All categories

3 years ago

Make 'y' the subject of the equation. x = 2y -7

Mathematics

1 answer:

geniusboy [140]3 years ago

5 0

Answer:

y = (x + 7)/2

here's your solution

=>. it is given. x = 2y - 7

=> we have to make y as subject

=> now, we can writ this equation in this way

=> 2y - 7 = x

=> now put all constant to right side

=> 2y = x +7

=>. y = x+ 7 /2

hope it helps

You might be interested in

Solve for h. 6+h+8=24

lozanna [386]

Answer:

the answer to your question is h = 10

8 0

3 years ago

Read 2 more answers

Maggie's brother is three years younger than twice her age. The sum of their ages is 24 how old is maggie?

Vanyuwa [196]

The variable for Maggie will be M and 2x-3 be her younger brothers age (Twice her age). We then would turn this into an algebraic problem. (m+2x -3=24).

3x-3=24, we would add 3 to both sides (-3 and 24). 24 + 3 equals 27, we now have to divide -3x and 27 on both sides, which equals 9. (x=9)

6 0

3 years ago

What is the product of the polynomials below? (4x^2-2x-4)(2x+4)

Travka [436]

Answer: 8x³ + 12x² - 16x - 16

Step-by-step explanation:

(4x² - 2x - 4)(2x + 4)

= (2x + 4)(4x² - 2x - 4)

= 2x(4x² - 2x - 4) + 4(4x² - 2x - 4)

= 8x³ - 4x² - 8x + 16x² - 8x - 16

= 8x³ + (-4x² + 16x²) + (-8x - 8x) - 16

= 8x³ + 12x² - 16x - 16

7 0

3 years ago

Ndicate the answer choice that best completes the statement or answers the question.

kirill115 [55]

Answer:

I think the answer is "adjacent, supplementary".

4 0

2 years ago

A study sampled 350 upperclassmen (Group 1) and 250 underclassmen (Group 2) at high schools around the city of Portland. The stu

iVinArrow [24]

Answer:

$-0.061 < P_1 -P_2< 0.025$

Step-by-step explanation:

Give data:

$n_1 = 350$

$n_2 =250$

$x_1 = 23$

$x_2 = 21$

$\hat{P} 1 = \frac{x_1}{n_1} = \frac{23}{350} = 0.066$

$\hat{P} 2 = \frac{x_2}{n_2} = \frac{21}{250} = 0.084$

for 95% confidence interval

$\alpha = 1 - 0.95 = 0.05 and \alpha/2 = 0.025$

$z_{\alpha/2} = 1.96$ from standard z- table

confidence interval for P_1 and P_2 is

$\hat{P} 1 - \hat{P} 2 \pm z_{\alpja/2} \sqrt{\frac{\hat{P} 1(1-\hat{P} 1)}{n_1} +\frac{\hat{P} 2(1-\hat{P} 2)}{n_2}}$

$(0.066 - 0.084) \pm 1.96 \sqrt{\frac{0.066(1-0.066)}{350} +\frac{0.084(1-0.084)}{250}}$

$-0.018 \pm 0.043$

confidence interval is

$-0.018 - 0.043 < P_1 -P_2$

$-0.061 < P_1 -P_2< 0.025$

7 0

3 years ago