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2 years ago

I don't know what 5+5 is can someone tell me what it is

Mathematics

2 answers:

Grace [21]2 years ago

4 0

Answer:

Wow!

It is 10

Step-by-step explanation:

SVEN [57.7K]2 years ago

3 0

Answer:5+5=10

Step-by-step explanation:

For this all you would do is add 5 units to another 5 units you would have 10 units. Or we could use your hands for example you should have 5 fingers on each hand and you should also have 2 hands so if you look at your hands you should see 5 fingers on each hand or ten fingers in total.

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3 years ago

Which equation is related to the given equation? column a column b 1. 61 + 18 = z 2. z = 61 – 18 3. –18 = 61 – z 4. 18 = 61 – z

Fofino [41]

Question 1.).

Column A.):

Answer ==> z = 61 – 18, is Related to z + 18 = 61, ==> z = 43

Answer: ===> 18 = 61 - z, is Related to z + 18 = 61, ===> z = 43

z + 18 = 61

Let's solve your equation step-by-step.

z + 18 = 61

Step 1: Subtract 18 from both sides.

z + 18 − 18 = 61 − 18

z = 43

Graph: ========> z + 18 = 61 =========> z =======> 43

Column (A), Answer: =======> z = 43

Question 2.).

Column. B.):

Answer: =====> 61 + 18 = z, Related to z - 18 = 61 ====> z = 79 Answer: ===> –18 = 61 – z, Related to z - 18 = 61 ====> z = 79

z - 18 = 61

Let's solve your equation step-by-step.

z − 18 = 61

Step 1: Add 18 to both sides.

z − 18 + 18 = 61 + 18

z = 79

Answer z = 79

Therefore, The answers to your Equation: Column (A),/Column (B)

Question 1.).

Column A.):

Answer ==> z = 61 – 18, is Related to z + 18 = 61, ==> z = 43

Answer: ===> 18 = 61 - z, is Related to z + 18 = 61, ===> z = 43

Question 2.).

Column. B.):

Answer: =====> 61 + 18 = z, Related to z - 18 = 61 ====> z = 79 Answer: ===> –18 = 61 – z, Related to z - 18 = 61 ====> z = 79

Hope that helps!!!!!!!! : )

7 0

3 years ago

Read 2 more answers

Pls help if pls do most important 15 number 20 number 26 number 28 number 30 number 17 number 25 number 22 number 24 number if u

Tamiku [17]

Answer:

Step-by-step explanation:

15. $\frac{cotx+1}{cotx-1} = \frac{cotx+cotx.tanx}{cotx-cotx.tanx} = \frac{cotx(1+tanx)}{cotx(1-tanx)} = \frac{1+tanx}{1-tanx} \\$

16. $\frac{1+cos\alpha }{sin\alpha } = \frac{sin\alpha }{1-cos\alpha }$

$=> (1+cos\alpha )(1-cos\alpha ) = sin^{2} \alpha \\=> 1-cos^{2}\alpha =sin^{2} \alpha \\=> sin^{2}\alpha +cos^{2}\alpha =1\\$

=> The clause is correct

17. Do the same (16)

18. $\frac{sinx}{1-cosx}+\frac{sinx}{1+cosx} = \frac{sinx(1+cosx) + sinx(1-cosx)}{(1+cosx)(1-cosx)} = \frac{sinx + sinx.cosx + sinx - sinx.cosx}{1-cos^{2}x} = \frac{2sinx}{sin^{2}x }= \frac{2}{sinx} = 2cosecx$

19.

$\frac{sinA}{1+cosA} + \frac{1+cosA}{sinA}=\frac{2}{sinA} \\=> \frac{sinA}{1+cosA} + \frac{1+cosA}{sinA} - \frac{2}{sinA} \\ = 0\\=> \frac{sinA}{1+cosA} + (\frac{1+cosA}{sinA} - \frac{2}{sinA} \\) = 0\\=> \frac{sinA}{1+cosA} + \frac{1+cosA-2}{sinA} = 0\\=> \frac{sinA}{1+cosA} +\frac{cosA-1}{sinA} = 0\\=> \frac{sin^{2} A+(cosA-1)(1+cosA)}{(1+cosA)sinA}=0\\=> \frac{sin^{2} A+cos^{2} A-1}{(1+cosA)sinA}=0\\=> \frac{0}{(1+cosA)sinA} =0\\$

=> The clause is correct

20. Do the same (18)

21. Left = $\frac{1}{cosecA+cotA} = \frac{1}{\frac{1}{sinA}+\frac{cosA}{sinA} } = \frac{1}{\frac{1+cosA}{sinA} }= \frac{sinA}{1+cosA}$

Right = $cosecA-cotA = \frac{1}{sinA}-\frac{cosA}{sinA}= \frac{1-cosA}{sinA} \\$

Same with (16) => Left = Right => The clause is correct

22. Do the same (21)

Too long, i'm so lazy :))))

5 0

3 years ago