All categories

3 years ago

HELP PLEASE!!!! solve for x

Mathematics

1 answer:

LenKa [72]3 years ago

7 0

Answer: how do you solve for this

Step-by-step explanation:

You might be interested in

the cos c of recycle Tin can varies directly as it's weight w in kilograms .if a junk shop pays 60 pesos per kilo of tin cans ho

liraira [26]

Answer:

Step-by-step explanation:

c = 60w

390 = 60w

w = 6.5 kilograms

3 0

2 years ago

Im to layz to do my test so ye all the questions worth 10 points

Arisa [49]

Answer:

subtract 6 from 27 the divide by 3

Step-by-step explanation:

6 0

3 years ago

The temperature of a city changed by -20°C over a four week. If the temperature changed by the same amount each week write an ex

andre [41]

Answer:

A) $T=\frac{-20}{4}$

B) T = -5 °C

Step-by-step explanation:

A) Let T be the change in temperature of the week. T is determined as the total temperature change (-20°C) divided by the number of weeks that have passed (4 weeks). The expression is:

$T=\frac{-20}{4}$

B) In order to simplify the expression, simply perform the division.

$T=\frac{-20}{4}\\T=-5^oC$

This means that the temperature has dropped by 5 degrees each week.

8 0

3 years ago

F(x)=5x^5+ 24x^4+25x^3-86x^2-224x+48

Serjik [45]

I believe you graph this.. please let me know if wrong and I’ll help solve otherwise! The picture is the answer on how to graph

4 0

2 years ago

Consider the initial value function y given by

Nuetrik [128]

Answer:

y(s) = $\frac{5s-53}{s^{2} - 10s + 26}$

we will compare the denominator to the form $(s-a)^{2} +\beta ^{2}$

$s^{2} -10s+26 = (s-a)^{2} +\beta ^{2} = s^{2} -2as +a^{2} +\beta ^{2}$

comparing coefficients of terms in s

$s^{2} :$ 1

s: -2a = -10

a = -2/-10

a = 1/5

constant: $a^{2}+\beta ^{2} = 26$

$(\frac{1}{5} )^{2} + \beta ^{2} = 26\\\\\beta^{2} = 26 - \frac{1}{10} \\\\\beta =\sqrt{26 - \frac{1}{10}} =5.09$

hence the first answers are:

a = 1/5 = 0.2

β = 5.09

Given that y(s) = $A(s-a)+B((s-a)^{2} +\beta ^{2} )$

we insert the values of a and β

$\\5s-53$ = $A(s-0.2)+B((s-0.2)^{2} + 5.09^{2} )$

to obtain the constants A and B we equate the numerators and we substituting s = 0.2 on both side to eliminate A

5(0.2)-53 = A(0.2-0.2) + B((0.2-0.2)²+5.09²)

-52 = B(26)

B = -52/26 = -2

to get A lets substitute s=0.4

5(0.4)-53 = A(0.4-0.2) + (-2)((0.4 - 0.2)²+5.09²)

-51 = 0.2A - 52.08

0.2A = -51 + 52.08

A = -1.08/0.2 = 5.4

the constants are

a = 0.2

β = 5.09

A = 5.4

B = -2

Step-by-step explanation:

since the denominator has a complex root we compare with the standard form $s^{2} -10s+26 = (s-a)^{2} +\beta ^{2} = s^{2} -2as +a^{2} +\beta ^{2}$
Expand and compare coefficients to obtain the values of a and β as shown above
substitute the values gotten into the function
Now assume any value for 's' but the assumption should be guided to eliminate an unknown, just as we've use s=0.2 above to eliminate A
after obtaining the first constant, substitute the value back into the function and obtain the second just as we've shown clearly above

Thanks...

3 0

3 years ago