I'm confused on how to find x​

Answer:

bout 7

Step-by-step explanation:

Answer:3.049533e+44 to be exact

Step-by-step explanation:

Answer:

22:2, 55:5, 1:11

Step-by-step explanation:

Answer:

22:2 55:5 and no others

Step-by-step explanation:

Answer:

6n+1

Step-by-step explanation:

n=term#

Answer: 19 months

Step-by-step explanation:

given data:

amount paid initially when joining the club = $50.

monthly payments made = $10.

total amount paid since joining the club = $240.

Solution:

= 10X + 50 = 240

collect like terms

= 10X = 240 – 50

10X = 190

divide both sides by 10

10X/10 = 190/10

X = 19

isabella has been a member of the club for 19 months.

Answer:

#2

Even though the graph dips down, x is still increasing, and y will increase as will .

As x is decreasing, y is also decreasing.

Answer:

An increase of $0.94

Step-by-step explanation:

Answer:

I not see the number 2

I think that is number 2

Answer:

first and last one should be the answer (not sure about the first, cant see it too good)

Answer:

15

Step-by-step explanation:

Following PEMDAS you would do the equation in the parentheses first.

1 - 15 = -14

13 - 2  ⋅ (-14) divided by 14

-2 ⋅ (-14) = 28

13 + 28 divided by 14

28 divided by 14 = 2

13 + 2 = 15

Your answer is 15

Hope I helped :)

Please consider Brainliest :)

Answer:

As

Step-by-step explanation:

Aaa

Answer:

2.35 hamburgers

Step-by-step explanation:

H6.25 + H18.75 = 58.75

Answer:

Jenn runs faster at 9 minutes/mile.

Step-by-step explanation:

Jana runs 3 miles in 1/3 hour.

1/3 hour is equivalent to 1/3*60 = 20 minutes

The ratio time/distance is the reciprocal of the speed and is greater as the speed is lower and vice-versa.

This ratio for Jana is 20/3=16.7 minutes/mile

As for Jenn, she runs 5 miles in 3/4 hour.

3/4 hour is equivalent to 3/4*60=45 minutes

The ratio time/distance is 45/5=9 minutes/mile

Since Jenn has a lower ration, her speed is greater than Jana's, thus Jenn runs faster.

Answer:

Step-by-step explanation:

2g=7h-3f

g=[tex]\frac{7h-3f}{2}[/tex]

Answer:

g= [tex]g=\frac{7}{2}h- \frac{3}{2}f[/tex], f∈R, h∈R

Step-by-step explanation:

3f+2g=7h

2g=7h-3f

g=[tex]\frac{7}{2}h - \frac{3}{2} f[/tex]

Answer:

x=12

Step-by-step explanation:

multiply 2 from both sides to isolate x

then u should end up with 1x=12

Answer:

-4( 3a+4)

Step-by-step explanation:

-12a - 16

-4*3a -4*4

Factor out a -4

-4( 3a+4)

Answer:

269.75

Hope this helps:)

Step-by-step explanation:

Answer: D,C,B

Step-by-step explanation:These are the answers

Answer:

-3.21

Step-by-step explanation:

If you pay close attention the numbers are 5 points away from each other, so after -3.2 is -3.21! Hope that helped love :)

Answer:

21.97025026/250

Answer:

f(x) = x2 – p(x + 1) – c, then (α + 1)(β + 1)f(x) = x2 – p(x + 1) – c, then (α + 1)(β + 1)f(x) = x2 – p(x + 1) – c, then (α + 1)(β + 1)f(x) = x2 – p(x + 1) – c, then (α + 1)(β + 1)f(x) = x2 – p(x + 1) – c, then (α + 1)(β + 1)f(x) = x2 – p(x + 1) – c, then (α + 1)(β + 1)

Step-by-step explanation:

Answer:

x = 39

Step-by-step explanation:

A circle is made up of 360 degrees

235+ 54+ x+32 = 360

Combine like terms

x+321=360

Subtract 321 from each side

x+321-321 = 360-321

x = 39

Answer:

[tex]\mathbf{y(x) = \dfrac{(c_1 + c_2) \ cos (x)}{e^{6x} }+ \dfrac{i(c_1 -c_2) \ sin (x)}{e^{6x}}+ c_3e^{3\ x}}[/tex]

Step-by-step explanation:

The objective of this question is to solve:

[tex]\dfrac{dy(x)}{dx}+ 9\dfrac{d^2y(dx)}{dx^2}+\dfrac{d^3y(x)}{dx^3}-111y(x) = 0 :[/tex]

Suppose the general solution is proportional to [tex]e^{\lambda x}[/tex] for [tex]\lambda[/tex] is constant; Then:

Let's replace [tex]y(x) = e^{\lambda\ x}[/tex] into the above equation:

i.e.

[tex]\dfrac{d^3}{dx^3}(e ^{\lambda x} )+ 9 \dfrac{d^2}{dx^2}(e ^{\lambda x} ) + \dfrac{d}{dx}(e ^{\lambda x} )- 111 \ e ^{\lambda x} = 0[/tex]

To Replace:

[tex]\dfrac{d^3}{dx^3}(e ^{\lambda x} )[/tex] with [tex]\lambda^3 e ^{\lambda x }[/tex]

[tex]\dfrac{d^2}{dx^2}(e ^{\lambda x} ) \ with \ \lambda^2 e^{\lambda\ x}[/tex]

[tex]\dfrac{d}{dx}(e ^{\lambda x} ) \ with \ \lambda e ^{\lambda \ x}[/tex]

Thus;

[tex]\lambda^3 e ^{\lambda x }[/tex] + [tex]9 \lambda^2 e^{\lambda\ x}[/tex] + [tex]\lambda e ^{\lambda \ x}[/tex] - 111 [tex]e ^{\lambda \ x}[/tex] = 0

[tex]e ^{\lambda \ x} (\lambda ^3 + 9 \lambda ^2 + \lambda - 111 )= 0[/tex]

∴

In as much as [tex]e^{ \lambda x}\neq 0[/tex] for any finite [tex]\lambda[/tex]; Then:

[tex]\lambda ^3 + 9 \lambda ^2 + 111 = 0[/tex]

By Factorization:

[tex](\lambda - 3) ( \lambda ^2 + 12 \lambda + 37) = 0[/tex]

[tex]\lambda = -6 + i \ or\ \lambda = -6 - i \ or \ \lambda = 3[/tex]

However;

The root [tex]\lambda = -6 \pm i[/tex] yield;

[tex]y_1 = (x) = c_1 e ^ {(-6+i)x}[/tex]

[tex]y_2 (x) = c_2e^{(-6-i)x}[/tex]

The root [tex]\lambda = 3[/tex] yield;

[tex]y_3(x) = c_3 e^{3x}[/tex]

∴

The general solution is:

[tex]y(x) = y_1(x) + y_2(x) + y_3(x) = \dfrac{c_1}{c^{(6-i)}x}+\dfrac{c_2}{c^{(6+i)}x}+ c_3e^{3x}[/tex]

Using Euler's Identity ;

[tex]e^{\alpha+i \beta} = e^\alpha \ cos (\beta ) + i \ e^\alpha \ sin ( \beta)[/tex]

[tex]y(x) = c_1 ( \dfrac{cos (x) }{e^{6x}}+ \dfrac{i \ sin x }{e^{6x} }) + c_2 ( \dfrac{cos (x)}{e^{6x}}- \dfrac{-i \ sin (x)}{e^{6x}})+c_3 e^{3x}[/tex]

[tex]\mathbf{y(x) = \dfrac{(c_1 + c_2) \ cos (x)}{e^{6x} }+ \dfrac{i(c_1 -c_2) \ sin (x)}{e^{6x}}+ c_3e^{3\ x}}[/tex]

Answer:

The answer is 11

hence 5 + 6 = 11

Answer:

hi

Step-by-step explanation:

[tex]5 + 3(2) = 11[/tex]

probably…

Answer:

Paired samples t- test is appropriate for Dean's hypothesis

Step-by-step explanation:

Given Data :

Student           1     2     3      4      5      6      7      8

Marketing      82  86   74    93    90    76   87   100

Finance         76   91    70    79    96     70  85   81

We are supposed to find the test which is appropriate for Dean's hypothesis

We are given the grades of two different subjects for same student

We perform the paired t test because the final grades for the Marketing and Finance  is taken for the same student.

So, Option B is correct

Hence Paired samples t- test is appropriate for Dean's hypothesis