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NARA [144]
2 years ago
6

What is 34,375 x 20 will give brainliest

Mathematics
2 answers:
antoniya [11.8K]2 years ago
5 0
687,500 is the answer
yan [13]2 years ago
4 0

Answer:

687500

Step-by-step explanation:

3 4 3 7 5

x        2 0

----------------------

+      0 0  0

+  6 8 7 5 0

---------------------

687500

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How many pairs of whole numbers can you find that have a sum of 12?
Arte-miy333 [17]

There are 7 pairs of whole numbers that have a sum of 12 (12 and 0,1 and 11,2 and 10, 3 and 9, 4 and 8, 5 and 7, and 6 and 6.

5 0
3 years ago
What kind of graph is this
quester [9]
Bro what that looks like a face
3 0
2 years ago
ANSWER QUICK PLEASE<br> If f(x) = |x – 5| + 3, find f(2).
Ivanshal [37]

Answer:

SO if f(x)= |x-5|+3 then F(2)= 6

Step-by-step explanation:

These signs |  | mean that it is an absolute value equation. SO that means any negative will automatically change to positive values and positive values will stay at positive values. You are supposed to substitute 2 in for x and if you do 2-5 then that is negative 3 but it changes to positive 3 and add three and then f(2)= 6


6 0
3 years ago
Five computer program modules are ranked as M1, M2, M3, M4, and M5 according to the ascending order of effort required to debug
gulaghasi [49]

Answer:

Follows are the solution to this question:

Step-by-step explanation:

Technician selects three out of 5 systems  

In C(5,3)=10ways, this can be achieved  

In part a:

Space sample chooses 3 of a 5 systems  

(M_1, \ M_2,\ M_3),(M_1,M_2,M_4) \ (M_1,M_2,M_5) \ (M_1,M_3,M_4) \ (M_1,M_3,M_5),(M_1,M_4,M_5) \ (M_2,M_3,M_4)\ (M_2,M_3,M_5) \ (M_2,M_4,M_5),(M_3,M_4,M_5)}

In point b:

A =MODULE WHICH INCLUDE M1 minimal amount of effort  

Outcomes probable =

(M_1,M_2,M_3),\ (M_1,M_2,M_4) \ (M_1,M_2,M_5)\ (M_1,M_3,M_4)\\\\(M_1,M_3,M_5),\ (M_1,M_4,M)5)\ =\ 6

\to p(A)=\frac{6}{10}\\\\

            =0.6

In point c:

B = highest effort that is M_5

Potential result=

(M_1,M_2,M_5) \ (M_1,M_3,M_5) \ (M_2,M_3,M_5)\(M_2,M_4,M_5) \\ (M_2,M_4,M_5), \ (M_3,M_4,M_5) \ =\ 6  \\\\

\to B= \frac{6}{10} \\\\

        =0.6

\to P(B)=10

In point d:

\to \ A  \ intersection \ B=(M_1,M_2,M_5), \ (M_1,M_3,M_5) \ ,(M_1,M_4,M_5)

\to A (A \ intersection \ B) = \frac{3}{10} \\\\\ \ \ \ \ \

                                      =0.3

In point e:

\to (A \cup B) =  (M_1,M_2,M_3),\ (M_1,M_2,M_4)\ (M_1,M_2,M_5)(M_1,M_3,M_4)\ (M_1,M_3,M_5), \\ (M_1,M_4,M_5)\ (M_2,M_3,M_5) \ (M_2,M_4,M_5),(M_3,M_4,M_5) \ = \ 9\to P(A \cap B)=\frac{9}{10}

                    = 0.9

In point f:

\to (A\cap B) = \frac{3}{10}

                 = 0.3

In point g:

\to (A \cup B) = \frac{7}{10}

                 =0.7

In point h:

\to p(A \cap B) = 0.3 \neq 0

8 0
3 years ago
2x(x^4-x^2-6) what is the answer to this when foiled?
sp2606 [1]
When you foil it out you should get: 2x^5-2x^3-12x. When you factor it out completely you get: 2x(x^2-3)(x^2+2)
5 0
3 years ago
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