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2 years ago

Which ordered pair would form a proportional relationship with the point graphed below

Mathematics

2 answers:

butalik [34]2 years ago

8 0

Your answer should be b

mr_godi [17]2 years ago

6 0

Answer:

(15,-30) when y=-2x

Step-by-step explanation:

As you can see in the point in the graph the relationship between x and y would be y=-2x or y=x+60, but since proportional relationships require that y equals x times something that is not an option.

So we just have to evaluate the possibilities from the points given for this problem with the relationship found:

(-40,20)

y=-2x> y=-2(-40)> y=80> since y is not equal to 80 this one is wrong

(-10,-20)

y=-2x> y=-2(-10)> y=20 since y= 20 is not true this is wrong

(15,-30)

y=-2x> y=15(-2)> y=-30 since y=-30 this is correct

(5,-15)

y=-2x> y=-2(5)> y=-10 since y=-10 is not true, this is wrong

So after evaluating the options we know that the correct one that would create a proprotional relationship would be (15,-30)

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A rectangular portrait measures 50cm by 70cm. It is surrounded by a rectangular frame of uniform width. If the area of the frame

snow_tiger [21]

Let us assume uniform width = x cm wide.

Length of rectangular portrait = 50cm and width of rectangular portrait = 70cm.

Therefore, length of rectangle made by frame = 50 + x+x = (2x+50) cm.

And width of rectangle made by frame = 70+x+x = (2x+70) cm.

We know, the area of rectangular portrait = 50 × 70 = 3500 cm^2.

Total area of the rectangle made by frame would be = (2x+50) * (2x+70)

We know,

Actual area of frame = Area of rectangle made by frame - area of rectangular portrait.

We also given "the area of the frame is the same as the area of the portrait."

We can setup an equation now,

3500 = (2x+50) * (2x+70) - 3500.

Subtracting 3500 from both sides, we get

3500-3500 = (2x+50) * (2x+70) - 3500-3500.

0 = (2x+50) * (2x+70) -7000.

FOIL (2x+50) * (2x+70), we get

0 = 2x*2x +2x*70 + 50*2x +50*70 - 7000.

0 = 4x^2 +140x +100x +3500 -7000.

4x^2 +240x -3500 = 0.

Dividing whole equation by 4, we get

x^2 +60x - 875 =0

Applying quadratic formula $=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$ , we get

$=\frac{-60\pm \sqrt{60^2-4\cdot \:1\left(-875\right)}}{2\cdot \:1}$

$x=\frac{-60+\sqrt{60^2-4\cdot \:1\left(-875\right)}}{2\cdot \:1}=5\left(\sqrt{71}-6\right)$

$x=\frac{-60-\sqrt{60^2-4\cdot \:1\left(-875\right)}}{2\cdot \:1}:\quad -5\left(6+\sqrt{71}\right)$

We cant take negative value.

So, $x=5\left(\sqrt{71}-6\right)=12.13$

We could take approximately 12 cm.

7 0

2 years ago

Please tell me the answer and show the work please

Alexeev081 [22]

In case you can't find
(1 * 10^-6)*(7.5 * 10^18) = (1*7.5) * 10^(-6+18) = 7.5*10^12
in your head, your calculator can help.

6 0

2 years ago

Discuss the validity of the following statement. If the statement is always true, explain why. If not, give a counterexample.

Zarrin [17]

Correction:

Because F is not present in the statement, instead of working onP(E)P(F) = P(E∩F), I worked on

P(E∩E') = P(E)P(E').

Answer:

The case is not always true.

Step-by-step explanation:

Given that the odds for E equals the odds against E', then it is correct to say that the E and E' do not intersect.

And for any two mutually exclusive events, E and E',

P(E∩E') = 0

Suppose P(E) is not equal to zero, and P(E') is not equal to zero, then

P(E)P(E') cannot be equal to zero.

P(E)P(E') ≠ 0

This makes P(E∩E') different from P(E)P(E')

Therefore,

P(E∩E') ≠ P(E)P(E') in this case.

8 0

2 years ago

You and your friend are selling magazine subscriptions for a fundraiser. After w w weeks, you have sold (3w+4) subscriptions and

alukav5142 [94]

The answer is 7 cause it's 7

4 0

2 years ago

Read 2 more answers

The perimeter of a rectangular flower garden is 80 m and its area is 375 m2. Find the length of the garden.

Umnica [9.8K]

The answer is 25m.
Height = 15
Length = 25
Height+lengthx2= 80(perimeter)
Height x length=375m2(area)

5 0

2 years ago

Which ordered pair would form a proportional relationship with the point graphed below​

Which ordered pair would form a proportional relationship with the point graphed below