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Helen [10]
3 years ago
15

Please help me with this problem. I have tried many times and have gotten it wrong. I need help especially on the last part. Can

I also have a good explanation because I want to know how to do problems like this in the future. Also please answer at least the bottom two parts.

Mathematics
1 answer:
Arlecino [84]3 years ago
5 0

1) The domain is all the possible x values in the function so it would be [-4,4]

2) There are only 3 zeros shown on the graph and they are (-2, 0) (0, 0) (2, 0) the zeros are the value of x when y = 0.

3)The function is positive/Negative is asking for what x values make the y values positive aka interval notation. The function is positive if x = (0, 2) because 0 and 2 aren't included you use parentheses () instead of brackets []  

The function is negative if x [-4, -2), (-2,0), (2,4]

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Simpifly 6/30 <br> ..... ........
ollegr [7]

Answer:

1/5

Step-by-step explanation:

Divide both numerator and denominator by 6:

1/5

7 0
3 years ago
Read 2 more answers
Ye has his own business. He checks his sales receipts three times a day, his afternoon sales were 50$ more than his mourning sal
IRINA_888 [86]

Answer:

The answer is 150

Step-by-step explanation:

It's 150 because ye evening sales are three times of his afternoon sales.

SO that means you have to times 50x3=150 or you can add 50+50+50=150. there is your answer.

3 0
3 years ago
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Who's taking geometry and can help me?
Mariulka [41]
I seen your comment on the other persons thing, it should be 60° :) correct me if i’m wrong!
5 0
3 years ago
Questions attached as screenshot below:Please help me I need good explanations before final testI pay attention
Nikitich [7]

The acceleration of the particle is given by the formula mentioned below:

a=\frac{d^2s}{dt^2}

Differentiate the position vector with respect to t.

\begin{gathered} \frac{ds(t)}{dt}=\frac{d}{dt}\sqrt[]{\mleft(t^3+1\mright)} \\ =-\frac{1}{2}(t^3+1)^{-\frac{1}{2}}\times3t^2 \\ =\frac{3}{2}\frac{t^2}{\sqrt{(t^3+1)}} \end{gathered}

Differentiate both sides of the obtained equation with respect to t.

\begin{gathered} \frac{d^2s(t)}{dx^2}=\frac{3}{2}(\frac{2t}{\sqrt[]{(t^3+1)}}+t^2(-\frac{3}{2})\times\frac{1}{(t^3+1)^{\frac{3}{2}}}) \\ =\frac{3t}{\sqrt[]{(t^3+1)}}-\frac{9}{4}\frac{t^2}{(t^3+1)^{\frac{3}{2}}} \end{gathered}

Substitute t=2 in the above equation to obtain the acceleration of the particle at 2 seconds.

\begin{gathered} a(t=1)=\frac{3}{\sqrt[]{2}}-\frac{9}{4\times2^{\frac{3}{2}}} \\ =1.32ft/sec^2 \end{gathered}

The initial position is obtained at t=0. Substitute t=0 in the given position function.

\begin{gathered} s(0)=-23\times0+65 \\ =65 \end{gathered}

8 0
1 year ago
Edgar owns 234 shares of Cawh Consolidated Bank, which he bought for $21.38 apiece. Each share pays a yearly dividend of $3.15.
trapecia [35]

Answer: B. The stocks have a yield 6.84 percentage points greater than that of the bonds.

Step-by-step explanation:

Firstly, the yield for stocks will be calculated as:

= return/ investment cost

= $3.15/$ 21.38

= 0.14733395

= 14.73%

The yield for bonds will be calculated as:

= Return/Investment cost

Return = 1,000 x 8.3% = 83

Investment cost = 1,000 x 105.166/100 = 1051.66‬

Yield = 83/1051.66

= 0.07892284

= 7.89%

Then, the difference between the yield will be:

= 14.73% - 7.89%

= 6.84%

Therefore, the stocks have a yield 6.84 percentage points greater than that of the bonds.

8 0
3 years ago
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