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3 years ago

Que números primos son mayores que -6

Mathematics

1 answer:

kumpel [21]3 years ago

7 0

No sé, prueba 2,3,5,7,11,13....
Hay muchos números primos!

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You must be at least 42 inches tall to ride the bumper cars at an amusement park. Write an inequality that represents this situa

Anestetic [448]

Answer:

x is greater than or equal to 42 inches (I can't do the smybol for it)

Step-by-step explanation:

5 0

3 years ago

If x = 4, y = 3, z = -5, then evaluate: z - z + x - y

deff fn [24]

I got -15 as the answer, hope it works

7 0

3 years ago

Read 2 more answers

in the figure below trianglePQM and triangleQRP are right triangles. the measure of lineQM is 6 and the measure of lineQP is 8.

Kipish [7]

Answer:

Option 4 is correct. The length of PR is 6.4 units.

Step-by-step explanation:

From the given figure it is noticed that the triangle PQR and triangle MQR.

Let the length of PR be x.

Pythagoras formula

$hypotenuse^2=base^2+perpendicular^2$

Use pythagoras formula for triangle PQM.

$PM^2=QM^2+PQ^2$

$PM^2=(6)^2+(8)^2$

$PM^2=36+64$

$PM^2=100$

$PM=10$

The value of PM is 10. The length of PR is x, so the length of MR is (10-x).

Use pythagoras formula for triangle PQR.

$PQ^2=QR^2+PR^2$

$(8)^2=QR^2+x^2$

$64-x^2=QR^2$ .....(1)

Use pythagoras formula for triangle MQR.

$MQ^2=QR^2+MR^2$

$(6)^2=QR^2+(10-x)^2$

$36=QR^2+x^2-20x+100$

$36-x^2+20x-100=QR^2$ .... (2)

From equation (1) and (2) we get

$36-x^2+20x-100=64-x^2$

$20x-64=64$

$20x=128$

$x=6.4$

Therefore length of PR is 6.4 units and option 4 is correct.

3 0

3 years ago

Two cargo ships are sailing a direct course into harbor. Ship A is 22 miles from harbor. Ship Bis sailing into harbor on a cours

s344n2d4d5 [400]

Answer:

The distance of ship B from the Harbor is 32.26 miles

Step-by-step explanation:

Hello, great question. These types are questions are the beginning steps for learning more advanced Equations.

I have added a visual aid of the problem. As we can see the courses of both ship to the harbor forms a triangle with ship B being the hypotenuse. Since we are given the distance of the course for ship A and the angle between both ships, we can use this information with the <u>COSINE operator</u> to solve for the length of the course of ship B.

$cos(x) = \frac{adjacent}{hypotenuse}$