(a) The area of a rectangular field is 5336 m².If the width of the field is 58 m, what is its length?
Length of the field: 1 m

Answers

Answer 1
Answer:

Answer:

92m².

Step-by-step explanation:

well if you just divide 5336m² with 58m² you get 92m² I think The Height is 92m²


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72y^5 + 18y^2 what is the GCF? or is this prime?

Answers

GCF = 18y^2 because both can be divided by this 

Algebraically solve the system of equations shown below. Note that you can use either factoring or the quadratic formula to find the X – coordinates, but the quadratic formula is probably easier.

Answers

6x^2+19x-15=-12x+15\n6x^2+31x-30=0\n6x^2+36x-5x-30=0\n6x(x+6)-5(x+6)=0\n(6x-5)(x+6)=0\nx=(5)/(6) \vee x=-6\n\ny=-12\cdot(5)/(6) +15\vee y=-12\cdot(-6)+15\ny=-10 +15\vee y=72+15\ny=5 \vee y=87\n\nx=(5)/(6) \wedge y=5\nx=-6 \wedge y=87
y=6x^2+19x-15 \ny=-12x+15 \n \n6x^2+19x-15=-12x+15 \n6x^2+19x+12x-15-15=0 \n6x^2+31x-30=0 \n \n a=6 \n b=31 \n c=-30 \n \n x=(-b \pm √(b^2-4ac))/(2a)=(-31 \pm √(31^2-4 \cdot 6 \cdot (-30)))/(2 \cdot 6)=(-31 \pm √(1681))/(12)=(-31 \pm 41)/(12) \n x=(-31 -41)/(12) \ \hbox{or} \ x=(-31+41)/(12) \nx=-6 \ \hbox{or} \ x=(5)/(6)

y=-12 \cdot (-6)+15 \ \hbox{or} \ y=-12 \cdot (5)/(6)+15 \ny=87 \ \hbox{or} \ y=5 \n \n \hbox{the answer:} \n \boxed{x=-6, \ y=87} \ \hbox{or} \ \boxed{x=(5)/(6), \ y=5}

Chad washes windows after school to make some extra money. He charges $5.50 to wash each window. If the customer provides the supplies, Chad deducts $3.25 from the total cost. One customer paid a total of $27.50 and did not provide supplies. Which equation could be used to find the number of windows, w, that Chad washed for this customer? 5.5w + 3.25 = 27.5 5.5w - 3.25 = 27.5 5.5w = 27.5 5.5 - 3.25w = 27.5

Answers

Answer:

\$5.50w=\$27.50

Step-by-step explanation:

We know that Chad charges $5.50 per window washed, and he deducts $3.25 from the total cost if the costumer provides the supplies. However, in this case, the costumer doesn't provide the supplies, so Chad will charge $5.50 per window.

This relation can be expressed as

\$5.50w

Where w represents windows.

If he charges $27.50, the relation is

\$5.50w=\$27.50

Therefore, the right answer is the third choice \$5.50w=\$27.50

If you want to find the number of windows Chad washed, you just have to solve the expression

\$5.50w=\$27.50\nw=(\$27.50)/(\$5.50)\n w=5

Chad washed 5 windows.

3.25w=27.5 is the correct answer.

BEST ANSWER GETS BRAINLIESTT!! PLEASE HELPWhat conclusion can be determined from the dot plot below?

A dot plot showing five dots above 9, six dots above 10, three dots above 11, and one dot above 12.

The number of observations is 10.
The median of the data set is 10.
The mean of the data set is 15.
The range of the data set is 12.

Answers

the answer would be the median is 10 because if u actually take the time and draw it you x everything out and see whats in the middle and you will have 10 dots lolz hope this helped

From the numbers from the result had to be added it would be 47, but that is not the point all of the number are between the numbers 9 and 16. If the dot plot was set to a line going diagonally.

Write the equation of the line that has a y-intercept of (0,4) and a slope of 2.

Answers

Answer:

y=2x+4

Step-by-step explanation:

the y-intercept is the y value of the point given. in the slope form, 2 would be your m

The table shows data for a class's mid-term and final exams Which data set has the largest IQR?Mid-Term Final
100 98
100 95
100 93
95 91
95 88
92 82
92 78
88 78
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75 60

Answers

Putting the numbers in order;

Mid-term numbers: 100 100 100 95 95 92 92 88 85 75

Median = (95 + 92) / 2 = 93.5

First quarter = Q1 = 100 100 100 95

Median of first quarter = (100 + 100) / 2 = 100

Third quarter = Q3 = 92 88 85 75

Median of third quarter = (88 + 85) / 2 = 86.5

Interquartile range (IQR) can be calculated bysubtracting Q3 from Q1;

IQR of mid-term = 86.5 – 100 = -13.5

Now putting the final term numbers in order;

Final-term numbers: 98 95 93 91 88 82 78 78 65 60

Median = (88 + 82) / 2 = 85

First quarter = Q1 = 98 95 91 88

Median of first quarter = (95 + 91) / 2 = 93

Third quarter = Q3 = 78 78 65 60

Median of third quarter = (78 + 65) / 2 = 71.5

Interquartile range (IQR) can be calculated bysubtracting Q3 from Q1;

IQR of final-term = 71.5 – 93 = -21.5

Thus the IQR of mid-term is the largest.