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3 years ago

How do i do this plz help me

Mathematics

2 answers:

erastova [34]3 years ago

3 0

79,002 this is the answer

Aleksandr-060686 [28]3 years ago

3 0

You ALWAYS start math problems from right to left. WHENEVER you're multiplying you take the WHOLE top number and multiply the bottom row ONE number at a time. Step 2: Whenever your done multiplying one number you put a ZERO on the far left UNDER the first set of numbers you multiplied together. Step 3: You do it again for the next set. Step 4:When your done add up all numbers when finished and you have your answer

Your answer should be:
31,122

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How many solutions does the equation have, x1 + x2 + x3 = 10 , where x1 , x2, and x3 are non-negative integers?

MAVERICK [17]

Non-negative integers are positive integers or zero.

1. When $x_1=0,$ then there are such possible cases for $x_2$ and $x_3:$

$x_2=0,\ x_3=10;$
$x_2=1,\ x_3=9;$
$x_2=2,\ x_3=8;$
$x_2=3,\ x_3=7;$
$x_2=4,\ x_3=6;$
$x_2=5,\ x_3=5;$
$x_2=6,\ x_3=4;$
$x_2=7,\ x_3=3;$
$x_2=8,\ x_3=2;$
$x_2=9,\ x_3=1;$
$x_2=10,\ x_3=0.$

In total 11 solutions for $x_1=0.$

2. For $x_1=1,$ there are such possible cases for $x_2$ and $x_3:$

$x_2=0,\ x_3=9;$
$x_2=1,\ x_3=8;$
$x_2=2,\ x_3=7;$
$x_2=3,\ x_3=6;$
$x_2=4,\ x_3=5;$
$x_2=5,\ x_3=4;$
$x_2=6,\ x_3=3;$
$x_2=7,\ x_3=2;$
$x_2=8,\ x_3=1;$
$x_2=9,\ x_3=0.$

In total 10 solutions for $x_1=1.$

3. This process gives you

for $x_1=2$ - 9 solutions;
for $x_1=3$ - 8 solutions;
for $x_1=4$ - 7 solutions;
for $x_1=5$ - 6 solutions;
for $x_1=6$ - 5 solutions;
for $x_1=7$ - 4 solutions;
for $x_1=8$ - 3 solutions;
for $x_1=9$ - 2 solutions;
for $x_1=10$ - 1 solution.

4. Add all numbers of solutions:

$11+10+9+8+7+6+5+4+3+2+1=66.$

Answer: there are 66 possible solutions (with non-negative integer variables)

8 0

3 years ago

Read 2 more answers

Mike thought of a number. He added to it to two other numbers. One of these numbers was 12 less than his number, the other was 1

Inessa [10]

Answer:

number 1 + number 2 + number 3 = 15

Number 1 = x + 12

Number 2 = x - 12

Number 3 = x

(x + 12) + (x - 12) + x = 15

x + 12 + x - 12 + x = 15

x + x + x +12 -12 = 15

3x = 15

x = 15/3 = 5

Check:

Number 1 = 17

Number 2 = 5 -12 = -7

Number = 5

number 1 + number 2 + number 3 = 15

17 + (-7) + 5 = 15

10 + 5 = 15

15 = 15

Answer is 5

Step-by-step explanation:

5 0

3 years ago

Which equation is not equal to 6^3/6^6 ?: 1/6^2 6^-3 1/216 1/6^3

SCORPION-xisa [38]

The answer would be 60 because its multiplication all those numbers

4 0

3 years ago

Help pls- <br><br> I dont know it T^T

docker41 [41]

Answer:

Ans: H and A

Step-by-step explanation:

By the identity:

1-(cos(x)) $cos^{2}(x)+ sin^{2}(x)=1\\cos^{2}(x)=1- sin^{2}(x)\\\\cos(x)=\sqrt{1- sin^{2}(x)} \\sin(x)=\sqrt{1- cos^{2}(x)}\\\\\sqrt{1- cos^{2}(x)}/sinx=1\\\sqrt{1- sin^{2}(x)}/cosx=1\\\sqrt{1- cos^{2}(x)}/sinx+\sqrt{1- sin^{2}(x)}/cosx=2\\\\\\$

Ans: H

For second quesiton, the transformation of a graph is that:

f(x) + k means vertical translation up by k units

Two curve have the same maximum value, so the graph doesn't translate upwards or downwards, thus b=0

Ans: A

7 0

3 years ago

If sin=3/5 then what cos equal to

kap26 [50]

$\sin {}^{2} ( \alpha ) + \cos {}^{2} ( \alpha ) = 1$
$\frac{9}{25} + \cos {}^{2} ( \alpha ) = 1$
$\cos {}^{2} ( \alpha ) = \frac{16}{25}$
$\cos( \alpha ) = \frac{4}{5}$

8 0

3 years ago

Read 2 more answers