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3 years ago

1. Find the distance from point P to QS

Mathematics

1 answer:

12345 [234]3 years ago

4 0

Answer:

5.7 units

Step-by-step explanation:

The distance from point P to QS is the distance from point P (1, 1) to the point of interception R(-3, 5).

Use distance formula to calculate distance between P and R:

$PR = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$

Let,

$P (1, 1) = (x_1, y_1)$

$R(-3, 5) = (x_2, y_2)$

Plug in the values into the formula.

$PR = \sqrt{(-3 - 1)^2 + (5 - 1)^2}$

$PR = \sqrt{(-4)^2 + (4)^2}$

$PR = \sqrt{16 + 16}$

$PR = \sqrt{32}$

$PR = 5.7 units$ (to nearest tenth)

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Eight times the reciprocal of a number equals 4 times the reciprocal of 8. find the number

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64 = 4x
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3 years ago

a shop is having a sale. all items are reduced by 20%.work out the sale price of an item normally priced at £37

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Answer:

£29.6

Step-by-step explanation:

37/5=7.4

37-7.4=29.6

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The half-life of a radioactive isotope is the time it takes for a quantity of the isotope to be reduced to half its initial mass

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3 years ago

The graph below represents the solution set of which inequality?

natulia [17]

Answer:

option: B ( $x^2+2x-8$ ) is correct.

Step-by-step explanation:

We are given the solution set as seen from the graph as:

(-4,2)

On solving the first inequality we have:

$x^2-2x-8$

On using the method of splitting the middle term we have:

$x^2-4x+2x-8$

⇒ $x(x-4)+2(x-4)=0$

⇒ $(x+2)(x-4)$

And we know that the product of two quantities are negative if either one of them is negative so we have two cases:

case 1:

$x+2>0$ and $x-4$

i.e. x>-2 and x<4

so we have the region as:

(-2,4)

Case 2:

$x+2$ and $x-4>0$

i.e. x<-2 and x>4

Hence, we did not get a common region.

Hence from both the cases we did not get the required region.

Hence, option 1 is incorrect.

We are given the second inequality as:

$x^2+2x-8$

On using the method of splitting the middle term we have:

$x^2+4x-2x-8$

⇒ $x(x+4)-2(x+4)$

⇒ $(x-2)(x+4)$

And we know that the product of two quantities are negative if either one of them is negative so we have two cases:

case 1:

$x-2>0$ and $x+4$

i.e. x>2 and x<-4

Hence, we do not get a common region.

Case 2:

$x-2$ and $x+4>0$

i.e. x<2 and x>-4

Hence the common region is (-4,2) which is same as the given option.

Hence, option B is correct.

$x^2-2x-8>0$

On using the method of splitting the middle term we have:

$x^2-4x+2x-8>0$

⇒ $x(x-4)+2(x-4)>0$

⇒ $(x-4)(x+2)>0$

And we know that the product of two quantities are positive if either both of them are negative or both of them are positive so we have two cases:

Case 1:

$x+2>0$ and $x-4>0$

i.e. x>-2 and x>4

Hence, the common region is (4,∞)

Case 2:

$x+2$ and $x-4$

i.e. x<-2 and x<4

Hence, the common region is: (-∞,-2)

Hence, from both the cases we did not get the desired answer.

Hence, option C is incorrect.

$x^2+2x-8>0$

On using the method of splitting the middle term we have:

$x^2+4x-2x-8>0$

⇒ $x(x+4)-2(x+4)>0$

⇒ $(x-2)(X+4)>0$

And we know that the product of two quantities are positive if either both of them are negative or both of them are positive so we have two cases:

Case 1:

$x-2$ and $x+4$

i.e. x<2 and x<-4

Hence, the common region is: (-∞,-4)

Case 2:

$x-2>0$ and $x+4>0$

i.e. x>2 and x>-4.

Hence, the common region is: (2,∞)

Hence from both the case we do not have the desired region.

Hence, option D is incorrect.

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3 years ago

Sam is flying a kite. The length of the kite string is 80 meters, and it makes an angle of 75° with the ground. The height of th

olya-2409 [2.1K]

Answer:

THE HEIGHT OF THE KITE FROM THE GROUND=80mtr

Step-by-step explanation:

let H be the height of kite from ground

$\sin\theta$ = $\frac{height}{hypotenuse}$

$\sin\90$ = $\frac{H}{80}$

1= $\frac{H}{80}$

H=80meters

5 0

3 years ago