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3 years ago

Which expression are equivalent to q+p+p+q

Mathematics

2 answers:

Keith_Richards [23]3 years ago

7 0

Answer:

p^2 + q^2 is the equivalent term

Step-by-step explanation:

Dimas [21]3 years ago

6 0

Answer:

I think it's iq

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30 men repair a road in 56 days by working 6 hours daily. In how many days 45 men will repair the same road by working 7 hours d

svetoff [14.1K]

Step-by-step explanation:

(30 men, 56 days, 6 hours daily)

=> (30 men, 48 days, 7 hours daily)

=> (45 men, 32 days, 7 hours daily)

Hence it will take 32 days.

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2 years ago

Find the area A if b = 14 cm and h = 6 cm

kipiarov [429]

(14)(6)=84 (base)(height)

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3 years ago

for every three short horned lizards there are two fence lizards. If there are 54 short horned lizards how many fence lizards ar

Komok [63]

Answer: 36

Step-by-step explanation:

54/3 = 18 x 2 = 36

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3 years ago

2x-5) = 2(x-5) One Solution Infinitely Many Solutions No Solution 5x + 4) = 5(1-6) 5(x + 4) = 3(x-6) 4x - 2) = 4(x + 2) x + 2(x-

Gekata [30.6K]

9514 1404 393

Answer:

top down: ∞, 0, 1, 0, ∞

Step-by-step explanation:

The equation will have infinite solutions when the left side and right side simplify to the same expression. This is the case for the first and last expressions listed.

2(x -5) = 2(x -5) . . . . expressions are already identical

x +2(x -5) = 3(x -2) -4 ⇒ 3x -10 = 3x -10 . . . the same simplified expression

The equation will have no solutions when the x-coefficients are the same, but there are different added constants.

5(x +4) = 5(x -6) ⇒ x +4 = x -6 . . . not true for any x

4(x -2) = 4(x +2) ⇒ x -2 = x +2 . . . not true for any x

The equation will have one solution when coefficients of x are different.

5(x +4) = 3(x -6) ⇒ 2x = -38 ⇒ x = -19

4 0

2 years ago

In the △ABC, the height AN = 24 in, BN = 18 in, AC = 40 in. Find AB and BC. Consider all possible cases. Case 1 : N∈BC, AB = __,

aivan3 [116]

Answer:

Case 1:

$AB = 30$

$BC = 50$

Case 2:

$AB = 15.9$

$BC = 36.7$

Case 3: Not possible

Step-by-step explanation:

Given

See attachment for illustration of each case

Required

Find AB and BC

Case 1:

Using Pythagoras theorem in ANB, we have:

$AB^2 = AN^2 + BN^2$

This gives:

$AB^2 = 24^2 + 18^2$

$AB^2 = 576 + 324$

$AB^2 = 900$

Take square roots of both sides

$AB = \sqrt{900$

$AB = 30$

To calculate BC, we consider ANC, where:

$AC^2 = AN^2 + NC^2$

$40^2 = 24^2 + NC^2$

$1600 = 576 + NC^2$

Collect like terms

$NC^2 = 1600 - 576$

$NC^2 = 1024$

Take square roots

$NC = \sqrt{1024$

$NC = 32$

So:

$BC = NC + BN$

$BC = 32 + 18$

$BC = 50$

Case 2:

Using Pythagoras theorem in ANB, we have:

$AN^2 = AB^2 + BN^2$

This gives:

$24^2 = AB^2 + 18^2$

$576 = AB^2 + 324$

Collect like terms

$AB^2 = 576 - 324$

$AB^2 = 252$

Take square roots of both sides

$AB = \sqrt{252$

$AB = 15.9$

To calculate BC, we consider ABC, where:

$AC^2 = AB^2 + BC^2$

$40^2 = 252 + BC^2$

$1600 = 252 + BC^2$

Collect like terms

$BC^2 = 1600 - 252$

$BC^2 = 1348$

Take square roots

$BC = \sqrt{1348$

$BC = 36.7$

Case 3:

This is not possible because in ANC

The hypotenuse AN (24) is less than AC (40)

3 0

2 years ago