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Arisa [49]
3 years ago
8

A boat's crew rowed 15 kilometers downstream, with the current, in 1 hours. The return trip upstream, against the current, cover

ed the same distance, and it took 3 hours. Find the crew's rowing rate in still water.
Mathematics
1 answer:
Mandarinka [93]3 years ago
3 0
B and c are the speeds of the boat (in still water) and the current.

Traveling with the current, the boat goes b+c km per hour.
1 hour * (b+c km)/hour = 15 km
b+c = 15

Traveling against the current, the boat goes b-c km per hour.
3 hours * (b-c km)/hour = 15
b-c = 5

Add the two equations
b = 10 km per hour
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25ft+25ft+20ft+20ft+2.5ft+2.5ft+8ft+13ft
Anna71 [15]

we are given

25ft+25ft+20ft+20ft+2.5ft+2.5ft+8ft+13ft

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so, we can take out ft

=(25+25+20+20+2.5+2.5+8+13)ft

now, we can add them

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7 0
4 years ago
What is the quotient of 3/7
wel

Answer:

is 3/7

Step-by-step explanation:

4 0
3 years ago
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What is 4 divided by 298
OLga [1]
 The answer is 74.5 if you  saying 4 divided by 298 then that is the right answer

8 0
3 years ago
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Calculate the straight line distance between the points (-10, -7) and (-8,1).<br> G.2(B)
tia_tia [17]

Answer:

2√(17) or about 8.2462 units.

Step-by-step explanation:

We want to determine the distance between the two points (-10, -7) and (-8, 1).

We can use the distance formula. Recall that:

\displaystyle d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}

Substitute and evaluate:

\displaystyle \begin{aligned} d &= \sqrt{((-10)-(-8))^2 + ((-7) - (1))^2} \\ \\ &=\sqrt{(-2)^2+(-8)^2} \\ \\ &= \sqrt{(4) + (64)} \\ \\ &= \sqrt{68}\\ \\ &= 2\sqrt{17}\end{aligned}

Hence, the distance between (-10, -7) and (-8, 1) is 2√(17) units or about 8.2462 units.

4 0
3 years ago
Tomika heard that the diagonals of a rhombus are perpendicular to each other. Help her test her conjecture. Graph quadrilateral
Stella [2.4K]

Answer:

a. The four sides of the quadrilateral ABCD are equal, therefore, ABCD is a rhombus

b. The equation of the diagonal line AC is y = 5 - x

The equation of the diagonal line BD is y = 5 - x

c. The diagonal lines AC and BD of the quadrilateral ABCD are perpendicular to each other

Step-by-step explanation:

The vertices of the given quadrilateral are;

A(1, 4), B(6, 6), C(4, 1) and D(-1, -1)

a. The length, l, of the sides of the given quadrilateral are given as follows;

l = \sqrt{\left (y_{2}-y_{1}  \right )^{2}+\left (x_{2}-x_{1}  \right )^{2}}

The length of side AB, with A = (1, 4) and B = (6, 6) gives;

l_{AB} = \sqrt{\left (6-4  \right )^{2}+\left (6-1  \right )^{2}} = \sqrt{29}

The length of side BC, with B = (6, 6) and C = (4, 1) gives;

l_{BC} = \sqrt{\left (1-6  \right )^{2}+\left (4-6  \right )^{2}} = \sqrt{29}

The length of side CD, with C = (4, 1) and D = (-1, -1) gives;

l_{CD} = \sqrt{\left (-1-1  \right )^{2}+\left (-1-4  \right )^{2}} = \sqrt{29}

The length of side DA, with D = (-1, -1) and A = (1,4)   gives;

l_{DA} = \sqrt{\left (4-(-1)  \right )^{2}+\left (1-(-1)  \right )^{2}} = \sqrt{29}

Therefore, each of the lengths of the sides of the quadrilateral ABCD are equal to √(29), and the quadrilateral ABCD is a rhombus

b. The diagonals are AC and BD

The slope, m, of AC is given by the formula for the slope of a straight line as follows;

Slope, \, m =\dfrac{y_{2}-y_{1}}{x_{2}-x_{1}}

Therefore;

Slope, \, m_{AC} =\dfrac{1-4}{4-1} = -1

The equation of the diagonal AC in point and slope form is given as follows;

y - 4 = -1×(x - 1)

y = -x + 1 + 4

The equation of the diagonal AC is y = 5 - x

Slope, \, m_{BD} =\dfrac{-1-6}{-1-6} = 1

The equation of the diagonal BD in point and slope form is given as follows;

y - 6 = 1×(x - 6)

y = x - 6 + 6 = x

The equation of the diagonal BD is y = x

c. Comparing the lines AC and BD with equations, y = 5 - x and y = x, which are straight line equations of the form y = m·x + c, where m = the slope and c = the x intercept, we have;

The slope m for the diagonal AC = -1 and the slope m for the diagonal BD = 1, therefore, the slopes are opposite signs

The point of intersection of the two diagonals is given as follows;

5 - x = x

∴ x = 5/2 = 2.5

y = x = 2.5

The lines intersect at (2.5, 2.5), given that the slopes, m₁ = -1 and m₂ = 1 of the diagonals lines satisfy the condition for perpendicular lines m₁ = -1/m₂, therefore, the diagonals are perpendicular.

5 0
3 years ago
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