All categories

3 years ago

The sum of two numbers is 54. One number is 26 more than the other. Find the numbers.

Mathematics

1 answer:

Nikitich [7]3 years ago

6 0

Answer: a=52

b=2

Step-by-step explanation:

I hope the explanation below helps

A=52

B=2

You might be interested in

Ill give brainliest<br> Triangle Sum Theorem CLASSWORK

Viefleur [7K]

Answer:

$8x + 32 + 116 = 180 \\ 8x + 148 = 180 \\ 8x = 180 - 148 \\ 8x =32 \\ \boxed{x = 4}$

<h3>4 is the right answer.</h3>

$(7x + 5) + 52 + 67 = 180 \\ 7x + 124 = 180 \\ 7x = 56 \\ \boxed{x = 8}$

<h3>8 is the right answer.</h3>

3 0

3 years ago

Read 2 more answers

Write an equation that could be used to find the value of x using the interior angles of ∆BDC, then solve it.

Alecsey [184]

9514 1404 393

Answer:

x = 16

Step-by-step explanation:

Either or both of the right triangles can be used to find x. Or, triangle ABC could be used. All numbers are assumed to be degrees.

<u>Using ∆ABD</u>

55 +90 +2x+3 = 180

2x = 32 . . . . . . subtract 148

x = 16

<u>Using ∆BCD</u>

50 +90 +2x+8 = 180

2x = 32 . . . . . . subtract 148

x = 16

<u>Using ∆ABC</u>

55 +(2x +3) +50 +(2x +8) = 180

4x = 64 . . . . . . . subtract 116

x = 16

6 0

3 years ago

If two of a triangles angles measure 42 degrees and 48 degrees, how would you classify that triangle

grin007 [14]

I think that would be a right angle triangle or is that not what it is meaning

5 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.

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