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3 years ago

An angle measures 15 degrees more than its complement. Find the measures of the two angles

Mathematics

1 answer:

ale4655 [162]3 years ago

3 0

x = one angle

y = other angle

y = 15+x

complements add to 90 degrees

x+y = 90

substitute for y (15+x)

x + 15+x = 90

combine like terms

2x+ 15 = 90

subtract 15 from each side

2x = 75

divide each side by 2

x = 75/2

x = 37.5

y = x+15

y = 37.5 + 15

y =52.5

The two angles are 37.5 and 52.5

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The equation of the line that goes through the point (1,5) and is parallel to the line 3x+5y=2 can be written in the form y=Mx+b

Llana [10]

Answer:

$y=-\frac{3}{5}x+\frac{28}{5}$

Step-by-step explanation:

step 1

Find the slope of the given line

we have

$3x+5y=2$

isolate the variable y

$5y=-3x+2\\\\y=-\frac{3}{5}x+\frac{2}{5}$

The slope of the given line is

$m=-\frac{3}{5}$

step 2

Find the equation of the line that goes through the point (1,5) and is parallel to the given line

Remember that

If two lines are parallel then their slopes are equal

therefore

we have

$m=-\frac{3}{5}$

$point\ (1,5)$

substitute in the equation of a line in slope intercept form

$y=mx+b$

$5=-\frac{3}{5}(1)+b$

solve for b

$b=5+\frac{3}{5}\\\\b=\frac{28}{5}$

substitute

$y=-\frac{3}{5}x+\frac{28}{5}$

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Answer:

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Step-by-step explanation:

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3 years ago

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Step-by-step explanation:

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4 years ago

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The two lines, P and Q, are graphed below: Line P is drawn by joining ordered pairs negative 8,15 and 6, negative 12. Line Q is

andreyandreev [35.5K]

Solution:

Keep in mind ,

Equation of line joining two points (a,b) and (p,q) is given by :

$\frac{q-b}{p-a}=\frac{y-b}{x-a}$

Equation of line P which is obtained by joining (- 8,15) and (6, - 12) is given by:

$\frac{y-15}{x+8}=\frac{-12-15}{6+8}\\\\ 14(y-15)=-27(x+8)\\\\ 14 y -210= -27 x - 216\\\\ 27 x+14 y+6=0$

Equation of line Q which is obtained by joining (4,16) and (-9, 10) is given by:

$\frac{y-16}{x-4}=\frac{16-10}{4+9}\\\\ 13(y-16)=6(x-4)\\\\ 13 y -208= 6 x - 24\\\\ 6 x-13 y+184=0$

Equation of line P and Q are

27 x+14 y+6=0-------(1)× 2

6 x-13 y+184=0-------(2)× 9

54 x + 2 8 y+12=0---(1)

54 x -117 y +1656=0----(2)

(1) - (2)

145 y= 1644

y=11.33,

27 x+14 y+6=0-------(1)×13

6 x-13 y+184=0-------(1)×14

351 x + 182 y + 78=0-----(1)

84 x - 182 y +2576=0----(2)

(1) + (2)

435 x + 2654=0

x = - 6.11

So, solution set is (-6.11, 11.33)

Option (C) is true. (−2, 4), because this point makes both the equations incorrect.

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3 years ago

The following is an incomplete paragraph proving that the opposite sides of parallelogram ABCD are congruent:

Setler [38]

Answer:

The sentence which accurately completes the proof is: "Triangles BCA and DAC are congruent according to the Angle-Side-Angle (ASA) Theorem." ⇒ 2nd answer

Step-by-step explanation:

Let us revise the cases of congruence

SSS ⇒ 3 sides in the 1st Δ ≅ 3 sides in the 2nd Δ
SAS ⇒ 2 sides and including angle in the 1st Δ ≅ 2 sides and including angle in the 2nd Δ
ASA ⇒ 2 angles and the side whose joining them in the 1st Δ ≅ 2 angles and the side whose joining them in the 2nd Δ
AAS ⇒ 2 angles and one side in the 1st Δ ≅ 2 angles and one side in the 2nd Δ
HL ⇒ hypotenuse leg of the 1st right Δ ≅ hypotenuse leg of the 2nd right Δ

In Parallelogram ABCD

∵ Segment AB is parallel to segment DC

∵ Segment BC is parallel to segment AD

- Construct diagonal A C with a straightedge

In Δs BCA and DAC

∵ AC is congruent to itself ⇒ Reflexive Property of Equality

∵ ∠BAC and ∠DCA are congruent ⇒ Alternate Interior Angles

∵ ∠BCA and ∠DAC are congruent ⇒ Alternate Interior Angles

- AC is joining the congruent angles

∴ Δ BCA is congruent to Δ DAC by ASA Theorem of congruence

By CPCTC

∴ AB is congruent to CD

∴ BC is congruent to DA

The sentence which accurately completes the proof is: "Triangles BCA and DAC are congruent according to the Angle-Side-Angle (ASA) Theorem."

3 0

3 years ago