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Lelu [443]
3 years ago
7

Simplify the expression

Mathematics
1 answer:
kozerog [31]3 years ago
8 0
10[(6 + 4) / 2]
10[10/2]
10 * 5
50 <==
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A car travels a distance of 500 km in a time of 5 hours. <br> What is the car's average speed?
FrozenT [24]

Answer:

100km/h

Step-by-step explanation:

The formula for speed is:

v  =  x / d  =  500/5  =  100km/h

7 0
3 years ago
There is 1500 ft of fencing available to make 6 identical pens. Find the maximum area for EACH pen (meaning maximum of one pen).
hjlf

Answer:

The maximum area is therefore is 93750 ft²

Step-by-step explanation:

The given parameter are;

The a]length of fencing available = 1500 ft.

The perimeter of the figure = 9·x + 4·y

Therefore, 9·x + 4·y = 1500 ft.

The area of the figure = 6 × (x × y) = 3·x × 2·y

From the equation for the perimeter, we have;

9·x + 4·y = 1500

y = 1500/4 - 9/4·x = 375 - 9/4·x

y = 375 - 9/4·x

Substituting the value of y in the equation for the area gives;

Area = 3·x × 2·y = 3·x × 2·(375 - 9/4·x) = 2250·x - 27/2·x²

Area = 2250·x - 27/2·x²

The maximum area is found by taking the derivative and equating to zero as follows;

d(2250·x - 27/2·x²)/dx = 0

2250 - 27·x = 0

x = 2250/27 = 250/3

x = 250/3

y = 375 - 9/4·x = 375 - 9/4×250/3 = 187.5

The maximum area is therefore, 3·x × 2·y = 3 × 250/3 × 2 × 187.5 = 93750 ft²

The maximum area is therefore = 93750 ft.²

4 0
3 years ago
Use the substitution method to solve the following system of equations:
Morgarella [4.7K]
3x + 5y = 3
x + 2y = 0

3x + 5y = 3
-3x -6y = 0

-y = 3
y = -3

x + 2y = 0
x + -6 = 0
x = 6


The answer is B (6,-3).
8 0
3 years ago
Read 2 more answers
Find the vectors T, N, and B at the given point. r(t) = &lt; t^2, 2/3t^3, t &gt;, (1, 2/3 ,1)
maxonik [38]

Answer with Step-by-step explanation:

We are given that

r(t)=< t^2,\frac{2}{3}t^3,t >

We have to find T,N and B at the given point t > (1,2/3,1)

r'(t)=

\mid r'(t) \mid=\sqrt{(2t)^2+(2t^2)^2+1}=\sqrt{(2t^2+1)^2}=2t^2+1

T(t)=\frac{r'(t)}{\mid r'(t)\mid}=\frac{}{2t^2+1}

Now, substitute t=1

T(1)=\frac{}{2+1}=\frac{1}{3}

T'(t)=\frac{-4t}{(2t^2+1)^2} +\frac{1}{2t^2+1}

T'(1)=-\frac{4}{9}+\frac{1}{3}

T'(1)=\frac{1}{9}=

\mid T'(1)\mid=\sqrt{(\frac{-2}{9})^2+(\frac{4}{9})^2+(\frac{-4}{9})^2}=\sqrt{\frac{36}{81}}=\frac{2}{3}

N(1)=\frac{T'(1)}{\mid T'(1)\mid}

N(1)=\frac{}{\frac{2}{3}}=

N(1)=

B(1)=T(1)\times N(1)

B(1)=\begin{vmatrix}i&j&k\\\frac{2}{3}&\frac{2}{3}&\frac{1}{3}\\\frac{-1}{3}&\frac{2}{3}&\frac{-2}{3}\end{vmatrix}

B(1)=i(\frac{-4}{9}-\frac{2}{9})-j(\frac{-4}{9}+\frac{1}{3})+k(\frac{4}{9}+\frac{2}{9})

B(1)=-\frac{2}{3}i+\frac{1}{3}j+\frac{2}{3}k

B(1)=\frac{1}{3}

5 0
3 years ago
Find each sum or difference. Then use front-
rosijanka [135]
The answer is 1, 240.
8 0
3 years ago
Read 2 more answers
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