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3 years ago

Juan bought 1 1/9 gallons of black paint, 2 1/3 gallons of white paint, and a 3/4 inch paint brush.

Mathematics

2 answers:

stepan [7]3 years ago

5 0

it looks like C THAT looks good

jasenka [17]3 years ago

5 0

I believe that this wonderful questions answer is letter C

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Gala2k [10]

Number one. 897 2. Eight hundred eighty four and fourth seven hundredths. Number three. 17.818. 4. Two thousand forty four and four tenths. Number five. 22.766 6. Twelve thousand five hundred and twelve Number seven. 23.62

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2 years ago

What is 2km / 29/2minutes

Eduardwww [97]

Answer: simple 2.29885057 m / s

Step-by-step explanation:

8 0

3 years ago

What is the slope m and y-intercept point of the line y = 2x 1?

muminat

The slope is 2 and the y-intercept is 1.

8 0

3 years ago

Chris built a rectangular snow fort with a perimeter of 24 feet. The length of the fort was 8 feet less than 3 time the width. W

melomori [17]

The perimeter of a polygon is equal to the sum of all its sides. In a rectangle with dimensions length, x and width, y is calculated through the equation,

P = 2x + 2y

If width is y, then length is equal to 3y - 8. Substituting these to the equaiton,

P = 2(3y - 8) + 2y

Simplifying,
24 = 6y - 16 + 2y
24 = 8y - 16

The value of y from the equation is 5.

x = 3y - 8 = 3(5) - 8 = 7

Answers:
x = 7 feet
y = 5 feet

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3 years ago

There is a single sequence of integers $a_2$, $a_3$, $a_4$, $a_5$, $a_6$, $a_7$ such that \[\frac{5}{7} = \frac{a_2}{2!} + \frac

Nataliya [291]

You have a single sequence of integers $a_2,\ a_3,\ a_4,\ a_5,\ a_6,\ a_7$ such that

$\dfrac{a_2}{2!} + \dfrac{a_3}{3!} + \dfrac{a_4}{4!} + \dfrac{a_5}{5!} + \dfrac{a_6}{6!} + \dfrac{a_7}{7!}=\dfrac{5}{7},$

where $0 \le a_i < i$ for $i = 2, 3, \dots, 7.$

1. Multiply by 7! to get

$\dfrac{7!a_2}{2!} + \dfrac{7!a_3}{3!} + \dfrac{7!a_4}{4!} + \dfrac{7!a_5}{5!} + \dfrac{7!a_6}{6!} + \dfrac{7!a_7}{7!}=\dfrac{7!\cdot 5}{7},\\ \\7\cdot 6\cdot 5\cdot 4\cdot 3\cdoa a_2+7\cdot 6\cdot 5\cdot 4\cdot a_3+7\cdot 6\cdot 5\cdot a_4+7\cdot 6\cdot a_5+7\cdot a_6+a_7=6!\cdot 5,\\ \\7(6\cdot 5\cdot 4\cdot 3\cdoa a_2+6\cdot 5\cdot 4\cdot a_3+6\cdot 5\cdot a_4+6\cdot a_5+a_6)+a_7=3600.$

By Wilson's theorem,

$a_7+7\cdot (6\cdot 5\cdot 4\cdot 3\cdoa a_2+6\cdot 5\cdot 4\cdot a_3+6\cdot 5\cdot a_4+6\cdot a_5+a_6)\equiv 2(\mod 7)\Rightarrow a_7=2.$

2. Then write $a_7$ to the left and divide through by 7 to obtain

$6\cdot 5\cdot 4\cdot 3\cdoa a_2+6\cdot 5\cdot 4\cdot a_3+6\cdot 5\cdot a_4+6\cdot a_5+a_6=\dfrac{3600-2}{7}=514.$

Repeat this procedure by $\mod 6:$

$a_6+6(5\cdot 4\cdot 3\cdoa a_2+ 5\cdot 4\cdot a_3+5\cdot a_4+a_5)\equiv 4(\mod 6)\Rightarrow a_6=4.$

And so on:

$5\cdot 4\cdot 3\cdoa a_2+ 5\cdot 4\cdot a_3+5\cdot a_4+a_5=\dfrac{514-4}{6}=85,\\ \\a_5+5(4\cdot 3\cdoa a_2+ 4\cdot a_3+a_4)\equiv 0(\mod 5)\Rightarrow a_5=0,\\ \\4\cdot 3\cdoa a_2+ 4\cdot a_3+a_4=\dfrac{85-0}{5}=17,\\ \\a_4+4(3\cdoa a_2+ a_3)\equiv 1(\mod 4)\Rightarrow a_4=1,\\ \\3\cdoa a_2+ a_3=\dfrac{17-1}{4}=4,\\ \\a_3+3\cdot a_2\equiv 1(\mod 3)\Rightarrow a_3=1,\\ \\a_2=\dfrac{4-1}{3}=1.$

Answer: $a_2=1,\ a_3=1,\ a_4=1,\ a_5=0,\ a_6=4,\ a_7=2.$

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3 years ago