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Maksim231197 [3]

3 years ago

12

A truck carries a load of 50 boxes of 2types.if type one boxes weight 20kg each and type two boxes weigh 25 kg each ,and the tot

al weight of all boxes is 1175 kg, then how many of each type are there? type one is _____ and type two is _____

1 answer:

Law Incorporation [45]3 years ago

7 0

Answer:

Step-by-step explanation:

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A shipment contains 9 igneous, 7 sedimentary, and 7 metamorphic rocks. If 7 rocks are selected at random, find the probability t

yKpoI14uk [10]

Answer:

75 percent

Step-by-step explanation:

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7 0

2 years ago

Plz help me. will give brainliest!!

Nesterboy [21]

Answer:

1. 110

2. 84

Step-by-step explanation:

6 0

3 years ago

Solve for all values of x in simplest form. 9-4|3+5x|=5

Zinaida [17]

Answer:

$x=-2/5\text{ or } x=-4/5$

Step-by-step explanation:

So we have the equation:

$9-4|3+5x|=5$

First, subtract 9 from both sides:

$-4|3+5x|=-4$

Divide both sides by -4:

$|3+5x|=1$

Definition of Absolute Value:

$3+5x=1\text{ or } 3+5x=-1$

Subtract 3 from both sides:

$5x=-2\text{ or } 5x=-4$

Divide both sides by 5:

$x=-2/5\text{ or } x=-4/5$

And we're done!

5 0

4 years ago

Read 2 more answers

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 <em>c)</em>

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

This becomes the solution <em>m=α±βi</em> where <em>α=0</em> and <em>β=2</em>

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

Where <em>A</em> and <em>B</em> 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

Need help with this if you can explain please do if not then just and answer is okay

pshichka [43]

Step-by-step explanation:

f(x) = x / (x - 7) = 1 + 7 / (x - 7).

The domain of f(x) depends whether the denominator of the rational expression is non-zero.

When (x - 7) = 0, x = 7.

Hence domain of f(x)

= All real values of x excluding 7. (C)

5 0

3 years ago

Read 2 more answers