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Irina-Kira [14]
3 years ago
8

25 times what times 3 equals 435​

Mathematics
2 answers:
11111nata11111 [884]3 years ago
7 0
Let x be the number you want to find
435= x times 25 times 3
435= 75x
X= 435/75 =5.8 (ans)
Grace [21]3 years ago
7 0

Answer:

5⅘

Step-by-step explanation:

Divide 435 by 3 to get 145, then divide that by 25 to get 5⅘.

I am joyous to assist you anytime.

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How you know that the values of the digit in the numbers 1500,000 and 100,500 ate not the same
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In this number 1500,000,

1 is at million place and 5 is at hundred thousand place

So, the value of 1 = 1 (1,000,000) = 1 million

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In the second number 100,500

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The value of 1 = 1 (100,000) = 100,000 = Hundred thousand

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4 years ago
Which graph represents f(x)=4⋅5x ?
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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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