All categories

3 years ago

M<KGH=x+161,m<FGK=x+41,and m<FGH=180° Find x

Mathematics

1 answer:

Butoxors [25]3 years ago

4 0

Answer: x = -11 ∠FGK = 30° ∠KGH = 150°

Step-by-step explanation:

∠FGK + ∠KGH = ∠FGH Segment Addition Postulate

x + 41 + x + 161 = 180 Substitution

2x + 202 = 180 Simplify (added like terms)

2x = -22 Subtraction Property of Equality

x = -11 Division Property of Equality

∠FGK = x + 41 = (-11) + 41 = 30

∠KGH = x + 161 = (-11) + 161 = 150

You might be interested in

You plan to go to a technical school, and there are two options in your city. School A will charge you $750 per semester for tui

bixtya [17]

School A : 750 + 200 = 950 per semester

School B : if u get a 10% discount, then ur actually paying 90%
0.90(1000) = 900 per semester

The best deal would be school B...u would save $ 50 per semester by choosing B over A

8 0

3 years ago

Find the slope of the given line on the graph. REMEMBER: Rise over run! Pls show work!

Leviafan [203]

Answer:

the slope is $-\frac{1}{3}$

Step-by-step explanation:

Notice that you are given two points to use for the slope calculation:

$(x_1,y_1)=(-4,4) \,\,and\,\, (x_2,y_2)=(5,1)$

So we use the formula for the slope of a line given two points:

$slope\,=\,\frac{y_2-y_1}{x_2-x_1} \\slope\,=\,\frac{1-4}{5-(-4)} \\slope\,=\,\frac{-3}{9} \\slope\,=\,-\frac{1}{3} \\$

3 0

3 years ago

Tina drew a triangle with side lengths 6, 8, and 14. She claims it is a right triangle. Is she correct?

fomenos

Answer:

No such a triangle can not exist.

4 0

3 years ago

Which expression is equivalent to the given expression? (-5y + 21) (y-5)

lapo4ka [179]

Answer:

The Answer is -5(y - 3)(y - 7 )

<h3>hope this helps </h3>

5 0

3 years ago

The sum of three consecutive terms in an arithmetic sequence is 27, and their product is 585. find the three terms.

sammy [17]

$\text{the three consecutive terms in an arithmetic sequence}\\a;\ a+d;\ a+2d\\\\\text{system of equals}\\\\ \left\{\begin{array}{ccc}a+a+d+a+2d=27\\a(a+d)(a+2d)=585\end{array}\right\\ \left\{\begin{array}{ccc}3a+3d=27&|:3\\a(a+d)(a+2d)=585\end{array}\right\\ \left\{\begin{array}{ccc}a+d=9&\to d=9-a\\a(a+d)(a+2d)=585\end{array}\right\\$
$\text{substitute to the second equation}\\\\a(a+9-a)(a+2(9-a))=585\\\\a(9)(a+18-2a)=585\\9a(18-a)=585\ \ \ |:9\\a(18-a)=65\\18a-a^2=65\ \ \ |\text{change the signs}\\a^2-18a=-65\\a^2-2a\cdot9=-65\ \ \ |+9^2\\\underbrace{a^2-2a\cdot9+9^2}_{(a-b)^2=a^2-2ab+b^2}=-65+9^2\\(a-9)^2=16\to a-9\pm\sqrt{16}\\a-9=-4\ \vee\ a-9=4\ \ \ |+9\\a=5\ \vee\ a=13\\\\d=9-5=4\ \vee\ d=9-13=-4$
$Answer:\ a=5;\ a+d=9; a+2d=13\ vee\ a=13;\ a+d=9;\ a+2d=5$

5 0

4 years ago