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den301095 [7]
3 years ago
11

What is 12229 divided by 175? Please help.

Mathematics
1 answer:
alina1380 [7]3 years ago
3 0

Answer: 12229/175= 69.88

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Help me solve this question
irga5000 [103]
The correct answer is 1.

So, circle the 0 after the equals sign because the equation does not equal 0.

Anything raised by the power of 0 is 1.

This is known as the zero exponent rule :)
8 0
3 years ago
Find x and y part 4 for these problems​
Dvinal [7]

Answer:

Step-by-step explanation:

3) Sin30 = 11/x

x = 11/Sin30 = 11/0.5

x = 22

Tan 30 = 11/y

y = 11/tan30 = 11/0.5774

y = 19.1

4) Sin30 = 6/x

x = 6/Sin30 = 6/0.5

x = 12

Tan 30 = 6/y

y = 6/tan30 = 6/0.5774

y = 10.39

5) Sin45 = 9√2/y

y = 9√2/Sin45 = 9√2/(√2/2) =

9√2 × 2/√2 = 18

x = 18

Tan 45 = 9√2/x

x = 9√2/Tan 45 = 9√2/1

x = 9√2

6)

Sin60 = 9/x

x = 9/Sin60 = 9/0.866

x = 10.39

Tan 60 = 9/y/2 = 18/y

1.7321 = 18/y

y = 18/1.7321

y = 10.39

6 0
2 years ago
translate the sentence into an equation. the sum of 4 times a number and 8 equals 7. use the variable c for the unknown number
jek_recluse [69]

Answer: 4× 8c = 7

Step-by-step explanation:

3 0
3 years ago
A rectangle has dimensions of 15 centimeters long and 16 meters wide what is the area of the rectangle
Brrunno [24]
240 is the answer to your question just multiply them

6 0
3 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.
katovenus [111]
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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