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3 years ago

The measure of an angle is 1°. What is the measure of its supplementary angle?

Mathematics

1 answer:

snow_tiger [21]3 years ago

4 0

Answer:

179°

Step-by-step explanation:

Supplementary angles sum to 180° , thus

supplementary angle = 180° - 1° = 179°

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Bianca's bank offers a savings account with a 2.1% APR, compounded monthly. What is the actual annual percentage yield on this a

katen-ka-za [31]

Bianca's bank offers a savings account with a 2.1% APR, compounded monthly. The the actual annual percentage yield on this account can be calculated using i = (1+ r/m)^m, where i is the actual APR, r is the nominal interest rate and m the compounding period in a year. APR is equal to 2.12%

6 0

3 years ago

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What is the equation of the line that passes through the point of intersection of the lines y = 2x − 5 and y = −x + 1, and is al

castortr0y [4]

Answer:

$y = \frac{1}{2}x - 2$

Step-by-step explanation:

Given lines,

y = 2x - 5,

y = -x + 1

Subtracting these two equations,

0 = 3x - 6

$\implies 3x = 6$

$\implies x = \frac{6}{3}=2$

By first equation,

$y=2(2) -5=4-5 = -1$

Thus, point of intersecting would be (2, -1).

Now, the equation of a line is y = mx + c,

Where,

m = slope of the line,

So, the slope of the line $y=\frac{1}{2}x+4$ is 1/2.

∵ two parallel lines have same slope.

Hence,

Equation of the parallel line passes through (2, -1),

$y+1=\frac{1}{2}(x-2)$

$y+1=\frac{1}{2}x - 1$

$y = \frac{1}{2}x - 2$

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3 years ago

PLS HELP ME!!!! ASAP I WILL MARK BRAINLIEST!!!!!!

julsineya [31]

Answer:

(-3)^3 x^5

Step-by-step explanation:

x*x*x*x*x *(-3)*(-3)*(-3)

There are 5 x terms

x^5

There are 3 (-3) terms

(-3)^3

Multiply theses together

(-3)^3 x^5

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3 years ago

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Five times the difference of a number, x, and 3 plus 7 is equal to -45.

miskamm [114]

Answer:

5(x-3)+7=-45

Step-by-step explanation:

The only possibly way for 5 to multiply 2 numbers is if those numbers are in parenthesis

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3 years ago

Prove that H c G is a normal subgroup if and only if every left coset is a right coset, i.e., aH = Ha for all a e G

Kaylis [27]

$\Rightarrow$

Suppose first that $H\subset G$ is a normal subgroup. Then by definition we must have for all $a\in H$ , $xax^{-1} \in H$ for every $x\in G$ . Let $a\in G$ and choose $(ab)\in aH$ ( $b\in H$ ). By hypothesis we have $aba^{-1} =abbb^{-1}a^{-1}=(ab)b(ab)^{-1} \in H$ , i.e. $aba^{-1}=c$ for some $c\in H$ , thus $ab=ca \in Ha$ . So we have $aH\subset Ha$ . You can prove $Ha\subset aH$ in the same way.

$\Leftarrow$

Suppose $aH=Ha$ for all $a\in G$ . Let $h\in H$ , we have to prove $aha^{-1} \in H$ for every $a\in G$ . So, let $a\in G$ . We have that $ha^{-1} =a^{-1}h'$ for some $h'\in H$ (by the hypothesis). hence we have $aha^{-1}=h' \in H$ . Because $a$ was chosen arbitrarily we have the desired .

5 0

3 years ago