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2 years ago

11

Immigration lead to faster expansion

2 answers:

julsineya [31]2 years ago

7 0

If this is a T or F question the answer would be True

pav-90 [236]2 years ago

4 0

This is in fact true, if this is a true or false question I would put true :) hope this helps

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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

What is the missing value for the probability distribution table ?

Brums [2.3K]

The answer is 0.3 cause when you subtract all of it you can get it

3 0

2 years ago

I need helpp its my final exam

Mashcka [7]

Answer:

hey i believe that your answer is A. without having to do the math if you look at it you can tell that its bigger then options B and C so those are not it. and it is to small to be D. so with that said your answer is A have a nice dy and good luck!

Step-by-step explanation:

7 0

3 years ago

Read 2 more answers

What are the solutions of the quadratic equation? 4x^2+34x+60=0

insens350 [35]

Answer:

Option 3) -6,-5/2

Step-by-step explanation:

we know that

The formula to solve a quadratic equation of the form

$ax^{2} +bx+c=0$

is equal to

$x=\frac{-b(+/-)\sqrt{b^{2}-4ac}} {2a}$

in this problem we have

$4x^{2} +34x+60=0$

so

$a=4\\b=34\\c=60$

substitute in the formula

$x=\frac{-34(+/-)\sqrt{34^{2}-4(4)(60)}} {2(4)}$

$x=\frac{-34(+/-)\sqrt{196}} {8}$

$x=\frac{-34(+/-)14} {8}$

$x_1=\frac{-34(+)14} {8}=-\frac{20}{8}=-\frac{5}{2}$

$x_2=\frac{-34(-)14} {8}=-6$

6 0

3 years ago

What is (2+y)³ written out?

Aleksandr-060686 [28]

${ \left( 2+y \right) }^{ 3 }\\ \\ =\left( 2+y \right) \left( 2+y \right) \left( 2+y \right) \\ \\ =\left( 2+y \right) \left( 4+4y+{ y }^{ 2 } \right)$

$\\ \\ =2\left( 4+4y+{ y }^{ 2 } \right) +y\left( 4+4y+{ y }^{ 2 } \right) \\ \\ =8+8y+2{ y }^{ 2 }+4y+4{ y }^{ 2 }+{ y }^{ 3 }\\ \\ ={ y }^{ 3 }+6{ y }^{ 2 }+12y+8$

5 0

3 years ago