrado.

Si sabemos que m = 150\,kg y a = 4\,\frac{m}{s^{2}}, entonces la fuerza neta del ascensor es:

F = m\cdot a

F = 600\,N

La fuerza neta que experimenta un ascensor cuando asciende con una aceleración de 4 metros por segundo al cuadrado es 600 newtons hacia arriba.

b) Si sabemos que m = 150\,kg y a = -6\,\frac{m}{s^{2}}, entonces la fuerza neta del ascensor es:

F = m\cdot a

F = -900\,N

La fuerza neta que experimenta un ascensor cuando desciende con una aceleración de 6 metros por segundo al cuadrado es 900 newtons hacia abajo.

c) De manera simplificada, el ascensor experimenta dos fuerzas que definen la fuerza y aceleración netas: (i) La fuerza de tensión sobre el cable que traslada el ascensor y el peso total del ascensor, opuesta a la anterior y en función de la aceleración gravitacional. Puesto que la masa no varía en ningún caso, se concluye que el peso es constante, entonces la diferencia de valores se debe a la fuerza por tensión del cable. En el descenso, es fuerza es menor que en el ascenso.

8 0
3 years ago
Look at the proof. Name the postulate you would use to prove the two triangles are congruent.
vovikov84 [41]

Answer:

Option A, SSS Postulate

Step-by-step explanation:

<em>Since it gives you three sides, that means that it is a SSS.</em>

There are no angles given which eliminates 2 options.

Answer:  Option A, SSS Postulate

3 0
3 years ago
In the derivation of Newton’s method, to determine the formula for xi+1, the function f(x) is approximated using a first-order T
dimaraw [331]

Answer:

Part A.

Let f(x) = 0;

suppose x= a+h

such that f(x) =f(a+h) = 0

By second order Taylor approximation, we get

f(a) + hf'±(a) + \frac{h^{2} }{2!}f''(a) = 0

h = \frac{-f'(a) }{f''(a)} ± \frac{\sqrt[]{(f'(a))^{2}-2f(a)f''(a) } }{f''(a)}

So, we get the succeeding equation for Newton's method as

x_{i+1} = x_{i} + \frac{1}{f''x_{i}}  [-f'(x_{i}) ± \sqrt{f(x_{i})^{2}-2fx_{i}f''x_{i} } ]

Part B.

It is evident that Newton's method fails in two cases, as:

1.  if f''(x) = 0

2. if f'(x)² is less than 2f(x)f''(x)    

Part C.

In case  x_{i+1} is close to x_{i}, the choice that shouldbe made instead of ± in part A is:

f'(x) = \sqrt{f'(x)^{2} - 2f(x)f''(x)}  ⇔ x_{i+1} = x_{i}

Part D.

As given x_{i+1} = x_{i} = h

or                 h = x_{i+1} - x_{i}

We get,

f(a) + hf'(a) +(h²/2)f''(a) = 0

or h² = -hf(a)/f'(a)

Also,             (x_{i+1}-x_{i})² = -(x_{i+1}-x_{i})(f(x_{i})/f'(x_{i}))

So,                f(a) + hf'(a) - (f''(a)/2)(hf(a)/f'(a)) = 0

It becomes   h = -f(a)/f'(a) + (h/2)[f''(a)f(a)/(f(a))²]

Also,             x_{i+1} = x_{i} -f(x_{i})/f'(x_{i}) + [(x_{i+1} - x_{i})f''(x_{i})f(x_{i})]/[2(f'(x_{i}))²]

6 0
3 years ago
Classify the expression, name the operation(s) and identify any variables. 3s – 5
Elodia [21]
Numerical, subtraction, s
8 0
3 years ago
Please help me it’s really important to me
shepuryov [24]
1. b
2. e
3. a
4. c
5. d
6. f
7. g
8. h
hope this helps there is really no way of explaining you would have to study you theorems
7 0
3 years ago
Other questions:
  • PLZ HURRYYY 15 Points!!
    7·1 answer
  • John's monthly salary is 7,000. If John deposits 0.75 of his paycheck into a savings account. How much money does he put in his
    15·1 answer
  • Number 25 please help
    14·1 answer
  • After the fine arts booster club's soup supper, there were 120 pieces of pie left over. The leftover pie was distributed equally
    14·1 answer
  • A rectangular prism has a length of 4
    9·1 answer
  • Given the pattern 3,10,17,24..the nth term could be represented by which of the following?
    15·1 answer
  • HELP ASAP PLEASE AND SHOW WORK!!!
    9·1 answer
  • Find g(x), where g(x) is the reflection across the y-axis of f(x)= –9|x–10|+1.
    6·1 answer
  • How far can a cyclist travel in 1.5 hours if his average speed is 12 km/hr?
    14·1 answer
  • What could you multiply by to clear the fractions? Solve for x
    6·2 answers
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!