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myrzilka [38]
2 years ago
13

Triangle ABC has vertices A(–2, 3), B(0, 3), and C(–1, –1). Find the coordinates of the image after a reflection over the x-axis

.A’ B’ C’
Mathematics
1 answer:
slava [35]2 years ago
4 0

Answer:

(-2,-3), (0,-3), (-1,1)

Step-by-step explanation:

When reflecting over an axis, the axis you are reflecting over stays the same while the opposite axis becomes its opposite number, like reflect (12, -9) over the y axis, the y (-9) would stay the same but the x (12) would become (-12)

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1. Find 3 equivalent fraction for the following: a) 5/6 b) 7/11 c) 9/7
SOVA2 [1]

Answer:

Step-by-step explanation:

To find equivalent fractions, multiply the numerator and the denominator by the same table.

a) 5/6

\frac{5*2}{6*2}=\frac{10}{12}\\\\\frac{5*3}{6*3}=\frac{15}{18}\\\\\frac{5*4}{6*4}=\frac{20}{24}

b) 7/11

\frac{7*3}{11*3}=\frac{21}{33}\\\\\frac{7*4}{11*4}=\frac{28}{44}\\\\\frac{7*5}{11*5}=\frac{35}{55}

c)9/7

\frac{9*2}{7*2}=\frac{18}{14}\\\\\frac{9*5}{7*5}=\frac{45}{35}\\\\\frac{9*10}{7*10}=\frac{90}{70}

8 0
3 years ago
What is the y intercept of a line that passes through (1,-7) and (5,-25). How do I find it.
LUCKY_DIMON [66]

Answer:

y- intercept = - 2.5

Step-by-step explanation:

The equation of a line in slope- intercept form is

y = mx + c ( m is the slope and c the y- intercept )

Calculate m using the slope formula

m = \frac{y_{2}-y_{1}  }{x_{2}-x_{1}  }

with (x₁, y₁ ) = (1, - 7) and (x₂, y₂ ) = (5, - 25)

m = \frac{-25+7}{5-1} = \frac{-18}{4} = - 4.5 , then

y = - 4.5x + c

To find c substitute either of the 2 points into the equation

Using (1, - 7), then

- 7 = - 4.5 + c ⇒ c = - 7 + 4.5 = - 2.5

y- intercept c = - 2.5

3 0
3 years ago
42:28
gogolik [260]

Answer:

The statements about arcs and angles that are true in the figure are;

1) ∠EFD ≅ ∠EGD

2) \overline{ED}\cong \overline{FD}

3) mFD = 120°

Step-by-step explanation:

1) ∠ECD + ∠CEG + ∠CDG + ∠GDE = 360° (Sum of interior angle of a quadrilateral)

∠CEG = ∠CDG = 90° (Given)

∠GDE = 60° (Given)

∴ ∠ECD = 360° - (∠CEG + ∠CDG + ∠GDE)

∠ECD = 360° - (90° + 90° + 60°) = 120°

∠ECD = 2 × ∠EFD (Angle subtended is twice the angle subtended at the circumference)

120° = 2 × ∠EFD

∠EFD = 120°/2 = 60°

∠EFD ≅ ∠EGD

∠ECD = 120°

∠EGD = 60°

∴∠EGD ≠ ∠ECD

2) Given that arc mEF ≅ arc mFD

Therefore, ΔECF and ΔDCF are isosceles triangles having two sides (radii EC and CF in ΔECF and radii EF and CD in ΔDCF

∠ECF = mEF = mFD = ∠DCF (Given)

∴ ΔECF ≅ ΔDCF (Side Angle Side, SAS, rule of congruency)

\\ \overline{EF}\cong \overline{FD} (Corresponding Parts of Congruent Triangles are Congruent, CPCTC)

∠FED ≅ ∠FDE (base angles of isosceles triangle)

∠FED + ∠FDE + ∠EFD = 180° (sum of interior angles of a triangle)

∠FED + ∠FDE = 180° - ∠EFD = 180° - 60° = 120°

∠FED + ∠FDE = 120° = ∠FED + ∠FED (substitution)

2 × ∠FED  = 120°

∠FED = 120°/2 = 60° = ∠FDE

∴ ∠FED = ∠FDE = ∠EFD =  60°

ΔEFD  is an equilateral triangle as all interior angles are equal

\\ \overline{EF}\cong \overline{FD}\cong \overline{ED} (definition of equilateral triangle)

\overline{ED}\cong \overline{FD}

3) Having that ∠EFD = 60° and ∠CFE = ∠CFD (CPCTC)

Where, ∠EFD = ∠CFE + ∠CFD (Angle addition)

60° = ∠CFE + ∠CFD = ∠CFE + ∠CFE (substitution)

60° = 2 × ∠CFE

∠CFE =60°/2 = 30° = ∠CFD

\overline{CF}\cong \overline{CD} (radii of the same circle)

ΔFCD is an isosceles triangle (definition)

∠CFD ≅ ∠CDF (base angles of isosceles ΔFCD)

∠CFD + ∠CDF + ∠DCF = 180°

∠DCF = 180° - (∠CFD + ∠CDF) = 180° - (30° + 30°) = 120°

mFD = ∠DCF (definition)

mFD = 120°.

5 0
3 years ago
Suppose that a large mixing tank initially holds 100 gallons of water in which 50 pounds of salt have been dissolved. Another br
Serggg [28]

Answer:

dA/dt = 12 - 2A/(100 + t)

Step-by-step explanation:

The differential equation of this problem is;

dA/dt = R_in - R_out

Where;

R_in is the rate at which salt enters

R_out is the rate at which salt exits

R_in = (concentration of salt in inflow) × (input rate of brine)

We are given;

Concentration of salt in inflow = 4 lb/gal

Input rate of brine = 3 gal/min

Thus;

R_in = 4 × 3 = 12 lb/min

Due to the fact that solution is pumped out at a slower rate, thus it is accumulating at the rate of (3 - 2)gal/min = 1 gal/min

So, after t minutes, there will be (100 + t) gallons in the tank

Therefore;

R_out = (concentration of salt in outflow) × (output rate of brine)

R_out = [A(t)/(100 + t)]lb/gal × 2 gal/min

R_out = 2A(t)/(100 + t) lb/min

So, we substitute the values of R_in and R_out into the Differential equation to get;

dA/dt = 12 - 2A(t)/(100 + t)

Since we are to use A foe A(t), thus the Differential equation is now;

dA/dt = 12 - 2A/(100 + t)

5 0
3 years ago
Help me I need help this is stressful
Alisiya [41]

Answer:

"You could add each number one by one in your head, and the total is your answer."

5 0
3 years ago
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