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3 years ago

Need Help, Trying To Finish .

Mathematics

1 answer:

hammer [34]3 years ago

3 0

Answer:

8

Steps:

Plug in 2 for x in h(x)

h(2) = -(2)^2 + 6 * 2

h(2) = -4 + 12

h(2) = 8

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Select all expressions less than 5-square root of 135

belka [17]

It should be b im not sure

5 0

2 years ago

Will give <br> BRAINLY AND FIVE STARS and A THANKS

11Alexandr11 [23.1K]

Answer:

k = 12h

Step-by-step explanation:

Since each of the inputs of h can be mutiplied by 12 to get the output k, the equation is k = 12h
Hope this helps :)

3 0

2 years ago

Yuri purchased 8 trees to have planted at his house. The store charged a delivery fee of $5 per tree.

cestrela7 [59]

Answer:

$16.

Step-by-step explanation:

Let t represent the cost of each tree.

We have been given that Yuri purchased 8 trees to have planted at his house. So the cost of 8 trees will be 8t.

We are also told that the store charged a delivery fee of $5 per tree. So the delivery charges of 8 trees will be $8\times \$5=\$40$

The total cost of purchasing 8 trees will be $8t+40$ .

Since the total cost of the trees including the delivery was $168, so we can get an equation as:

$8t+40=168$

$8t+40-40=168-40$

$8t=128$

Upon dividing both sides of our given equation by 8, we will get:

$\frac{8t}{8}=\frac{128}{8}$

$t=16$

Therefore, the cost of each tree is $16.

5 0

2 years ago

Sherman has two sequences . The first sequence is described by the explicit rule f(n) = 15n + 4 and the second sequence is descr

Viktor [21]

Answer:

Step-by-step explanation:

Given the explicit function as

f(n) = 15n+4

The first term of the sequence is at when n= 1

f(1) = 15(1)+4

f(1) = 19

a = 19

Common difference d = f(2)-f(1)

f(2) = 15(2)+4

f(2) = 34

d = 34-19

d = 15

Sum of nth term of an AP = n/2{2a+(n-1)d}

S20 = 20/2{2(19)+(20-1)15)

S20 = 10(38+19(15))

S20 = 10(38+285)

S20 = 10(323)

S20 = 3230.

Sum of the 20th term is 3230

For the explicit function

f(n) = 4n+15

f(1) = 4(1)+15

f(1) = 19

a = 19

Common difference d = f(2)-f(1)

f(2) = 4(2)+15

f(2) = 23

d = 23-19

d = 4

Sum of nth term of an AP = n/2{2a+(n-1)d}

S20 = 20/2{2(19)+(20-1)4)

S20 = 10(38+19(4))

S20 = 10(38+76)

S20 = 10(114)

S20 = 1140

Sum of the 20th terms is 1140

7 0

2 years ago

1. (a) Solve the differential equation (x + 1)Dy/dx= xy, = given that y = 2 when x = 0. (b) Find the area between the two curves

erastova [34]

(a) The differential equation is separable, so we separate the variables and integrate:

$(x+1)\dfrac{dy}{dx} = xy \implies \dfrac{dy}y = \dfrac x{x+1} \, dx = \left(1-\dfrac1{x+1}\right) \, dx$

$\displaystyle \frac{dy}y = \int \left(1-\frac1{x+1}\right) \, dx$

$\ln|y| = x - \ln|x+1| + C$

When x = 0, we have y = 2, so we solve for the constant C :

$\ln|2| = 0 - \ln|0 + 1| + C \implies C = \ln(2)$

Then the particular solution to the DE is

$\ln|y| = x - \ln|x+1| + \ln(2)$

We can go on to solve explicitly for y in terms of x :

$e^{\ln|y|} = e^{x - \ln|x+1| + \ln(2)} \implies \boxed{y = \dfrac{2e^x}{x+1}}$

(b) The curves y = x² and y = 2x - x² intersect for

$x^2 = 2x - x^2 \implies 2x^2 - 2x = 2x (x - 1) = 0 \implies x = 0 \text{ or } x = 1$

and the bounded region is the set

$\left\{(x,y) ~:~ 0 \le x \le 1 \text{ and } x^2 \le y \le 2x - x^2\right\}$

The area of this region is

$\displaystyle \int_0^1 ((2x-x^2)-x^2) \, dx = 2 \int_0^1 (x-x^2) \, dx = 2 \left(\frac{x^2}2 - \frac{x^3}3\right)\bigg|_0^1 = 2\left(\frac12 - \frac13\right) = \boxed{\frac13}$

7 0

2 years ago