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IRINA_888 [86]
3 years ago
8

one of the requirements of becoming a court reporter is the ability to type 225 words per minute. Donald can currently type 180

words per minute, and believes that with practice he can increase his typing speed by 5 words per minute each month. Which of the following represents the number of words per minute that don't believe he'll be able to type M months from now?<br /> A) 5+180m<br /> B) 225+5m<br /> C) 180+5m<br /> D) 180-5m
Mathematics
1 answer:
Sedaia [141]3 years ago
7 0
D because it is always subtract 5 words every month
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Please help, i need to figure out the price of bag, not proportional ($)
Lilit [14]

Answer:

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Step-by-step explanation:

5 0
3 years ago
You are creating bracelets and then selling them to your friends at a 85% markup rate. If it costs you $1.80 to create the brace
posledela

Answer:

you will add $1.53 to every bracelet you make

Step-by-step explanation:

figure out what 85 percent of 1.80 is and then add that to 1.80 for the total price

85% x 1.8 = 1.53 (markup ammount)

1.80 + 1.53 = 3.33 (total cost)

5 0
3 years ago
Based on the given segment lengths, determine whether the segments can form a triangle.
murzikaleks [220]

For 3 sides to form a triangle two sides added together need to be greater then the 3rd side. You need to check all 3 combinations:

1. 11.5 + 10.5 = 22, 22> 6.1

11.5 + 6.1 = 17.6 17.6 > 10.5

10.5 + 6.1 = 16.6, 16.6 > 11.5

This one can make a triangle.

2. 9.5 + 3.5 > 4.5

9.5 + 4.5 > 3.5

4.5 + 3.5 < 9.5

Cannot make a triangle.

3. 8 + 6.7 > 1

6.7 + 1 < 8

Cannot make a triangle.

4. Can make a triangle

5. Can

6. Can

4 0
3 years ago
99 POINT QUESTION, PLUS BRAINLIEST!!!
VladimirAG [237]
First, we have to convert our function (of x) into a function of y (we revolve the curve around the y-axis). So:


y=100-x^2\\\\x^2=100-y\qquad\bold{(1)}\\\\\boxed{x=\sqrt{100-y}}\qquad\bold{(2)} \\\\\\0\leq x\leq10\\\\y=100-0^2=100\qquad\wedge\qquad y=100-10^2=100-100=0\\\\\boxed{0\leq y\leq100}

And the derivative of x:

x'=\left(\sqrt{100-y}\right)'=\Big((100-y)^\frac{1}{2}\Big)'=\dfrac{1}{2}(100-y)^{-\frac{1}{2}}\cdot(100-y)'=\\\\\\=\dfrac{1}{2\sqrt{100-y}}\cdot(-1)=\boxed{-\dfrac{1}{2\sqrt{100-y}}}\qquad\bold{(3)}

Now, we can calculate the area of the surface:

A=2\pi\int\limits_0^{100}\sqrt{100-y}\sqrt{1+\left(-\dfrac{1}{2\sqrt{100-y}}\right)^2}\,\,dy=\\\\\\= 2\pi\int\limits_0^{100}\sqrt{100-y}\sqrt{1+\dfrac{1}{4(100-y)}}\,\,dy=(\star)

We could calculate this integral (not very hard, but long), or use (1), (2) and (3) to get:

(\star)=2\pi\int\limits_0^{100}1\cdot\sqrt{100-y}\sqrt{1+\dfrac{1}{4(100-y)}}\,\,dy=\left|\begin{array}{c}1=\dfrac{-2\sqrt{100-y}}{-2\sqrt{100-y}}\end{array}\right|= \\\\\\= 2\pi\int\limits_0^{100}\dfrac{-2\sqrt{100-y}}{-2\sqrt{100-y}}\cdot\sqrt{100-y}\cdot\sqrt{1+\dfrac{1}{4(100-y)}}\,\,dy=\\\\\\ 2\pi\int\limits_0^{100}-2\sqrt{100-y}\cdot\sqrt{100-y}\cdot\sqrt{1+\dfrac{1}{4(100-y)}}\cdot\dfrac{dy}{-2\sqrt{100-y}}=\\\\\\

=2\pi\int\limits_0^{100}-2\big(100-y\big)\cdot\sqrt{1+\dfrac{1}{4(100-y)}}\cdot\left(-\dfrac{1}{2\sqrt{100-y}}\, dy\right)\stackrel{\bold{(1)}\bold{(2)}\bold{(3)}}{=}\\\\\\= \left|\begin{array}{c}x=\sqrt{100-y}\\\\x^2=100-y\\\\dx=-\dfrac{1}{2\sqrt{100-y}}\, \,dy\\\\a=0\implies a'=\sqrt{100-0}=10\\\\b=100\implies b'=\sqrt{100-100}=0\end{array}\right|=\\\\\\= 2\pi\int\limits_{10}^0-2x^2\cdot\sqrt{1+\dfrac{1}{4x^2}}\,\,dx=(\text{swap limits})=\\\\\\

=2\pi\int\limits_0^{10}2x^2\cdot\sqrt{1+\dfrac{1}{4x^2}}\,\,dx= 4\pi\int\limits_0^{10}\sqrt{x^4}\cdot\sqrt{1+\dfrac{1}{4x^2}}\,\,dx=\\\\\\= 4\pi\int\limits_0^{10}\sqrt{x^4+\dfrac{x^4}{4x^2}}\,\,dx= 4\pi\int\limits_0^{10}\sqrt{x^4+\dfrac{x^2}{4}}\,\,dx=\\\\\\= 4\pi\int\limits_0^{10}\sqrt{\dfrac{x^2}{4}\left(4x^2+1\right)}\,\,dx= 4\pi\int\limits_0^{10}\dfrac{x}{2}\sqrt{4x^2+1}\,\,dx=\\\\\\=\boxed{2\pi\int\limits_0^{10}x\sqrt{4x^2+1}\,dx}

Calculate indefinite integral:

\int x\sqrt{4x^2+1}\,dx=\int\sqrt{4x^2+1}\cdot x\,dx=\left|\begin{array}{c}t=4x^2+1\\\\dt=8x\,dx\\\\\dfrac{dt}{8}=x\,dx\end{array}\right|=\int\sqrt{t}\cdot\dfrac{dt}{8}=\\\\\\=\dfrac{1}{8}\int t^\frac{1}{2}\,dt=\dfrac{1}{8}\cdot\dfrac{t^{\frac{1}{2}+1}}{\frac{1}{2}+1}=\dfrac{1}{8}\cdot\dfrac{t^\frac{3}{2}}{\frac{3}{2}}=\dfrac{2}{8\cdot3}\cdot t^\frac{3}{2}=\boxed{\dfrac{1}{12}\left(4x^2+1\right)^\frac{3}{2}}

And the area:

A=2\pi\int\limits_0^{10}x\sqrt{4x^2+1}\,dx=2\pi\cdot\dfrac{1}{12}\bigg[\left(4x^2+1\right)^\frac{3}{2}\bigg]_0^{10}=\\\\\\= \dfrac{\pi}{6}\left[\big(4\cdot10^2+1\big)^\frac{3}{2}-\big(4\cdot0^2+1\big)^\frac{3}{2}\right]=\dfrac{\pi}{6}\Big(\big401^\frac{3}{2}-1^\frac{3}{2}\Big)=\boxed{\dfrac{401^\frac{3}{2}-1}{6}\pi}

Answer D.
6 0
3 years ago
Read 2 more answers
Among all positive numbers , whose sum is 12, find those for which the product of the two numbers and their difference is larges
Karolina [17]

Answer: The numbers whose sum equals 12 and whose product and difference is largest is 6 and 6

Step-by-step explanation:

numbers whose sum is 12

Sum =12         Product       Difference               Product- Difference  

11 +1                   11                    10                                1

10 +2                20                   8                                 12

9 +3                  27                   6                                 21

 8 +4                 32                   4                                 28

7  +5                  35                   2                                33

6+6                    36                   0                                36

The numbers whose sum equals 12 and whose product and difference is largest is 6 and 6

4 0
3 years ago
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