Negative 6 times the sum of a number and 4 is 2

NEED HELP ASAP, WILL MARK BRAINLIEST!!!

EastWind [94]

Answer:

7. -

a. 3.5 or 7/2 hours

b. 24.5

c. After 7 hours (and at the beginning when t=0)

8. -

a. 4 seconds

b. 1 second

c. 54 meters

Step-by-step explanation:

7. M(t)=-2t^2+14t

M’(t)=-4t+14

a)

-4t+14=0

4t=14

2t=7

t=7/2

b) M(7/2)=24.5

c)

M(t)=0

-2t^2+14t=0

t^2-7t=0

t(t-7)=0

t=0, 7

8.

a)

h(t)=0

-6t^2+12t+48=0

t^2-2t-8=0

(t-4)(t+2)=0

t=-2, 4

t=4

b)

h(t)=-6t^2+12t+48

h’(t)=-12t+12

h’(t)=0

-12t+12=0

12t=12

t=1

c) h(1)=54

I hope this helps; please vote brainliest!

3 0

2 years ago

Given the table below for selected values of f(x), use 6 left rectangles to estimate the value of

S_A_V [24]

Answer:

Step-by-step explanation:

Given is a table of x, f(x) as below

x 1 3 4 6 7 9 10

f(x) 4 8 6 10 10 12 16

Mid pt 6 7 8 10 11 14

width of

interval 2 1 2 1 2 1

dA1 12 7 16 10 22 14

area = area of all rectangles= 81 sq units.

8 0

2 years ago

7. Use the quadratic formula to find the solution(s). x² + 2x - 8 = 0

morpeh [17]

Hey there!

Use the quadratic formula to find the solution(s). x² + 2x - 8 = 0

Answer :

x = -4 or x = 2 ✅

Explanation :

Quadratic formula : ax² + bx + c = 0 where a ≠ 0

The number of real-number solutions (roots) is determined by the discriminant (b² - 4ac) :

If b² - 4ac > 0 , There are 2 real-number solutions

If b² - 4ac = 0 , There is 1 real-number solution.

If b² - 4ac < 0 , There is no real-number solution.

The roots of the equation are determined by the following calculation:

$x = \frac{ - b \pm \sqrt{ {b}^{2} - 4ac } }{2a}$

Here, we have :

a = 1
b = 2
c = -8

1) Calculate the discriminant :

b² - 4ac ⇔ 2² - 4(1)(-8) ⇔ 4 - (-32) ⇔ 36

b² - 4ac = 36 > 0 ; The equation admits two real-number solutions

2) Calculate the roots of the equation:

▪️ (1)

$x_1 = \frac{ - b - \sqrt{ {b}^{2} - 4ac} }{2a} \\ \\ x_1 = \frac{ - 2 - \sqrt{36} }{2(1) } \\ \\ x_1 = \frac{ - 2 - 6}{2} \\ \\ x_1 = \frac{ - 8}{2} \\ \\ \blue{\boxed{\red{x_1 = -4}}}$

▪️ (2)

$x_2 = \frac{ - b + \sqrt{ {b}^{2} - 4ac } }{2a} \\ \\ x_2 = \frac{ - 2 + \sqrt{36} }{2(1)} \\ \\ x_2 = \frac{ - 2 + 6}{2} \\ \\ x_2 = \frac{4}{2} \\ \\ \red{\boxed{\blue{x_2 = 2}}}$