All categories

2 years ago

Find the measure of an angle if its supplement measures 20° less than 3 times its complement.

Mathematics

1 answer:

Alex777 [14]2 years ago

4 0

53.3 ° approxiametly
In the given question:
an angle which is three times of its compliment and less than 20°
based on this statement , the equation formed
3x-20° = 180° (
sum of supplement)
3x =180-20
3x= 160
x=53.3
complement angle is a sum of two angles which is equal to 90° , if two angles add up to form a right angle, then these angles are referred to as complementary angles
supplement angle is sum of two angles which is equal to 180° , if two angles add up, to form a straight angle, then those angles are referred to as supplementary angles.

To know more about supplement angles visit :

brainly.com/question/25889161

#SPJ9

You might be interested in

Irene knows there are 2 cups in 1 pint. She writes the equation c=2p to find the number of cups, c, in any number of pints, p. U

Natasha2012 [34]

Answer:

See Explanation

Step-by-step explanation:

The question is incomplete, as there are no options to select from.

However, a general explanation is as follows.

Given

$2\ cups = 1\ pint$

Required

Is $c = 2p$ a correct expression?

Let

$c = cups$

$p = pint$

So:

$2\ cups = 1\ pint$

becomes

$2 * c = 1 * p$

$2c = p$

Divide both sides by 2

$c =\frac{1}{2}p$

Hence:

$c = 2p$ is incorrect

4 0

3 years ago

The sum of two consecutive integers is – 7. Find the numbers

Y_Kistochka [10]

$n,n+1$ - two consecutive integers

$n+n+1=-7\\2n=-8\\n=-4\\n+1=-3$

-4 and -3

7 0

4 years ago

13 Pts PLEASE SOMEONE HELP ME WITH THIS QUESTION PLZ EXPLAIN THANK U

AlexFokin [52]

It's the third one and the last one are the only ones that make immediate sense.

4 0

3 years ago

Read 2 more answers

Maritza remembers her PIN because it is between 1,000 and 1,500 and it is the product of two consecutive prime numbers. What is

valentina_108 [34]

Answer:

The PIN is 1147

Step-by-step explanation:

Prime numbers

These are integer numbers that only divide exactly by themselves and by 1. To find the first possible prime number between 1,000 and 1,500, calculate the square root of 1,000.

$\sqrt{1000}\approx 31.6$

Thus, the first prime to consider is 31. Write the consecutive primes from 31:

31 , 37 , 41 , 43

We can see the only two consecutive primes whose product lies in the given interval are 31 and 37. Thus, the PIN is 31*37=1147

4 0

3 years ago

Prove A-(BnC) = (A-B)U(A-C), explain with an example

NikAS [45]

Answer:

Prove set equality by showing that for any element $x$ , $x \in (A \backslash (B \cap C))$ if and only if $x \in ((A \backslash B) \cup (A \backslash C))$ .

Example:

$A = \lbrace 0,\, 1,\, 2,\, 3 \rbrace$ .

$B = \lbrace0,\, 1 \rbrace$ .

$C = \lbrace0,\, 2 \rbrace$ .

$\begin{aligned} & A \backslash (B \cap C) \\ =\; & \lbrace 0,\, 1,\, 2,\, 3 \rbrace \backslash \lbrace 0 \rbrace \\ =\; & \lbrace 1,\, 2,\, 3 \rbrace \end{aligned}$ .

$\begin{aligned}& (A \backslash B) \cup (A \backslash C) \\ =\; & \lbrace 2,\, 3\rbrace \cup \lbrace 1,\, 3 \rbrace \\ =\; & \lbrace 1,\, 2,\, 3 \rbrace\end{aligned}$ .

Step-by-step explanation:

Proof for $[x \in (A \backslash (B \cap C))] \implies [x \in ((A \backslash B) \cup (A \backslash C))]$ for any element $x$ :

Assume that $x \in (A \backslash (B \cap C))$ . Thus, $x \in A$ and $x \not \in (B \cap C)$ .

Since $x \not \in (B \cap C)$ , either $x \not \in B$ or $x \not \in C$ (or both.)

If $x \not \in B$ , then combined with $x \in A$ , $x \in (A \backslash B)$ .
Similarly, if $x \not \in C$ , then combined with $x \in A$ , $x \in (A \backslash C)$ .

Thus, either $x \in (A \backslash B)$ or $x \in (A \backslash C)$ (or both.)

Therefore, $x \in ((A \backslash B) \cup (A \backslash C))$ as required.

Proof for $[x \in ((A \backslash B) \cup (A \backslash C))] \implies [x \in (A \backslash (B \cap C))]$ :

Assume that $x \in ((A \backslash B) \cup (A \backslash C))$ . Thus, either $x \in (A \backslash B)$ or $x \in (A \backslash C)$ (or both.)

If $x \in (A \backslash B)$ , then $x \in A$ and $x \not \in B$ . Notice that $(x \not \in B) \implies (x \not \in (B \cap C))$ since the contrapositive of that statement, $(x \in (B \cap C)) \implies (x \in B)$ , is true. Therefore, $x \not \in (B \cap C)$ and thus $x \in A \backslash (B \cap C)$ .
Otherwise, if $x \in A \backslash C$ , then $x \in A$ and $x \not \in C$ . Similarly, $x \not \in C \!$ implies $x \not \in (B \cap C)$ . Therefore, $x \in A \backslash (B \cap C)$ .

Either way, $x \in A \backslash (B \cap C)$ .

Therefore, $x \in ((A \backslash B) \cup (A \backslash C))$ implies $x \in A \backslash (B \cap C)$ , as required.

8 0

3 years ago