Solve 5x + 14 = k for x​

Answer:

I say it's TRUE

Step-by-step explanation:

I hope this help you

Answer: See explanation

Step-by-step explanation:

Your question isn't well written, the store contains $227 and not $277.

let c be the number of $1 bills

let f be the number of $5 bills

let t be the number of $10 bills.

Therefore, the equation will be:

1c + 5f + 10t = 227

Since there are six more $5 bills than $10 bills, therefore,

f = t+6

Also, the number of $1 bills is two more than 24 times the number of $10 bills and this will be:

c = 24t + 2

Now, we substitute the equation for f and c into the main equation and this will be:

1c + 5f + 10t = 277

1(24t + 2) + 5(t + 6) + 10t = 227

24t + 2 + 5t + 30 + 10t = 227

39t + 32 = 227

39t = 227 - 32

39t = 195

t = 195/39

t = 5

Therefore, there are 5 $10 bills

Since c = 24t + 2

c = 24(5) + 2

c = 120 + 2

c = 122

Therefore, there are 122 $1 bills

Since f = t+6

f = 5+6

f = 11

Therefore, there are 11 $5 bills.

9514 1404 393

Answer:

Step-by-step explanation:

1) The slope is given by the formula ...

  m = (y2 -y1)/(x2 -x1)

  m = (4 -(-3))/(3 -0) = 7/3

__

2) The y-intercept is the point where x=0:

  (x, y) = (0, -3) . . . . . y = -3

__

3) Then the slope-intercept equation is ...

  y = mx + b . . . . . . . line with slope m and y-intercept b

  y = 7/3x -3

Answer:

[tex]{ \tt{ {x}^{2} + {y}^{2} - 6x - 2y + 4 = 0 }} [/tex]

Center:

[tex]{ \tt{(3, \: 1)}}[/tex]

Radius:

[tex]{ \tt{radius = \sqrt{ {g}^{2} + {f}^{2} - c } }} \\ = \sqrt{36 + 4 - 4} \\ = 6 \: units[/tex]

Answer:

y = 3x+4

Step-by-step explanation:

The slope intercept form of a line is

y = mx+b  where m is the slope and b is the y intercept

y = 3x+4

Answer:

I think ∆5 is the correct Ans

Answer:

The answer is 12

Step-by-step explanation:

The range is the difference between the lowest and highest values. In this case, this means 142 - 130 which equals 12.

Answer:

12

Step-by-step explanation:

Range = Largest number - Smallest number

Larger number = 142 ; Smallest number = 130

142-130 = 12

Range = 12

Answer:

dl5fgtrrfhrruf ofender fkufgj jjgglr

Step-by-step explanation:

gjlhcohuguo

y.krgkyfjfhcxhhhxhhxfucuchhcjvgjjgvjvkvkx

Answer:

x = 19

Step-by-step explanation:

6x+4 + 4x - 14 = 180

10x    - 10 = 180

10 X = 190

x = 19

Answer:

Value of given expression is 3⁻¹⁴

Step-by-step explanation:

Given expression:

3⁻⁸ x 3⁻⁴ x 3⁻²

Find:

Value of given expression

Computation:

⇒ 3⁻⁸ x 3⁻⁴ x 3⁻²

⇒ 3[⁻⁸⁻⁴⁻²]

⇒ 3⁻¹⁴

Value of given expression is 3⁻¹⁴

Answer:

25%

Step-by-step explanation:

4+3+5=12

3/12 ×100=25

Answer:

[tex]f(x + 3) = {(x + 3)}^{3} [/tex]

Shifted left with x-intercept at (-3,0) and the y-intercept at (0,27)

Step-by-step explanation:

here's the answer to your question

9514 1404 393

Answer:

  A.  $674.43

Step-by-step explanation:

The account balance is given by ...

  A = P(1 +r/12)^(12t) . . . . . principal P invested at annual rate r for t years

  A = $500(1 +0.06/12)^(12·5) = $500(1.005^60)

  A ≈ $674.43

Answer:

3.803% Profit

Step-by-step explanation:

First let us calculate total sales, then calculate the total cost, and then divide the two.

Sale: 120*1.1=132 and 52*.9=46.8

Cost: 120 and 52

Profit equation:

[tex]\frac{Net sales-COGS}{Net Sales} =Profit[/tex]

[tex]\frac{(132+46.8)-(120+52)}{(132+46.8)}[/tex]

Simplify:

[tex]\frac{6.8}{178.8} =.03803[/tex] or 3.803% Profit

Answer:

dddjshsbs zhzusushwbsgzyzhsbssbs s hsbd d dhd ssbshebebs shyshs s sbshss s s

Step-by-step explanation:

shshshsnsnsjzj shsjs s sjsn s s snsnsbzbznnz ssjs s snsnsmkzkxnz

Answer:

30-5ab-14+6ab

16-ab ans

Answer:

[tex] \boxed{ab + 16} [/tex]

Step-by-step explanation:

5 (6 – ab) + 2 (-7 + 3ab)

[distribution property]

(6)(5) – (ab)(5) + (-7)(2) + (3ab)(2)

30 – 5ab – 14 + 6ab

[group like terms]

(6ab – 5ab) + (30 – 14)

ab + 16

Step-by-step explanation:

L.H.S

tanA * cosecA

=sinA/cosA * 1/sinA

now cancel sinA and sinA

then,

1/cosA

= secA

= RHS

hence proved

Answer:

D; 2 * 3; -4

Step-by-step explanation:

The dimension of the matrix refers to the number of elements in a row multiplied by the number of elements is a column

In a row, we have 2 elements, in a column , we have 3 elements

So the matrix is of 2 * 3 dimension

a21, refers to the element in the matrix that is in the row 2 column 1 and that is -4

Answer:

LP = 15

Step-by-step explanation:

Since the triangles are similar then the ratios of corresponding sides are equal, that is

[tex]\frac{LM}{LN}[/tex] = [tex]\frac{LQ}{LP}[/tex] , substitute values

[tex]\frac{4}{10}[/tex] = [tex]\frac{LQ}{LQ+QP}[/tex]

[tex]\frac{2}{5}[/tex] = [tex]\frac{LQ}{LQ+9 }[/tex] ( cross- multiply )

5LQ = 2(LQ + 9) = 2LQ + 18 ( subtract 2LQ from both sides )

3LQ = 18 ( divide both sides by 3 )

LQ = 6

Then

LP = LQ + QP = 6 + 9 = 15

Answer:

65°

Step-by-step explanation:

The sum of angles in a triangle =180

53°+62°+?°=180°

115°+?°=180°

?°=180°-115°

Rate the answer If you love everything I did n explanation

Answer:

65 degrees

Step-by-step explanation:

Angles in a triangle add up to 180, so 180 - 53 - 62 = 65.

Answer:

Below

Step-by-step explanation:

From the graph the first local minimum is -16.18

The local maximum is 3.75

The second local minimum is -3.

Answer:

----------------------

Hope it helps..

Have a great day!!

Step-by-step explanation:

Solution

First equation x ² - 4

= (x+2) (x-2)

Second equation = x ³ +8 + = x³+2^3

= (x-2) (x² + x₁2 +2²) = (x-2) (x² + 2x+4)

3 Third equation = x ² + 5x + 6

= x² + 6x= x +6 = x(x+6)-1(x+6)

(x-1) (x+6)

H.C. F= x-2

Given:

The given equation is:

[tex]16-3p=\dfrac{2}{3}p+5[/tex]

To find:

The value of p.

Solution:

We have,

[tex]16-3p=\dfrac{2}{3}p+5[/tex]

Multiply both sides by 3.

[tex]3(16-3p)=3\left(\dfrac{2}{3}p+5\right)[/tex]

[tex]48-9p=2p+15[/tex]

Isolating the variable terms, we get

[tex]48-15=2p+9p[/tex]

[tex]33=11p[/tex]

Divide both sides by 11, we get

[tex]\dfrac{33}{11}=p[/tex]

[tex]3=p[/tex]

Therefore, the required solution is [tex]p=3[/tex].

Answer:

p=3

Step-by-step explanation:

i got it right on khan

Answer:

Solution: To determine mass of the block we can use second Newton' law \vec F=m\vec a

F

=m

a

. The force and acceleration according the problem is directed along a horizontal surface, and we can omit the vector sign in Newton's law. The force we know F=45NF=45N, thus we should deduce the acceleration. The problem does not specify the initial speed at which time began to count, so for the first time interval, we may write the kinematics equation in the form

(1) S_1=v_1\cdot t_1+a\frac {t_1^2}{2}S

1

=v

1

⋅t

1

+a

2

t

1

2

, where S_1=8m, t_1=2s S

1

=8m,t

1

=2s , other quantities we don't know. The similar equation we can write for next time interval

(2) S_2=v_2\cdot t_2+ a\frac{t_2^2}{2}S

2

=v

2

⋅t

2

+a

2

t

2

2

. where S_2=8.5m, t_2=1s S

2

=8.5m,t

2

=1s

Note that during the first time interval, the speed of the block increased in accordance with the law of equidistant motion and it became the initial speed of the second interval, i.e.

(3) v_2=v_1+a\cdot t_1v

2

=v

1

+a⋅t

1

Substitute (3) to (2) we get

(4) S_2=(v_1+a\cdot t_1)\cdot t_2+ a\frac{t_2^2}{2}=v_1\cdot t_2+a\cdot t_1\cdot t_2+a\frac{t_2^2}{2}S

2

=(v

1

+a⋅t

1

)⋅t

2

+a

2

t

2

2

=v

1

⋅t

2

+a⋅t

1

⋅t

2

+a

2

t

2

2

From equation (1) and (4) we can exclude unknown quantity v_1v

1

, then remain only one unknown aa. For determine aa we dived (1) by t_1t

1

, (4) by t_2t

2

to find the average speed at time intervals and subtract (1) from (4).

(5) \frac {S_2}{t_2}-\frac {S_1}{t_1}=v_1+a\cdot t_1 +a\frac {t_2}{2}-(v_1+a\frac{t_1}{2})=a\frac{t_1+t_2}{2}-

t

2

S

2

−

t

1

S

1

=v

1

+a⋅t

1

+a

2

t

2

−(v

1

+a

2

t

1

)=a

2

t

1

+t

2

− For acceleration we get

(6) a=2\cdot ( {\frac{S_2}{t_2}-\frac{S_1}{t_1})/(t_1+t_2)}=2\cdot \frac{(8.5m/s-4m/s)}{3s}=3ms^{-2}a=2⋅(

t

2

S

2

−

t

1

S

1

)/(t

1

+t

2

)=2⋅

3s

(8.5m/s−4m/s)

=3ms

−2

For mass from second Newton's law we get

(7) m=\frac{F}{a}=\frac{45N}{3ms^{-2}}=15kgm=

a

F

=

3ms

−2

45N

=15kg

Answer: The mass of the block is 15 kg