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3 years ago

Solve the equation for y 5 ( y + 1 ) = 30 y = ??

2 answers:

5 0

Hi again :)

5(y + 1) = 30
Once again, use the distributive property
(5)(y) + (5)(1) = 30
5y + 5 = 30
Add -5 to both sides
5y + 5 - 5 = 30 - 5
5y = 25
Divide both sides by 5
5y/5 = 25/5
y = 5

Best of luck!

enot [183]3 years ago

5 0

The answer for the following problem is 5

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Alborosie

Answer:

its 65. bcuz you add it all together

Step-by-step explanation:

6 0

3 years ago

<img src="https://tex.z-dn.net/?f=%20%5Cfrac%7B%20%7B2%7D%5E%7Bx%20%2B%202%7D%20-%20%20%7B2%7D%5E%7B%20x%20%2B%203%7D%20%7D%7B%2

weeeeeb [17]

I suppose you just have to simplify this expression.

(2ˣ⁺² - 2ˣ⁺³) / (2ˣ⁺¹ - 2ˣ⁺²)

Divide through every term by the lowest power of 2, which would be x + 1 :

… = (2ˣ⁺²/2ˣ⁺¹ - 2ˣ⁺³/2ˣ⁺¹) / (2ˣ⁺¹/2ˣ⁺¹ - 2ˣ⁺²/2ˣ⁺¹)

Recall that nª / nᵇ = nª⁻ᵇ, so that we have

… = (2⁽ˣ⁺²⁾ ⁻ ⁽ˣ⁺¹⁾ - 2⁽ˣ⁺³⁾ ⁻ ⁽ˣ⁺¹⁾) / (2⁽ˣ⁺¹⁾ ⁻ ⁽ˣ⁺¹⁾ - 2⁽ˣ⁺²⁾ ⁻ ⁽ˣ⁺¹⁾)

… = (2¹ - 2²) / (2⁰ - 2¹)

… = (2 - 4) / (1 - 2)

… = (-2) / (-1)

… = 2

Another way to get the same result: rewrite every term as a multiple of y = 2ˣ :

… = (2²×2ˣ - 2³×2ˣ) / (2×2ˣ - 2²×2ˣ)

… = (4×2ˣ - 8×2ˣ) / (2×2ˣ - 4×2ˣ)

… = (4y - 8y) / (2y - 4y)

… = (-4y) / (-2y)

… = 2

8 0

3 years ago

PLEASEEE HELP FAST

Viktor [21]

Answer:

1. 4 as that is when he is closest.

2. 3 is when he is waiting as the graph is horizontal showing no movement

3. 4 would change as that is showing his pace walking home if it increased the slope would become steep but if it decreased it would level out more.

Step-by-step explanation:

4 0

3 years ago

Stella graphs the equation y=1/3x – 2

faltersainse [42]

we have

$y=\frac{1}{3}x-2$

Statements

case A) The graph is a straight line.

The statement is True

Because, this is a linear equation (see the attached figure)

case B) The line passes through the origin.

The statement is False

Because the point $(0,0)$ is not a solution of the equation

Verify

Substitute the value of x and y in the equation

$0=\frac{1}{3}*0-2$

$0=-2$ ------> is not true

the point $(0,0)$ is not a solution

therefore

The line does not pass through the origin

case C) The line passes through the point $(0,-2)$

The statement is True

Because the point $(0,-2)$ is a solution of the equation

Verify

Substitute the value of x and y in the equation

$-2=\frac{1}{3}*0-2$

$-2=-2$ ------> is true

the point $(0,-2)$ is a solution

therefore

The line passes through the point $(0,-2)$

case D) The slope of the line is $3$

The statement is False

Because, the slope of the line is $m=\frac{1}{3}$

case E) The y-intercept of the line is $2$

The statement is False

we know that

The y-intercept is the value of y when the value of x is equal to zero

For $x=0$

find the value of y

$y=\frac{1}{3}*0-2=-2$

the y-intercept is equal to $-2$

5 0

3 years ago

Read 2 more answers

The local branch of the Internal Revenue Service spent an average of 21 minutes helping each of 10 people prepare their tax retu

AURORKA [14]

Answer:

The confidence Interval is [- 0.7053 10.4521]

a: The hypotheses are

H0: μ1=μ2 against the claim Ha :μ1≠μ2

b. The critical value for t∝/2 for 17 d.f t > 2.508 and t < -2.111

c. t= -2.8422

d. The calculated value of t= -2.8422 is less than t < -2.11 the critical value therefore we reject H0 and conclude there is a difference between the two means.

Step-by-step explanation:

When the standard deviations are not the same then the confidence intervals for mean differences are calculated as

(x1`-x2`)- t∝/2 √s1²/n1 + s2²/n2 < u1-u2 < (x1`-x2`)+ t∝/2 √s1²/n1 + s2²/n2

x1`= 21 x2`= 27

n1= 10 n2= 14

s1= 5.6 s2= 4.3

The degrees of freedom is calculated using

υ = [s₁²/n1 + s₂²/n2]²/ (s₁²/n1 )²/ n1-1 + (s₂²/n2)²/n2-1

= 17

The t∝/2 for 17 d.f = 2.11

Putting the values

(x1`-x2`)- t∝/2 √s1²/n1 + s2²/n2 < u1-u2 < (x1`-x2`)+ t∝/2 √s1²/n1 + s2²/n2

(21-27) - 2.11√5.6²/10+ 4.3²/14 < u1-u2 <(21-27) +2.11√5.6²/10+4.3²/14

6- 2.11*2.111 < u1-u2 < ( 6 ) +2.11*2.111

6- 4.4521 < u1-u2 < ( 6 ) +5.294

- 1.5479 < u1-u2 < 10.4521

The confidence Interval is [- 0.7053 10.4521]

a: The hypotheses are

H0: μ1=μ2 against the claim Ha :μ1≠μ2

The claim is that there is a difference in the average time spent by the two services

b. The critical value for t∝/2 for 17 d.f t > 2.508 and t < -2.111

The degrees of freedom is calculated using

υ = [s₁²/n1 + s₂²/n2]²/ (s₁²/n1 )²/ n1-1 + (s₂²/n2)²/n2-1

= 17

c. The test statistic is

t= (x1`-x2`) /√s1²/n1 + s2²/n2

t= (21-27) /√5.6²/10+ 4.3²/14

t= -6/2.111

t= -2.8422

d. The calculated value of t= -2.8422 is less than t < -2.11 the critical value therefore we reject H0 and conclude there is a difference between the two means.

5 0

3 years ago