Which equation below can you use to solve for the width, w, of this rectangle?

Segments UV and WZ are parallel with line ST intersecting both at points Q and R respectively

Zanzabum

The correct answer is:

Corresponding (for both blanks)

Explanation:

Corresponding angles are angles that are in the same relative position to both the parallel line they are on and the transversal. In this case, SQU and WRS are both above the parallel line they are on and on the same side of the transversal; this makes them corresponding angles. This also means it is the corresponding angles postulate.

3 0

3 years ago

Read 2 more answers

What is 2 5/12 -(-2 1/6) Express the answer in simplest form.

VladimirAG [237]

Answer:
= 29/12 − -13/6

= ((29 × 6) − (-13 × 12)) / (12 × 6) 

= (174 - -156) / 72

= 330/72

= 55/12

= 4 7/12

3 0

3 years ago

Factories 54m3n + 81m4n2 b) 15x2y3z + 25x3y2z + 35x2y2z

Oliga [24]

Part a)

Given the expression

$54m^3n\:+\:81m^4n^2$

Apply exponent rule: $a^{b+c}=a^ba^c$

$54m^3n\:+\:81m^4n^2=54m^3n+81m^3mnn$ ∵ $m^4n^2=m^3mnn$

Rewrite 81 as 3 · 27

Rewrite 54 as 2 · 27

$=2\cdot \:27m^3n+3\cdot \:27m^3mnn$

Factor out the common term: 27m³n

$=27m^3n\left(2+3mn\right)$

Therefore,

$54m^3n\:+\:81m^4n^2=54m^3n+81m^3mnn=27m^3n\left(2+3mn\right)$

Part B)

Given the expression

$15x^2y^3z+25x^3y^2z+35x^2y^2z$

Apply exponent rule: $a^{b+c}=a^ba^c$

$15x^2y^3z+25x^3y^2z+35x^2y^2z=15x^2y^2yz+25x^2xy^2z+35x^2y^2z$

Rewrite as

$=3\cdot \:5y^2x^2zy+5\cdot \:5y^2x^2zx+7\cdot \:5y^2x^2z$

Factor out common term 5y²x²z

$=5y^2x^2z\left(3y+5x+7\right)$

Therefore,

$15x^2y^3z+25x^3y^2z+35x^2y^2z=5y^2x^2z\left(3y+5x+7\right)$

8 0

3 years ago

Every prime number greater than 10 has a digit in the ones place that is include in wich set of numbers A- 1,3,7,9 B-1,3,5,9 D-1

marta [7]

That's very interesting. I had never thought about it before.
Let's look through all of the ten possible digits in that place,
and see what we can tell:

-- 0:
   A number greater than 10 with a 0 in the units place is a multiple of
   either 5 or 10, so it's not a prime number.

-- 1:
    A number greater than 10 with a 1 in the units place could be
    a prime (11, 31 etc.) but it doesn't have to be (21, 51).

-- 2:
   A number greater than 10 with a 2 in the units place has 2 as a factor
   (it's an even number), so it's not a prime number.

-- 3:
   A number greater than 10 with a 3 in the units place could be
   a prime (13, 23 etc.) but it doesn't have to be (33, 63) .

-- 4:
   A number greater than 10 with a 4 in the units place is an even
   number, and has 2 as a factor, so it's not a prime number.

-- 5:
   A number greater than 10 with a 5 in the units place is a multiple
   of either 5 or 10, so it's not a prime number.

-- 6:
   A number greater than 10 with a 6 in the units place is an even
   number, and has 2 as a factor, so it's not a prime number.

-- 7:
   A number greater than 10 with a 7 in the units place could be
   a prime (17, 37 etc.) but it doesn't have to be (27, 57) .

-- 8:
   A number greater than 10 with a 8 in the units place is an even
   number, and has 2 as a factor, so it's not a prime number.

-- 9:
   A number greater than 10 with a 9 in the units place could be
   a prime (19, 29 etc.) but it doesn't have to be (39, 69) .

So a number greater than 10 that IS a prime number COULD have
any of the digits 1, 3, 7, or 9 in its units place.

It CAN't have a 0, 2, 4, 5, 6, or 8 .

The only choice that includes all of the possibilities is 'A' .

4 0

3 years ago

The denominator of a rational numbers is twice it’s numerator if the numerator is decreased by 3 and the denominator increased b

ASHA 777 [7]

Answer:

$\frac{29}{58}$