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Lisa [10]
3 years ago
5

Solve 7x + 1 < 4(x - 2).

Mathematics
1 answer:
luda_lava [24]3 years ago
8 0

The general approach to solving any such inequality is to get the variable by itself on one side of the relationship symbol. The instructions for this can vary, but here is a set of instructions that will always work.

1. Subtract the expression on the right side of the relationship symbol.

... 7x + 1 < 4(x - 2) . . . . starting inequality

... 7x +1 - 4(x -2) < 0 . . . after step 1

2. Eliminate parentheses and collect terms.

... 7x +1 -4x +8 < 0

... 3x +9 < 0 . . . . . . . . . after step 2

3. Divide by the coefficient of the variable. Here, divide by 3. If the coefficient of the variable is negative, also reverse the relationship symbol (≥ becomes ≤, for example).

... x + 3 < 0 . . . . . . . . . we divided by 3

4. Add the opposite of the constant, if any.

... x < -3 . . . . . . . . . . . . after adding -3

This solution matches your 4th selection:

... {x | x < -3}

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14.9

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8 0
2 years ago
How many even 5-digit numbers can be formed from the numbers 1, 2, 4, 7, 8 if no digit is repeated in a number? A. 72 B. 120 C.
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Hello,

Let's place the last digit: it must be 2 or 4 or 8 (3 possibilities)

It remainds 4 digits and the number of permutations fo 4 numbers is 4!=4*3*2*1=24

Thus there are 3*24=72 possibilities.

Answer A

If you do'nt believe run this programm

DIM n(5) AS INTEGER, i1 AS INTEGER, i2 AS INTEGER, i3 AS INTEGER, i4 AS INTEGER, i5 AS INTEGER, nb AS LONG, tot AS LONG
tot = 0
n(1) = 1
n(2) = 2
n(3) = 4
n(4) = 7
n(5) = 8
FOR i1 = 1 TO 5
    FOR i2 = 1 TO 5
        IF i2 <> i1 THEN
            FOR i3 = 1 TO 5
                IF i3 <> i2 AND i3 <> i1 THEN
                    FOR i4 = 1 TO 5
                        IF i4 <> i3 AND i4 <> i2 AND i4 <> i1 THEN
                            FOR i5 = 1 TO 5
                                IF i5 <> i4 AND i5 <> i3 AND i5 <> i2 AND i5 <> i1 THEN
                                    nb = ((((n(i1) * 10) + n(i2)) * 10 + n(i3)) * 10 + n(i4)) * 10 + n(i5)
                                    IF nb MOD 2 = 0 THEN
                                        tot = tot + 1
                                    END IF
                                END IF
                            NEXT i5
                        END IF
                    NEXT i4
                END IF
            NEXT i3
        END IF
    NEXT i2
NEXT i1
PRINT "tot="; tot
END


7 0
3 years ago
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