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2 years ago

1,write the successor of-5 and predecessor of 6 then add them

Mathematics

2 answers:

Burka [1]2 years ago

6 0

<u>Step by Step Solution for Part-1:</u>

The successor of a number defines one to be added to a number
The predecessor of a number defines one being subtracted to a number

<u>Solution:</u>

⇒ (-5 + 1) + (6 - 1)
⇒ (-4) + (5)
⇒ 1

<u>Step by Step Solution for Part-2:</u>

b - 1 = a
To solve: a - b

<u>Solution:</u>

⇒ a - b
⇒ (b - 1) - b
⇒ (-b + 1) + b
⇒ -b + 1 + b
⇒ 1

<u>Step by Step Solution for Part-3:</u>

(42 - 66) - (100 - 24)
=> (-24) - (76)
=> -100

Alona [7]2 years ago

3 0

Successor of -5=-4
Predecessor of 6=5

$\\ \tt\hookrightarrow -4+5=1$

Let one be x and other be x-1

$\\ \tt\hookrightarrow a-b=x-x+1=+1$

$\\ \tt\hookrightarrow 42-66-(100-24)$

$\\ \tt\hookrightarrow -24-76$

$\\ \tt\hookrightarrow -100$

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Conjugate/Rational Number?

hram777 [196]

Answer:

1) $\dfrac{2}{\sqrt{5} } = \dfrac{2 \cdot \sqrt{5} }{5}$

2) $-\dfrac{5}{\sqrt{3} } = -\dfrac{5 \cdot \sqrt{3} }{3}$

3) $\dfrac{\sqrt{2} + \sqrt{5} }{\sqrt{10} } =\dfrac{\sqrt{5} }{5} + \dfrac{ \sqrt{2} }{2}$

4) $\dfrac{3 + \sqrt{2} }{\sqrt{3} } \times \dfrac{\sqrt{3} }{\sqrt{3} } = \sqrt{3} + \dfrac{\sqrt{6} }{3}$

5) $\dfrac{\sqrt{3} }{\sqrt{5} + \sqrt{2} }= \dfrac{\sqrt{15} - \sqrt{6} }{3}$

Step-by-step explanation:

The rationalization of the denominator of the surds are found as follows;

1) $\dfrac{2}{\sqrt{5} }$

$\dfrac{2}{\sqrt{5} } \times \dfrac{\sqrt{5} }{\sqrt{5} } = \dfrac{2 \cdot \sqrt{5} }{5}$

$\dfrac{2}{\sqrt{5} } = \dfrac{2 \cdot \sqrt{5} }{5}$

2) $-\dfrac{5}{\sqrt{3} }$

$-\dfrac{5}{\sqrt{3} } \times \dfrac{\sqrt{3} }{\sqrt{3} } = -\dfrac{5 \cdot \sqrt{3} }{3}$

$-\dfrac{5}{\sqrt{3} } = -\dfrac{5 \cdot \sqrt{3} }{3}$

3) $\dfrac{\sqrt{2} + \sqrt{5} }{\sqrt{10} }$

$\dfrac{\sqrt{2} + \sqrt{5} }{\sqrt{10} } \times \dfrac{ \sqrt{10} }{\sqrt{10} } = \dfrac{\sqrt{20} + \sqrt{50} }{10 } = \dfrac{2\cdot \sqrt{5} + 5 \cdot \sqrt{2} }{10} = \dfrac{\sqrt{5} }{5} + \dfrac{ \sqrt{2} }{2}$

$\dfrac{\sqrt{2} + \sqrt{5} }{\sqrt{10} } =\dfrac{\sqrt{5} }{5} + \dfrac{ \sqrt{2} }{2}$

4) $\dfrac{3 + \sqrt{2} }{\sqrt{3} }$

$\dfrac{3 + \sqrt{2} }{\sqrt{3} } \times \dfrac{\sqrt{3} }{\sqrt{3} } = \dfrac{3 \cdot \sqrt{3}+\sqrt{6} }{3 } = \sqrt{3} + \dfrac{\sqrt{6} }{3}$

$\dfrac{3 + \sqrt{2} }{\sqrt{3} } \times \dfrac{\sqrt{3} }{\sqrt{3} } = \sqrt{3} + \dfrac{\sqrt{6} }{3}$

5) $\dfrac{\sqrt{3} }{\sqrt{5} + \sqrt{2} }$

$\dfrac{\sqrt{3} }{\sqrt{5} + \sqrt{2} } = \dfrac{\sqrt{5} - \sqrt{2} }{\sqrt{5} - \sqrt{2} } = \dfrac{\sqrt{15} -\sqrt{6} }{5 - 2} = \dfrac{\sqrt{15} - \sqrt{6} }{3}$

$\dfrac{\sqrt{3} }{\sqrt{5} + \sqrt{2} }= \dfrac{\sqrt{15} - \sqrt{6} }{3}$

6) $\dfrac{\sqrt{7} }{\sqrt{3} - \sqrt{5} }$

$\dfrac{\sqrt{7} }{\sqrt{3} - \sqrt{5} } \times \dfrac{\sqrt{3} + \sqrt{5}}{\sqrt{3} + \sqrt{5}} = \dfrac{\sqrt{21} + \sqrt{35}}{{3} + {5}} = \dfrac{\sqrt{21} + \sqrt{35}}{8}$

$\dfrac{\sqrt{7} }{\sqrt{3} - \sqrt{5} } \times \dfrac{\sqrt{3} + \sqrt{5}}{\sqrt{3} + \sqrt{5}} =\dfrac{\sqrt{21} + \sqrt{35}}{8}$

8 0

3 years ago

The solution of the equation 16 m + 16 = 32 is<br><br><br>1<br><br>3<br><br>6<br><br>12

blsea [12.9K]

16(1)=16
16(1)+16
16+16=32

7 0

3 years ago

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Fofino [41]

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Last Monday, two law students met up at café Literatura after school to read the pages they were assigned in the legal methods c

jonny [76]

Answer:

$N_{a} = T_{a}$

$N_{c} = 2 \times T_{c}$

Step-by-step explanation:

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If the variable N represents the number of pages read by the readers and T in minutes represents the time spent by the reader to read this number of pages, then

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3 years ago

4z to the third power minus six when z is equal to 2

antiseptic1488 [7]

4z^3 - 6
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3 years ago