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mr_godi [17]
3 years ago
14

A class of sixth grade students is given 5 homework problems each day . The class has completed 35 homework problems. If x is th

e number of days, which equation should be used to determine how many days the class had homework
Mathematics
2 answers:
Cloud [144]3 years ago
7 0

Answer:

X=7

Step-by-step explanation:

MissTica3 years ago
6 0

Answer:

x=35/5

Step-by-step explanation:

5x=35, so 35/5=x. 35/5=7 so x=7.

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Half of sarans wire is equal to 2/5 of daniel's. chris has 3 times as much as sarah. in all, their wire is 6 feet. how long is s
Pepsi [2]
You need to make a series of equations from what you are given first. I am going to use the first letter of each of the names to represent the length of that persons wire.
1/2s=2/5d
3c=s
s+d+c=6 ft

Okay. Now you can combine the first two equations knowing what s equals:
1/2(3c)=2/5d
d=15c/4

Now you have d=15c/4 and s=3c, so you can replace d and s in the third equation.

3c+15c/4+c=6

Then solve for c and plug it into the equation 3c=s to find the length of sarah's wire.
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3 years ago
Please help!!!!!!!!!
Andrews [41]
Since sides kl and kj are congruent that means that angles klj and kjl are congruent. if angle klj is 58 then that means kjl is also 58. so you take the total number of degrees in a triangle (180) and subtract 58 and 58 (116) which leaves you with 64 ( the degree measure of lkj) 
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3 years ago
Field book of an land is given in the figure. It is divided into 4 plots . Plot I is a right triangle , plot II is an equilatera
Kazeer [188]

Answer:

Total area = 237.09 cm²

Step-by-step explanation:

Given question is incomplete; here is the complete question.

Field book of an agricultural land is given in the figure. It is divided into 4 plots. Plot I is a right triangle, plot II is an equilateral triangle, plot III is a rectangle and plot IV is a trapezium, Find the area of each plot and the total area of the field. ( use √3 =1.73)

From the figure attached,

Area of the right triangle I = \frac{1}{2}(\text{Base})\times (\text{Height})

Area of ΔADC = \frac{1}{2}(\text{CD})(\text{AD})

                        = \frac{1}{2}(\sqrt{(AC)^2-(AD)^2})(\text{AD})

                        = \frac{1}{2}(\sqrt{(13)^2-(19-7)^2} )(19-7)

                        = \frac{1}{2}(\sqrt{169-144})(12)

                        = \frac{1}{2}(5)(12)

                        = 30 cm²

Area of equilateral triangle II = \frac{\sqrt{3} }{4}(\text{Side})^2

Area of equilateral triangle II = \frac{\sqrt{3}}{4}(13)^2

                                                = \frac{(1.73)(169)}{4}

                                                = 73.0925

                                                ≈ 73.09 cm²

Area of rectangle III = Length × width

                                 = CF × CD

                                 = 7 × 5

                                 = 35 cm²

Area of trapezium EFGH = \frac{1}{2}(\text{EF}+\text{GH})(\text{FJ})

Since, GH = GJ + JK + KH

17 = \sqrt{9^{2}-x^{2}}+5+\sqrt{(15)^2-x^{2}}

12 = \sqrt{81-x^2}+\sqrt{225-x^2}

144 = (81 - x²) + (225 - x²) + 2\sqrt{(81-x^2)(225-x^2)}

144 - 306 = -2x² + 2\sqrt{(81-x^2)(225-x^2)}

-81 = -x² + \sqrt{(81-x^2)(225-x^2)}

(x² - 81)² = (81 - x²)(225 - x²)

x⁴ + 6561 - 162x² = 18225 - 306x² + x⁴

144x² - 11664 = 0

x² = 81

x = 9 cm

Now area of plot IV = \frac{1}{2}(5+17)(9)

                                = 99 cm²

Total Area of the land = 30 + 73.09 + 35 + 99

                                    = 237.09 cm²

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What is subtraction?
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Wen you deduct an amount from something
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