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3 years ago

What is the area of rhombus ABCD? do not round any steps

Mathematics

1 answer:

katrin2010 [14]3 years ago

3 0

Because ABCD is a rhombus, ΔABC ~ ΔADC.

The length of AC = 4 (the base of ΔABC.
The height of ΔABC = 7 - 1 = 6
The area of ΔABC is
(1/2)*4*6 = 12

Therefore the area of ABCD = 2*12 = 24

Answer: 24

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Use quadratic regression to find

Rashid [163]

The quadratic equation is given by:

y = 3x² + 10x - 8

The standard equation of a parabola is given by:

y = ax² + bx + c

Where a, b, c are constants

At point (4, 80):

80 = a(4)² + b(4) + c

16a + 4b + c = 80 (1)

At point (-3, -11):

-11 = a(-3)² + b(-3) + c

9a - 3b + c = -11 (2)

At point (-1, -15):

-15 = a(-1)² + b(-1) + c

a - b + c = -15 (3)

Solving equations 1, 2 and 3 simultaneously gives:

a = 3, b = 10, c = -8

Therefore the quadratic equation becomes:

y = 3x² + 10x - 8

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2 years ago

FIND THE GCF. Mathematics

Rom4ik [11]

The G.C.F of the given algebraic expression is; ⁴⁹/₆a xy

<h3>What is the G.C.F (Greatest Common Factor)?</h3>

The greatest common factor (GCF or GCD or HCF) of a set of whole numbers is the largest positive integer that divides evenly into all numbers with zero remainder. For example, for the set of numbers 18, 30 and 42 the GCF = 6.

We are given the algebraic expression;

(3 x y * 4x²y * ¹/₁ * 5x³y ^ z * 4)7a * 1/1 * 7*6

Expanding this further gives;

⁴⁹/₆a((3xy * 4x²y * 20x³y ^ z )

Now, the GCF of the terms inside the bracket would be x y. Thus, expanding the GCF we have;

⁴⁹/₆a xy((3 * 4x * 20x²y^(z - 1))

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1 year ago

What is -3y+12=-48? Ueuehehhsh

Marina CMI [18]

Answer:

-12

Step-by-step explanation:-3y+12=48

48-12= 36

36/ (-3) = -12

7 0

3 years ago

Read 2 more answers

When you buy a new car it is a depreciating asset. Which means the moment you purchase your car it begins losing value. If you

Arturiano [62]

I can't see the graph but I'll try anyway

Answer:

x y

0 $20,000

1 $18,000

2 $16,200

3 $14,580

4 $13,122

5 $11,809.80

4 0

2 years ago

She is serving hot coffee and cold coffee in two rectangular-shaped tanks, each with dimensions width 8.75 inches, length 8.75 i

Blababa [14]

The volume of the coffee in each tank will be 1760. in³. The mass of cold coffee and hot coffee will be 6163.29 oz and 10213.45 oz. Then the number of the tank placed on a table is one only.

<h3>How to find the density and volume?</h3>

She is serving hot coffee and cold coffee in two rectangular-shaped tanks, each with dimensions width of 8.75 inches, a length of 8.75 inches, and a height of 23 inches.

The hot coffee has a density of 5.8 oz/in³ and the cold coffee has a density of 3.5 oz/ in³.

Then the volume of the coffee in each tank will be

We know that the dimension of both tanks is equal. Then the volume will remains equal.

Volume = L × B × H

Volume = 8.75 × 8.75 × 23

Volume = 1760.9375 ≅ 1760.94 in³

Then the mass of the coffee in each the tank will be

The mass of hot coffee will be

Mass = Density × Volume

Mass = 5.8 × 1760.94

Mass = 10213.45 oz

The mass of cold coffee will be

Mass = Density × Volume

Mass = 3.5 × 1760.94

Mass = 6163.29 oz

Then the number of the tank placed on a table is one only.

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2 years ago