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3 years ago

4) A rectangular court is 4 feet wide. The length is twice as long as the width. What is the area of the rectangular court?

Mathematics

1 answer:

inn [45]3 years ago

8 0

Answer:

B: 32

Step-by-step explanation:

Because the Length is twice the width, and the width is 4, that means the length is 8. Area= length×width, so the area is 4×8 which is 32

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Find the point on the directed segment from (−3, −2) to (4, 8) that divides it into a ratio of 3: 2

White raven [17]

Answer:

I think this is the correct answer

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3 years ago

Which ordered pair would form a proportional relationship with the point graphed below? On a coordinate plane, a line with negat

avanturin [10]

Answer:

Answer:

(5, 20)

Step-by-step explanation:

We can find the slope of the line

m = (y2-y1)/(x2-x1)

= (40-0)/(10-0)

= 40/10

Since the line goes through (0,0) the y intercept is 0

y = mx+b

y = 4x

Lets check the points by substituting into the equation

40, 10) 10 = 4*40 10 =160 no

(–5, –10) -10 = 4*-5 10 = -20 no

(5, 20) 20 = 4*5 20 = 20 yes

(–10, –20) -20 = 4*-10 -20 = -40 no

Step-by-step explanation:

Which ordered pair would form a proportional relationship with the point graphed below? On a coordinate plane, a line with negative slope goes through points (negative 20, 10), (0, 0), and (20, negative 10). (10, –20) (–30, 20) (–10, 5) (35, –20)

4 0

3 years ago

Read 2 more answers

Which property or properties of arithmetic do you use when you calculate 3×70 by first calculating 3×7 = 21 and then putting a z

ANEK [815]

The property used is the Associative property

3 × 70 = 3(70) = 3 (7×10) = (3×7) ×10

The formula is

a × bc = a (bc) = a (b×c) = (a×b) ×c

3 0

3 years ago

Pat has nine cards. Frank has two more cards than Steve. Steve has 3 cards how many more cards does Pat have than Frank?

n200080 [17]

4. Pat 9. Frank 5. Steve 3. 9-5=4.

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3 years ago

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Tom [10]

$\begin{array}{c|c}\underline{Statement}&\underline{Reason}\\ 1.AY=BX&\text{1. Given}\\ 2.AB \cong AB&\text{2. Reflexive Property}\\ 3. AD || BC&\text{3. Property of a square}\\ 4. \angle ABE \cong \angle AXB&\text{4. Alternate Interior Angles}\\ 5. \angle BAY \cong \angle BYA&\text{5. Alternate Interior Angles}\\6. \triangle BAX \cong \triangle ABY&\text{6. Angle-Side-Angle Theorem}\\ 7. AX \cong BY&\text{7. CPCTC}\\\end{array}$

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$\begin{array}{c|c}\underline{Statement}&\underline{Reason}\\1. AB=CF&\text{1. Given}\\2.AB+BF=A'F&\text{2. Segment Addition Postulate}\\3.CF+BF=A'F&\text{3. Substitution Property}\\4.CF+BF+BC&\text{4. Segment Addition Postulate}\\5.A'F=BC&\text{5. Transitive Property}\\6. \angle AFE = \angle DBC&\text{6. Given}\\7. EF = BD&\text{7. Given}\\8. \triangle AFE \cong \triangle CBD&\text{8. Side-Angle-Side Theorem}\\\end{array}$

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$\begin{array}{c|c}\underline{Statement}&\underline{Reason}\\\text{1.AC bisects }\angle BAD&\text{1. Given}\\2. \angle BAC \cong \angle DAC&\text{2. Property of angle bisector}\\3.AC = AC&\text{3. Reflexive Property}&4. \angle ACB \cong \angle ACD&\text{4. Property of angle bisector}\\5. \triangle ABC \cong \triangle ADC&\text{5. Angle-Side-Angle Theorem}\\6.BC=CD&\text{6. CPCTC}\\\end{array}$

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3 years ago