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g100num [7]
3 years ago
7

Please give the correct answer and the solution if you can

Mathematics
1 answer:
Klio2033 [76]3 years ago
4 0

Answer:

the answer is b the solution is 6

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1What is 1-1(0)-55+67
saveliy_v [14]

Answer:

13

Step-by-step explanation:

Questions with multiple signs are solved using the rule of BODMAS. Which means, Bracket Of Division, Multiplication, Addition and Subtraction.

In the question, we have Bracket, Addition and Subtraction, and will be answered in that order

1-1(0)-55+67

1-0-55+67

1-0+(-55+67)

1-0+(12)

1+12

13

7 0
3 years ago
URGENT! No links please!
Lelu [443]

Answer:

1) 17 units

Step-by-step explanation:

using pythagoean theorm you need to square 8 and 15 and add them and then take the sqaure root and finally you will get 17

5 0
3 years ago
Find the sum and product of the hexadecimal values, a01 and 20cba.
dybincka [34]
We have that
<span>a01
and
20cba
</span>
step 1
convert to decimal values
a01-----> [10*16²+0*16+1*16<span>^0]---> [2560+1]------> 2561
</span>
20cba--> [ 2*16^4+0*16^3+12*16^2+11*16^1+10*16^0]
=[131072+3072+176+10]
=134330

step 2
find the sum
2561+134330-----> 136891

step 3
find the product
2561*134330----> 344019130

step 4
convert to hexadecimal------> using a Hexadecimal calculator
the sum
136891---------> 216BB

the product
344019130-----> <span>148150BA</span>
6 0
4 years ago
Please solve this problem please ASAP !
torisob [31]
Ok done. Thank to me :>

3 0
2 years ago
Point M is the midpoint of AB. The coordinates of point A are(-6, 3) and the coordinates of M are
densk [106]

\bf ~~~~~~~~~~~~\textit{middle point of 2 points } \\\\ A(\stackrel{x_1}{-6}~,~\stackrel{y_1}{3})\qquad B(\stackrel{x_2}{x}~,~\stackrel{y_2}{y}) \qquad \left(\cfrac{ x_2 + x_1}{2}~~~ ,~~~ \cfrac{ y_2 + y_1}{2} \right) \\\\\\ \left( \cfrac{x-6}{2}~~,~~\cfrac{y+3}{2} \right)~~=~~\stackrel{\stackrel{Midpoint}{M}}{(-2~,~3)}\implies \begin{cases} \cfrac{x-6}{2}=-2\\[1em] x-6=-4\\ \boxed{x=2}\\ \cline{1-1} \cfrac{y+3}{2}=3\\[1em] y+3=6\\ \boxed{y=3} \end{cases}

7 0
4 years ago
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