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2 years ago

Where y=5x-7 and -3x-2y=-12 Write your answer as x,y

Mathematics

1 answer:

Mariulka [41]2 years ago

5 0

Answer:

(x,y) = (2,3)

Step-by-step explanation:

y=5x-7 (i)

-3x-2y= -12

-(3x+2y)= -12

3x+2y= 12 (ii)

By putting the value of y from (i) in (ii)

3x+2(5x-7)=12

3x+10x-14=12

13x=12+14

13x= 26

x = $\frac{26}{13}$

x= 2

By putting the value of x in (i)

y= 5(2)-7

y= 10-7

y= 3

(x,y) = (2,3)

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Find an equation of the set of all points equidistant from the points A(−3, 6, 3) and B(4, 1, −1). Describe the set.

valentinak56 [21]

Answer:

See below

Step-by-step explanation:

I will describe this set in R³. Let P=(x,y,z) be a point equidistant to A and B, that is, the distance from P to A is equal to the distance from P to B.

First, using the usual distance formula, the distance from P to A is equal to $d(P,A)=\sqrt{(x-(-3))^2+(y-6)^2+(z-3)^2}=\sqrt{(x+3)^2+(y-6)^2+(z-3)^2}$

On the other hand, the distance form P to B is equal to $d(P,B)=\sqrt{(x-4)^2+(y-1)^2+(z-(-1))^2}=\sqrt{(x-4)^2+(y-1)^2+(z+1)^2}$

P is equidistant from A and B if and only if P satisfies the equation d(P,A)=d(P,B), that is,

$\sqrt{(x+3)^2+(y-6)^2+(z-3)^2}=\sqrt{(x-4)^2+(y-1)^2+(z+-1)^2}$

Take the square in both sides of this equation to get

$(x+3)^2+(y-6)^2+(z-3)^2=(x-4)^2+(y-1)^2+(z+1)^2$

$(x+3)^2-(x-4)^2+(y-6)^2-(y-1)^2+(z-3)^2-(z+1)^2=0$

You can simplify using difference of squares and multiplying like this:

$(7)(2x-1)+(-5)(2y-7)+(-4)(2z-2)=0$

$14x-10y-8z+36=0$

which is the equation of a plane.

4 0

3 years ago

Which of the following sets are equal to {x | x > 9 and x < 2}

Arada [10]

Answer:

{ }

Step-by-step explanation:

There are no numbers that are both greater than 9 and less than 2. The expression describes the empty set.

7 0

2 years ago

A hospital claims that the proportion, p, of full-term babies born in their hospital that weigh more than 7 pounds is 36%. In a

Natali [406]

Answer:

Claim :The proportion, p, of full-term babies born in their hospital that weigh more than 7 pounds is 36%.

n = 170

x = 56

We will use one sample proportion test

$\widehat{p}=\frac{x}{n}$

$\widehat{p}=\frac{56}{170}$

$\widehat{p}=0.3294$

The proportion, p, of full-term babies born in their hospital that weigh more than 7 pounds is 36%.

$H_0:p \neq 0.36 \\H_a:p= 0.36$

Formula of test statistic = $\frac{\widehat{p}-p}{\sqrt{\frac{p(1-p)}{n}}}$

$=\frac{0.3294-0.36}{\sqrt{\frac{0.36(1-0.36)}{170}}}$

=−0.8311

Now refer the p value from the z table

P-Value is .202987 (Calculated by online calculator)

Level of significance α = 0.05

Since p value < α

So we reject the null hypothesis .

Hence the claim is true

6 0

2 years ago

3 (x + 1) - 2y= 0

Rina8888 [55]

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