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Len [333]
4 years ago
10

Solve the expression R/2 greater than or equal to 2.5 ty .

Mathematics
1 answer:
Step2247 [10]4 years ago
4 0

Answer:

R ≥ 5

Step-by-step explanation:

\frac{R}{2} ≥ 2.5              To solve, multiply both sides by 2

2 x \frac{R}{2} ≥ 2.5 x 2

R ≥ 5

If this answer is correct, please make me Brainliest.

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Faith, Sarah, and Guadalupe take French, Spanish, and German. No person's language begins with the same letter as their first na
ipn [44]
Sarah- German
Guadalupe- French
Faith- Spanish

I figures this out by seeing the fact that they gave which was that Sarah’s best friend takes French and it can’t be Faith because they learn a language that doesn’t start with the same letter as their name. If Sarah’s friend takes French then it can’t be Faith. What’s left is Spanish and German and Sarah and Faith. Faith can’t take French and Sarah can’t take Spanish so what’s left is Spanish for Faith and German for Sarah.
7 0
3 years ago
Nerissa paid a total of $8 for 4 packs of pencils she purchased from the convenience store. Let p represent the cost of one pack
BARSIC [14]

Answer: 4*p=8

Step-by-step explanation:

<em>You can use inverse operation to answer this equation.</em>

  • <em>write out you problem: 4 x p =8</em>
  • <em>The inverse of multiplication is division</em>
  • <em>you do 4/4 which gives you one but the 4 will cancel itself out</em>
  • <em>Do 8/4 which gives you 2</em>
  • <em>under the equation write p = 2</em>
  • <em>And be sure to line the p up with the p, the equal sign with the equal sign, and  the 2  stays where it is ( which should be already lined up with the 8)</em>
3 0
3 years ago
A) Compute the sum
avanturin [10]
A)

To calculate this sum, we could use trigonometric identity:

\arcsin(x)-\arcsin(y)=\arcsin\left(x\sqrt{1-y^2}-y\sqrt{1-x^2}\right)

We have:

\sum\limits_{k=1}^n\arcsin\left[\dfrac{\sqrt{k^2+2k}-\sqrt{k^2-1}}{k(k+1)}\right]=\\\\\\=&#10;\sum\limits_{k=1}^n\arcsin\left[\dfrac{\sqrt{k^2+2k}}{k(k+1)}-\dfrac{\sqrt{k^2-1}}{k(k+1)}\right]=\\\\\\=&#10;\sum\limits_{k=1}^n\arcsin\left[\dfrac{\sqrt{k^2+2k+1-1}}{k(k+1)}-\dfrac{\sqrt{k^2-1}}{k(k+1)}\right]=\\\\\\=&#10;\sum\limits_{k=1}^n\arcsin\left[\dfrac{\sqrt{(k+1)^2-1}}{k(k+1)}-\dfrac{\sqrt{k^2-1}}{k(k+1)}\right]=\\\\\\&#10;

=\sum\limits_{k=1}^n\arcsin\left[\dfrac{1}{k}\cdot\dfrac{\sqrt{(k+1)^2-1}}{\sqrt{(k+1)^2}}-\dfrac{1}{k+1}\cdot\dfrac{\sqrt{k^2-1}}{\sqrt{k^2}}\right]=\\\\\\=&#10;\sum\limits_{k=1}^n\arcsin\left[\dfrac{1}{k}\cdot\sqrt{\dfrac{(k+1)&^2-1}{(k+1)^2}}-\dfrac{1}{k+1}\cdot\sqrt{\dfrac{k^2-1}{k^2}}\right]=\\\\\\=&#10;\sum\limits_{k=1}^n\arcsin\left[\dfrac{1}{k}\cdot\sqrt{1-\dfrac{1}{(k+1)^2}}-\dfrac{1}{k+1}\cdot\sqrt{1-\dfrac{1}{k^2}}\right]=\\\\\\&#10;

=\sum\limits_{k=1}^n\arcsin\left[\dfrac{1}{k}\cdot\sqrt{1-\left(\dfrac{1}{k+1}\right)^2}-\dfrac{1}{k+1}\cdot\sqrt{1-\left(\dfrac{1}{k}\right)^2}\right]=\\\\\\=&#10;\sum\limits_{k=1}^n\left[\arcsin\left(\dfrac{1}{k}\right)-\arcsin\left(\dfrac{1}{k+1}\right)\right]=\\\\\\

=\bigg[\arcsin(1)-\arcsin\left(\frac{1}{2}\right)\bigg]+\bigg[\arcsin\left(\frac{1}{2}\right)-\arcsin\left(\frac{1}{3}\right)\bigg]+\\\\\\+&#10;\bigg[\arcsin\left(\frac{1}{3}\right)-\arcsin\left(\frac{1}{4}\right)\bigg]+\dots+&#10;\bigg[\arcsin\left(\frac{1}{n}\right)-\arcsin\left(\frac{1}{n+1}\right)\bigg]=\\\\\\

=\arcsin(1)-\arcsin\left(\frac{1}{2}\right)+\arcsin\left(\frac{1}{2}\right)-\arcsin\left(\frac{1}{3}\right)+\arcsin\left(\frac{1}{3}\right)-\\\\\\-\arcsin\left(\frac{1}{4}\right)+\dots+\arcsin\left(\frac{1}{n}\right)-\arcsin\left(\frac{1}{n+1}\right)=\\\\\\=&#10;\arcsin(1)-\arcsin\left(\frac{1}{n+1}\right)=\dfrac{\pi}{2}-\arcsin\left(\frac{1}{n+1}\right)

So the answer is:

\sum\limits_{k=1}^n\arcsin\left[\dfrac{\sqrt{k^2+2k}-\sqrt{k^2-1}}{k(k+1)}\right]=\dfrac{\pi}{2}-\arcsin\left(\dfrac{1}{n+1}\right)}

B)

\sum\limits_{k=1}^\infty\arcsin\left[\dfrac{\sqrt{k^2+2k}-\sqrt{k^2-1}}{k(k+1)}\right]=\\\\\\=&#10;\lim\limits_{n\to\infty}\sum\limits_{k=1}^n\arcsin\left[\dfrac{\sqrt{k^2+2k}-\sqrt{k^2-1}}{k(k+1)}\right]=\\\\\\=&#10;\lim\limits_{n\to\infty}\Bigg(\dfrac{\pi}{2}-\arcsin\left(\dfrac{1}{n+1}\right)\Bigg)=\dfrac{\pi}{2}-\lim\limits_{n\to\infty}\arcsin\left(\dfrac{1}{n+1}\right)=\\\\\\=&#10;\Bigg\{\dfrac{1}{n+1}\xrightarrow{n\to\infty}0\Bigg\}=\dfrac{\pi}{2}-\arcsin(0)=\dfrac{\pi}{2}-0=\dfrac{\pi}{2}

So we prove that:

\sum\limits_{k=1}^\infty\arcsin\left[\dfrac{\sqrt{k^2+2k}-\sqrt{k^2-1}}{k(k+1)}\right]=\dfrac{\pi}{2}
7 0
3 years ago
It takes Aaron five minutes to squeeze the juice from three grapefruits. At this rate how many minutes will it take him to squee
Nitella [24]
Your answer would be 20 minutes.

12/3=4 grapefruits. 4*5=20
6 0
4 years ago
Can anyone help with this problem??
marishachu [46]

Answer:

blue line : y = 3

black horizontal line : y = 0

black vertical line : x = 0

Step-by-step explanation:

it is really simple.

blue line : no matter what value we pick for x, the functional value for y is always 3. and that defines the equation y = 3 + 0*x or simply y = 3.

the same with the black horizontal line. the difference is that y is now always 0. so, y = 0.

7 0
3 years ago
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