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4 years ago

Mathematics question in the screenshot below.

Mathematics

1 answer:

irga5000 [103]4 years ago

4 0

Step-by-step explanation:

1/6×100%=50/3%

3/24×100%=25/4%

3/8×100%=75/2%

=>50/3%+25/4%+75/2%+x=100%

=>100%-725/12%=x

=>x=59 and 5/12

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(k/3^k)(x-6)^k

enot [183]

$\displaystyle\sum_{k\ge0}\frac k{3^k}(x-6)^k$

converges by the ratio test for

$\displaystyle\lim_{k\to\infty}\left|\frac{\frac{k+1}{3^{k+1}}(x-6)^{k+1}}{\frac k{3^k}(x-6)^k}\right|=\frac{|x-6|}3\lim_{k\to\infty}\frac{k+1}k$

The limit is 1, so the series converges as long as

$\dfrac{|x-6|}3$

which indicates a radius of convergence of 3.

5 0

3 years ago

Your cousin borrows $1125 to repair her car. The simple annual interest rate is 10%. She makes equal monthly payments of $25. Ho

Elena-2011 [213]

Answer:

neat

Step-by-step explanation:

8 0

3 years ago

Randy is walking home from school. According to the diagram above, what is his total distance from school to home? Show your wor

In-s [12.5K]

Answer:

First, when he walks, we can see in the image that between the school and his house he must walk 4 times a distance of 0.5km, so this is a total of 4¨*0.5km = 2km.

Then he needs to walk 2km.

Now if he has a jet-pack, he can ignore the buildings and just take the shorter path, here we can draw a triangle rectangle, in such a way that the hypotenuse of this triangle is the distance between the home and the school.

One of the catheti is the vertical distance (two blocks of 0.5km, so this catheti has a length of 2*0.5km = 1km), and the other one is the horizontal distance, also 1km.

The actual distance of this path is given by the Pythagorean's theorem:

A^2 + B^2 = H^2

Where A and B are the cathetus, and H is the hypotenuse, then:

H^2 = 1km^2 + 1km^2

H = (√2)km = 1.41km.

Now, in the case that he has a jet-pack, he can actually go to the school using this hypotenuse line as his path, in this case the distance and the displacement would be the same.

This is because the definitions of distance and displacement are:

Distance: "how much ground an object has covered"

Displacement: "Difference between the final position and the initial position"

When he walks, the distance is 2km and the displacement is 1.41km , but when he uses the jet pack, the distance is equal to the displacement, both are 1.41km.

7 0

3 years ago

Marcie bought a total of 20 used books and cds during a yard sale for a total of 54.50$. of books cost 1.50$ each and cds 5$ eac

melomori [17]

Let numbers of books be 'b' and numbers of CDs be 'c'

We can set up two equations:
Equation [1] ⇒ $b+c=20$
Equation [2] ⇒ $1.50b+5c=54.50$

We are solving for the number of books and the number of CDs bought

When we have two equations in terms of two different variables; $b$ and $c$ , that we need to solve, then this becomes a simultaneous equation problem.

First, rearrange Equation [1] to make either $b$ or $c$ the subject:
$b+c=20$
$b=20-c$

Then we substitute $b=20-c$ into Equation [2]
$1.50b+5c=54.50$
$1.50(20-c)+5c=54.50$
$30-1.50c+5c=54.50$
$5c-1.5c=54.50-30$
$3.5c=24.50$
$c=7$

Now we know the value of $c$ which is $c=7$ , substitute this value into $b=20-c$ we have $b=20-7=13$

Answer:
Numbers of books = 13
Numbers of CDs = 7

5 0

4 years ago

Two forces with magnitudes of 150 and 100 pounds act on an object at angles of 30° and 120°, respectively. Find the direction an

lyudmila [28]

<h2>Answer:</h2>

$Magnitude=180.27 \ lbf \\ \\ Direction=63.69 \ degrees$

<h2>Step-by-step explanation:</h2>

We have two forces as follows:

<u>First force:</u>

Magnitude: 150 pounds

Angle: 30°

<u>First force:</u>

Magnitude: 100 pounds

Angle: 120°

So the components can be found as follows:

$F_{1x}=150cos(30)=129.90 \ lbf\\F_{1y}=150sin(30) = 75 \ lbf \\ \\ F_{2x}=100cos(120)=-50 \ lbf\\F_{2y}=100sin(120) = 86.60 \ lbf$

So the components of the resultant force can be found by adding each component of the individual forces as follows:

$R_{x}=\Sigma F_{x} \\ R_{y}=\Sigma F_{y} \\ \\ R_{x}=129.90-50=79.90 \ lbf \\ R_{y}=75+86.60=161.6 \ lbf$

Finally, the magnitude and direction of the resultant force is:

$Magnitude \rightarrow R=\sqrt{R_{x}^2+R_{y}^2}=\sqrt{79.90^2+161.6^2}=180.27 \ lbf \\ \\ Direction \rightarrow \theta=tan^{-1}(\frac{R_{y}}{R_{x}})=tan^{-1}(\frac{161.6}{79.90})=63.69 \ degrees$

6 0

3 years ago

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