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3 years ago

Cans someone please help me I’m really confused?? and I’ll mark brainlist

Mathematics

2 answers:

cestrela7 [59]3 years ago

3 0

Answer:

A. 1/2

Step-by-step explanation:

3/16 / 3/8

3/3 = 1

16/8 = 2

1/2

nata0808 [166]3 years ago

3 0

Answer: 1/2

Step-by-step explanation:

3/16 divided by 3/8

To divide by a fraction multiply by the reciprocal of that fraction

3/16 x 8/3

Reduce the numbers with the GCF (3)

1/16 x 8

Reduce the numbers with the GCF (8)

1/2 or 0.5

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A container contains 10 diesel engines. the company chooses 8 engines at random, and will not ship the container if any of the

lions [1.4K]

Answer:

16/25

Step-by-step explanation:

We just have to find the probability that the 8 non-defects are chosen.

1. Chance first defect is undetected: 8/10 = 4/5

2. Chance second defect is undetected: 4/5

3. Total probability': 4/5 * 4/5 = 16/25

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3 years ago

kaitlin is on a long car trip every time she stops she loses 15 minutes of travel time is she stops 5 times how late will she be

yanalaym [24]

5(15)= 75 minutes or an hour and 15 minutes lost

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3 years ago

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Elise won 40 lollipops and the school fair. she gave 2 to every student in her math class and has at least 7 left. what is the m

geniusboy [140]

40-7= 33
33/2= 16.5
The maximum would be 17 students

8 0

3 years ago

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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:

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

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3 years ago

How to factor by grouping

Ostrovityanka [42]

Step1: group the first two terms together
next step 2:factor out a GFC from each separate binomical
next step 3: factor out the common binomial

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3 years ago

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