All categories

3 years ago

4. Solve each equation for x. (8 points: 2 points each)

Mathematics

1 answer:

maksim [4K]3 years ago

8 0

Answer:

a. x = 9 + 4y

b. x = (7y-2)/6

c. x = (17 + k)/4

d. x = (9-c)/k

Step-by-step explanation:

In each of the equations, we want to make x the formula subject

a) 3x-12y = 27

3x = 27 + 12y

3x = 3(9 + 4y)

x = 9 + 4y

b) 6x-7y = -2

6x = 7y-2

x = (7y-2)/6

c) 4x - k = 17

4x = 17 + k

x = (17 + k)/4

d) kx + c = 9

kx = 9-c

x = (9-c)/k

You might be interested in

Solve for h in 9=h-5

Misha Larkins [42]

Answer:

h = 14

Step-by-step explanation:

9=h-5

9 + 5 = h -5 +5

14 = h

3 0

3 years ago

Suppose you know that Mr. and Mrs. Vincent have at most eight children. One day you happen to meet two girls who were the Vincen

lianna [129]

Answer:

The number of children are 4 out of which 3 are girls

Step-by-step explanation:

Data provided in the question:

P(Two randomly selected children are girls) = $\frac{1}{2}$

now,

let the number of children be 'n'

the number of girls be 'x'

thus,

P(Two randomly selected children are girls) = $\frac{^xC_2}{^nC_2}$ = $\frac{1}{2}$

also,

$^nC_r$ = $\frac{n!r!}{(n-r)!}$

thus,

$\frac{\frac{x!2!}{(x-2)!}}{\frac{n!2!}{(n-2)!}}$ = $\frac{1}{2}$

$\frac{x(x-1)}{n(n-1)}$ = $\frac{1}{2}$

2x(x-1) = n(n-1)

now

for x = 3 and n = 4

i.e

2(3)(3-1) = 4(4-1)

12 = 12

hence, the relation is justified

therefore,

The number of children are 4 out of which 3 are girls

3 0

3 years ago

The quadratic function h(t)=-16.1t^2+150 models a balls height in feet over time in seconds after it is dropped from a 15 story

Allushta [10]

someone please answer

3 0

4 years ago

Multiple-Choice Integration, Picture Included, Please Include Work

ValentinkaMS [17]

$\displaystyle\int_0^2\sqrt{4-x^2}\,\mathrm dx$

Recall that a circle of radius 2 centered at the origin has equation

$x^2+y^2=4\implies y=\pm\sqrt{4-x^2}$

where the positive root gives the top half of the circle in the x-y plane. The definite integral corresponds to the area of the right half of this top half. Since the area of a circle with radius $r$ is $\pi r^2$ , it follows that the area of a quarter-circle would be $\dfrac{\pi r^2}4$ .

You have $r=2$ , so the definite integral is equal to $\dfrac{2^2\pi}4=\pi$ .

Another way to verify this is to actually compute the integral. Let $x=2\sin u$ , so that $\mathrm dx=2\cos u\,\mathrm du$ . Now

$\displaystyle\int_0^2\sqrt{4-x^2}\,\mathrm dx=\int_0^{\pi/2}\sqrt{4-(2\sin u)^2}(2\cos u)\,\mathrm du=4\int_0^{\pi/2}\cos^2u\,\mathrm du$

Recall the half-angle identity for cosine:

$\cos^2u=\dfrac{1+\cos2u}2$

This means the integral is equivalent to

$\displaystyle2\int_0^{\pi/2}(1+\cos 2u)\,\mathrm du=2u+\sin2u\bigg|_{u=0}^{u=\pi/2}=\pi$

4 0

3 years ago

4. Al restar 5 x ᶺ 2 – 8 x + 9 con – 9 + 16 x – 25 x ᶺ 2 se obtiene como resultado (Procedimiento)

Ierofanga [76]

Step-by-step explanation:

SE RESTAN ENTRE LOS NÚMEROS QUE TIENEN LA MISMA VARIAVLE.

$(5x^{2} + 25x^{2})$ - (8x - 16x) + (9 + 9) = $30x^{2}$ - 24x + 18

4 0

3 years ago