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3 years ago

Can you Help me? With B

Mathematics

1 answer:

coldgirl [10]3 years ago

4 0

She would need to buy 10 additional bags.

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Price of 1kg sugar is Rs. 100 calculate the price of 150g sugar?

Wewaii [24]

Answer:Rs15

Step-by-step explanation:

150g=150/1000=0.15kg

1kg cost Rs100

0.15kg cost =100 x 0.15=15

150g cost Rs15

3 0

2 years ago

Which equation represents the line with a slope of 5/6 that passes through the point (-2, 1)?

andreyandreev [35.5K]

Answer:

Step-by-step explanation:

Slope m = 5/6

The points (-2,1)

So; y1 = 1 and x1 = -2

The equation is y - y1 = m(x - x1)

y - 1 = 5/6(x + 2)

Multiply each term by 6

6y - 6 = 5(x + 2)

6y - 6 = 5x + 10

6y = 5x + 10 + 6

6y = 5x + 16

- 5x + 6y = 16

Multiplying by minus

5x - 6y = -16

5 0

2 years ago

Read 2 more answers

10 POINTS + BRAINLIEST TO FIRST CORRECT ANSWER!!! THANK YOU!

Fofino [41]

Answer:

The answer is A (Look at the graph of the relationship. Find the y-value of the point that corresponds to x = 1. That value is the unit rate)

Step-by-step explanation:

Because the proportional relation is defined by k = $\frac{y}{x}$ . If ratio is $\frac{units}{1unit}$ then it is a rate. Unit rate means per 1 unit such as $\frac{amountofmoney}{1kg}$ or $\frac{km}{1hour}$ thus the correct answer is A.

3 0

2 years ago

Write down the explicit solution for each of the following: a) x’=t–sin(t); x(0)=1

Kay [80]

Answer:

a) x=(t^2)/2+cos(t), b) x=2+3e^(-2t), c) x=(1/2)sin(2t)

Step-by-step explanation:

Let's solve by separating variables:

$x'=\frac{dx}{dt}$

a) x’=t–sin(t), x(0)=1

$dx=(t-sint)dt$

Apply integral both sides:

$\int {} \, dx=\int {(t-sint)} \, dt\\\\x=\frac{t^2}{2}+cost +k$

where k is a constant due to integration. With x(0)=1, substitute:

$1=0+cos0+k\\\\1=1+k\\k=0$

Finally:

$x=\frac{t^2}{2} +cos(t)$

b) x’+2x=4; x(0)=5

$dx=(4-2x)dt\\\\\frac{dx}{4-2x}=dt \\\\\int {\frac{dx}{4-2x}}= \int {dt}\\$

Completing the integral:

$-\frac{1}{2} \int{\frac{(-2)dx}{4-2x}}= \int {dt}$

Solving the operator:

$-\frac{1}{2}ln(4-2x)=t+k$

Using algebra, it becomes explicit:

$x=2+ke^{-2t}$

With x(0)=5, substitute:

$5=2+ke^{-2(0)}=2+k(1)\\\\k=3$

Finally:

$x=2+3e^{-2t}$

c) x’’+4x=0; x(0)=0; x’(0)=1

Let $x=e^{mt}$ be the solution for the equation, then:

$x'=me^{mt}\\x''=m^{2}e^{mt}$

Substituting these equations in c)

$m^{2}e^{mt}+4(e^{mt})=0\\\\m^{2}+4=0\\\\m^{2}=-4\\\\m=2i$

This becomes the solution m=α±βi where α=0 and β=2

$x=e^{\alpha t}[Asin\beta t+Bcos\beta t]\\\\x=e^{0}[Asin((2)t)+Bcos((2)t)]\\\\x=Asin((2)t)+Bcos((2)t)$

Where A and B are constants. With x(0)=0; x’(0)=1:

$x=Asin(2t)+Bcos(2t)\\\\x'=2Acos(2t)-2Bsin(2t)\\\\0=Asin(2(0))+Bcos(2(0))\\\\0=0+B(1)\\\\B=0\\\\1=2Acos(2(0))\\\\1=2A\\\\A=\frac{1}{2}$

Finally:

$x=\frac{1}{2} sin(2t)$

7 0

3 years ago

What is the missing value?

lorasvet [3.4K]

Answer:

The missing value is 19.

Step-by-step explanation:

The equation is increasing by 2 and 17+2 equals 19.

4 0

2 years ago