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3 years ago

I need help with these questions

Mathematics

1 answer:

m_a_m_a [10]3 years ago

6 0

Answer:

1. -14

2. v=10

Step-by-step explanation:

- 22= - 8+k

-k-22=-8

-k=-8+22

-k=14/(-1)

k=-14

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Find the missing value to the nearest whole number. sin17∘=x/7

wlad13 [49]

Answer:

Step-by-step explanation:

sin 17° = x/7

(sin 17°) 7 = (x/7)7

x = sin 17° (7)

x = 0.292371704722737 (7)

x = 2.046601933059157

x ≈ 2

6 0

3 years ago

Sean bought 8 slurpees at $1.50 each and 6 popsicles. He gave the cashier $25.00 and received $4.00 change. How much did each Po

Lina20 [59]

Answer:

1.50

Step-by-step explanation:

8 x 1.50=12

25-12=13

13-4=9

9/6=1.5

8 0

3 years ago

Durning Saturday’s thunderstorms a total of 500 mm of rain fell in fell in 20 minutes.How many mm fell per minute

Sonja [21]

Answer

25mm of rain fell each second

Step-by-step explanation:

we know that 500 mm fell in 20 minutes

so, we have to divide 500 by 20 giving us the amount of rain that fell each minute:

500/20 = 25

therefore, 25mm of rain fell each minute

hope this helped:)

5 0

3 years ago

Read 2 more answers

Find the solution of the given initial value problem:<br><br> y''- y = 0, y(0) = 2, y'(0) = -1/2

igor_vitrenko [27]

Answer: The required solution of the given IVP is

$y(x)=\dfrac{3}{4}e^x+\dfrac{5}{4}e^{-x}.$

Step-by-step explanation: We are given to find the solution of the following initial value problem :

$y^{\prime\prime}-y=0,~~~y(0)=2,~~y^\prime(0)=-\dfrac{1}{2}.$

Let $y=e^{mx}$ be an auxiliary solution of the given differential equation.

Then, we have

$y^\prime=me^{mx},~~~~~y^{\prime\prime}=m^2e^{mx}.$

Substituting these values in the given differential equation, we have

$m^2e^{mx}-e^{mx}=0\\\\\Rightarrow (m^2-1)e^{mx}=0\\\\\Rightarrow m^2-1=0~~~~~~~~~~~~~~~~~~~~~~~~~~[\textup{since }e^{mx}\neq0]\\\\\Rightarrow m^2=1\\\\\Rightarrow m=\pm1.$

So, the general solution of the given equation is

$y(x)=Ae^x+Be^{-x},$ where A and B are constants.

This gives, after differentiating with respect to x that

$y^\prime(x)=Ae^x-Be^{-x}.$

The given conditions implies that

$y(0)=2\\\\\Rightarrow A+B=2~~~~~~~~~~~~~~~~~~~~~~~~~~~(i)$

and

$y^\prime(0)=-\dfrac{1}{2}\\\\\\\Rightarrow A-B=-\dfrac{1}{2}~~~~~~~~~~~~~~~~~~~~~~~~(ii)$

Adding equations (i) and (ii), we get

$2A=2-\dfrac{1}{2}\\\\\\\Rightarrow 2A=\dfrac{3}{2}\\\\\\\Rightarrow A=\dfrac{3}{4}.$

From equation (i), we get

$\dfrac{3}{4}+B=2\\\\\\\Rightarrow B=2-\dfrac{3}{4}\\\\\\\Rightarrow B=\dfrac{5}{4}.$

Substituting the values of A and B in the general solution, we get

$y(x)=\dfrac{3}{4}e^x+\dfrac{5}{4}e^{-x}.$

Thus, the required solution of the given IVP is

$y(x)=\dfrac{3}{4}e^x+\dfrac{5}{4}e^{-x}.$

4 0

3 years ago

(a)

12345 [234]

Answer:

a) 30.60

b) 1010/33 = 30.60

Step-by-step explanation:

8 0

2 years ago