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gulaghasi [49]
2 years ago
12

Sasha's mom is ordering food trays for Sasha's elementary school graduation party. Banquet Brothers sells a tray with 48 serving

s for $60.00. Cookin' Catering sells a tray with 30 servings of pasta for $72.00. Which company has the better deal?
Mathematics
1 answer:
alexgriva [62]2 years ago
3 0

Answer:

Banquet brothers

Step-by-step explanation:

banquet brothers 48 servings cost 60 dollars making it

$1.25 per serving

cookin' catering 30 servings cost $72.00 making it

$2.40 per serving

$1.25 < 2.40

thus banquet brothers is the answer.

PLEASE MARK BRAINLIEST AND GIVE 5 STARS.

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Evaluate the expression when a =7 and y=-3.<br> y-9a
11111nata11111 [884]

Answer:

-6

Step-by-step explanation:

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Ross and Gabby sold boxes of cookies for their soccer team fundraiser. Ross sold b boxes of cookies, and Gabby, g, sold 35 boxes
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7 0
3 years ago
ΔXYZ ≅ΔFED where m∠X = 50°, m∠Y=30°, m∠D=2y+10, XY =9 and EF=3x−12. Solve for x and y.
Delicious77 [7]

Answer:

The value of x is 7 and value of y is 45

Step-by-step explanation:

In ΔXYZ

∠X = 50°, ∠Y=30°

To Find ∠Z we will use angle sum property of triangle (Sum of all angles of triangles is 180°)

So,∠X + ∠Y + ∠Z = 180°

So,50°+30°+ ∠Z = 180°

∠Z=180°-80°

∠Z=100°

Now we are given that ΔXYZ ≅ΔFED

So, the corresponding sides are equal and the corresponding angles are equal.

So, ∠Z= ∠D

So, 100=2y+10

100-10=2y

90=2y

45=y

Now XY = EF

So, 9 = 3x-12

21=3x

7=x

Hence The value of x is 7 and value of y is 45

5 0
3 years ago
Part B
alisha [4.7K]

Question:

Consider ΔABC, whose vertices are A (2, 1), B (3, 3), and C (1, 6); let the line segment  AC represent the base of the triangle.

(a)  Find the equation of the line passing through B and perpendicular to the line AC

(b)  Let the point of intersection of line AC with the line you found in part A be point D. Find the coordinates of point D.

Answer:

y = \frac{1}{5}x + \frac{12}{5}

D = (\frac{43}{26},\frac{71}{26})

Step-by-step explanation:

Given

\triangle ABC

A = (2,1)

B = (3,3)

C = (1,6)

Solving (a): Line that passes through B, perpendicular to AC.

First, calculate the slope of AC

m = \frac{y_2 - y_1}{x_2 - x_1}

Where:

A = (2,1) --- (x_1,y_1)

C = (1,6) --- (x_2,y_2)

The slope is:

m = \frac{6- 1}{1 - 2}

m = \frac{5}{-1}

m = -5

The slope of the line that passes through B is calculated as:

m_2 = -\frac{1}{m} --- because it is perpendicular to AC.

So, we have:

m_2 = -\frac{1}{-5}

m_2 = \frac{1}{5}

The equation of the line is the calculated using:

m_2 = \frac{y_2 - y_1}{x_2 - x_1}

Where:

m_2 = \frac{1}{5}

B = (3,3) --- (x_1,y_1)

(x_2,y_2) = (x,y)

So, we have:

\frac{1}{5} = \frac{y - 3}{x - 3}

Cross multiply

5(y-3) = 1(x - 3)

5y - 15 = x - 3

5y  = x - 3 + 15

5y  = x +12

Make y the subject

y = \frac{1}{5}x + \frac{12}{5}

Solving (b): Point of intersection between AC and y = \frac{1}{5}x + \frac{12}{5}

First, calculate the equation of AC using:

y = m(x - x_1) + y_1

Where:

A = (2,1) --- (x_1,y_1)

m = -5

So:

y=-5(x - 2) + 1

y=-5x + 10 + 1

y=-5x + 11

So, we have:

y=-5x + 11 and y = \frac{1}{5}x + \frac{12}{5}

Equate both to solve for x

i.e.

y = y

-5x + 11 = \frac{1}{5}x + \frac{12}{5}

Collect like terms

-5x -\frac{1}{5}x = \frac{12}{5} - 11

Multiply through by 5

-25x-x = 12 - 55

Collect like terms

-26x = -43

Solve for x

x = \frac{-43}{-26}

x = \frac{43}{26}

Substitute x = \frac{43}{26} in y=-5x + 11

y = -5 * \frac{43}{26} + 11

y =  \frac{-5 *43}{26} + 11

Take LCM

y =  \frac{-5 *43+11 * 26}{26}

y =  \frac{71}{26}

Hence, the coordinates of D is:

D = (\frac{43}{26},\frac{71}{26})

3 0
2 years ago
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