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stealth61 [152]

3 years ago

10

How many triangles can be constructed with angles measuring 10º, 80º, and 90º?

2 answers:

charle [14.2K]3 years ago

6 0

C! 80 + 10 is 90, 90+90 is 180. There are 180 degrees in one triangle

Sphinxa [80]3 years ago

5 0

More than one because lengths were not given for any side

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Identify angleA and angleB

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Will crown the correct answers brainliest 3x

Solnce55 [7]

Answer :

(1) $\frac{1}{x}+\frac{1}{y}=\frac{y+x}{xy}$

(2) $\frac{1}{5x}+\frac{1}{3y}=\frac{3y+5x}{15xy}$

(3) $\frac{1}{7x}-\frac{1}{2y}=\frac{2y-7x}{14xy}$

(4) $\frac{2}{5x}+\frac{3}{7y}=\frac{6y+15x}{21xy}$

(5) $\frac{7}{11x}-\frac{1}{33y}=\frac{231y-11x}{363xy}$

(6) $\frac{1}{2x}+\frac{3}{x}-\frac{1}{7y}=\frac{7y+42y-2x}{14xy}$

Step-by-step explanation :

(1) The given expression is: $\frac{1}{x}+\frac{1}{y}$

$\frac{1}{x}+\frac{1}{y}=\frac{y+x}{xy}$

(2) The given expression is: $\frac{1}{5x}+\frac{1}{3y}$

$\frac{1}{5x}+\frac{1}{3y}=\frac{3y+5x}{(5x)\times (3y)}=\frac{3y+5x}{15xy}$

(3) The given expression is: $\frac{1}{7x}-\frac{1}{2y}$

$\frac{1}{7x}-\frac{1}{2y}=\frac{2y-7x}{(7x)\times (2y)}=\frac{2y-7x}{14xy}$

(4) The given expression is: $\frac{2}{5x}+\frac{3}{7y}$

$\frac{2}{5x}+\frac{3}{7y}=\frac{(2\times 3y)+(3\times 5x)}{(5x)\times (7y)}=\frac{6y+15x}{21xy}$

(5) The given expression is: $\frac{7}{11x}-\frac{1}{33y}$

$\frac{7}{11x}-\frac{1}{33y}=\frac{(7\times 33y)-(1\times 11x)}{(11x)\times (33y)}=\frac{231y-11x}{363xy}$

(6) The given expression is: $\frac{1}{2x}+\frac{3}{x}-\frac{1}{7y}$

$\frac{1}{2x}+\frac{3}{x}-\frac{1}{7y}=\frac{(x\times 7y)+(3\times 2x\times 7y)-(2x\times x)}{(2x)\times (x)\times (7y)}=\frac{7xy+42xy-2x^2}{14x^2y}=\frac{7y+42y-2x}{14xy}$

7 0

3 years ago

MARINA, DO U SEE MY QUESTION? 8*9/5+3

andrey2020 [161]

$\frac{8*9}{5+3} = \frac{72}{8} = 9$

Also, Marina might be offline.

8 0

3 years ago

Read 2 more answers

a small acting club has 6 members. 3 of the members are to be chosen for a trip to see a Broadway play. how many different 3-mem

nignag [31]

A group of k elements can be chosen from a group of n elements in $\frac{n!}{k!(n-k)!}$ ways.

$(^6 _3)=\frac{6!}{3!(6-3)!}=\frac{6!}{3! \times 3!}=\frac{3! \times 4 \times 5 \times 6}{3! \times 6}=4 \times 5=20$

There are 20 different 3-member groups.

8 0

3 years ago