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Marianna [84]
3 years ago
6

When is a quadrilateral a parallelogram?

Mathematics
1 answer:
brilliants [131]3 years ago
6 0
C. when opposite sides are congruent and parallel
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Peter wants to purchase pizza pies and breadsticks for a party. The cashier tells him that pizza pies are $8 each and breadstick
Dmitry [639]
The answer is A.) 5x+8y≤120

To get this answer the first thing I did was notice what it says the x and y variables stand for. The "x variable represents the number of breadsticks purchased" and "the y variable represents the number of pizza pies purchased". It also says that each breadstick is $5 and each pizza pie is $8. Accordingly, we need to match our variables with what we're buying. So, the y be with 8 and the x be with 5.

So, our expression will have a 5x and an 8y in it.

Now if we notice, it says he can spend no more than $120, so that means he can spend $120 or less. The less than or equal to sign is ≤.

Now we can find our answer. The only answer with 5x and 8y with a ≤120 is answer choice A

Hope this helped!! :))

8 0
3 years ago
Read 2 more answers
I need help with both of them please!( Sample questions 13 and 14)
zaharov [31]
First~exercise! \\ We~ have `there ~some~ triangles.(2) \\ 1)~In~ a ~triangle~,the~degree~of~all~angles~is~180! \\ We~have~2~angles~known,so: \\ 180=x+45+56 \\ 180=101+x \\ x=79 \\  \\ Second~triangle! \\ We~have~to~find~2~angles! \\ An ~elongated`~angle~have~180~degrees! \\ 180=79+50+y \\ y=51. \\ In~the~second~triangle~,we~have~to~find~the`third~angle. \\ We~have~a~right~angle,so--\ \textgreater \ 90~degrees! \\ 180=90+51+z \\ z=39! \\ We~found~all~the~angles!(We~are`happy~now!!!~:))) \\ Second~exercise! \\ With~the~Pythagoras' Theorem,we~can~find~the~diagonal~in~this~ \\ rectangle! \\ c ^{2} =10 ^{2} +14 ^{2}  \\ c ^{2} =100+196 \\ c= \sqrt{296}  \\ c=17,20...
8 0
3 years ago
Which system of equations below has infinitely many solutions? y = –3x 4 and y = –3x – 4 y = –3x 4 and 3y = –9x 12 y = –3x 4 and
Flura [38]

the equations y = –3x + 4 and 3y = –9x + 12 have infinitely many solutions. option B is correct.

<h3>What is the linear system?</h3>

It is a system of an equation in which the highest power of the variable is always 1. A one-dimension figure that has no width. It is a combination of infinite points side by side.

Condition for the parallel lines.

L1,  ax + bx + c = 0

L2, dx + ey + f = 0

If \rm \dfrac{a}{d} = \dfrac{b}{e} = \dfrac{c}{f} then lines have infinitely many solutions.

<h3>Which system of equations below has infinitely many solutions?</h3>

y = –3x + 4 and 3y = –9x + 12

On comparing we have

a = -3 , b = 1, and c = 4

d = -9 , e = 3, and f = 12

Then their ratio will be

\rm \dfrac{1}{3} = \dfrac{-3}{-9} = \dfrac{4}{12}\\\\\rm \dfrac{1}{3} = \dfrac{1}{3} = \dfrac{1}{3}

Hence  y = –3x + 4 and 3y = –9x + 12 have infinitely many solutions.

Thus the option B is correct.

More about the linear system link is given below.

brainly.com/question/20379472

7 0
2 years ago
A weir is used to measure water flow in a channel. For a rectangular broad crested weir, the flow Q in cubic feet per second is
NikAS [45]

Answer:Find the water height for a weir that is 3 feet long and has flow of 38.4 cubic feet per second. SOLUTION: Therefore, the water height for the weir is 4 feet.

6 0
3 years ago
Read 2 more answers
Write one sine and one cosine equation for each graph below.
pychu [463]

Answer:

Q13. y = sin(2x – π/2); y = - 2cos2x  

Q14. y = 2sin2x -1; y = -2cos(2x – π/2) -1

Step-by-step explanation:

Question 13

(A) Sine function

y = a sin[b(x - h)] + k

y = a sin(bx - bh) + k; bh = phase shift

(1) Amp = 1; a = 1

(2) The graph is symmetrical about the x-axis. k = 0.

(3) Per = π. b = 2

(4) Phase shift = π/2.  

2h =π/2

h = π/4

The equation is

y = sin[2(x – π/4)} or

y = sin(2x – π/2)

B. Cosine function

y = a cos[b(x - h)] + k

y = a cos(bx - bh) + k; bh = phase shift

(1) Amp = 1; a = 1

(2) The graph is symmetrical about the x-axis. k = 0.

(3) Per = π. b = 2

(4) Reflected across x-axis, y ⟶ -y

The equation is y = - 2cos2x  

Question 14

(A) Sine function

(1) Amp = 2; a = 2

(2) Shifted down 1; k = -1

(3) Per = π; b = 2

(4) Phase shift = 0; h = 0

The equation is y = 2sin2x -1

(B) Cosine function

a = 2, b = -1; b = 2

Phase shift = π/2; h = π/4

The equation is

y = -2cos[2(x – π/4)] – 1 or

y = -2cos(2x – π/2) - 1

6 0
3 years ago
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