Is it ok you guys maybe do this one too sorry... (−6)(−0.4)(−0.5)

Give your answer as a mixed number. 2 6/7 ÷ 2/3 =?

AfilCa [17]

2 6/7 ÷ 2/3

= 20/7 ÷ 2/3

20/7 × 3/2

= 60/14

reduce to it's lowest term

30/7 = 4 2/7

8 0

3 years ago

In a triangle the length of two sides are 1.9m and 0.7m. What is the length of the third side, if you know that it should be a w

monitta

Answer:

their sum: 2.8m

their difference: 1.2m

the third side's length should be smaller than their sum, larger than their difference

it could be: 1.3-1.4-1.5-1.6-1.7-1.8-1.9-2-2.1-2.2-2.3-2.4-2.5-2.6-2.7 but since it has to be a whole number, 2m is the only eligible answer.

6 0

3 years ago

Verify that the points are the vertices of a parallelogram, and find its area. A(1, 1, 3), B(2, −5, 6), C(4, −2, −1), D(3, 4, −4

almond37 [142]

Answer:

Area = 71.3 sq unit

Step-by-step explanation:

a) We will use properties of parallelogram to verify the given vertices.

Property: Have a pair of parallel opposite sides

Hence, vectors AB AD BD BC CD AC. A pair should be parallel:

vectors

AB = A - B = (1, 1 , 3) - ( 2 , -5 , 6 ) = (-1 , 6 , -3)

AD = A - D = (1, 1 , 3) - (3, 4 , -4) = (-2 , -3 , 7)

BC = B - C = (2 , -5 , 6) - (4, -2 , -1) = (-2 ,-3 , 7)

CD = C - D = (4, -2 , -1) - (3, 4 , -4) = (1 , -6 , 3)

Hence we can see that AB //CD and AD // BC as their unit vector co-efficents are identical/scalar multiple of each other.

b)

Equation of line AB = (1,1,3) + t*(-1 , 6 , -3 )

Point E = (1-t , 1+6t , 3 -3t) ... denotes an position of point E on line AB.

Choose a point on line CD, lets suppose C = ( 4 , -2 , -1 )

Vector EC = ( 1-t , 1+6t , 3-3t ) - (4 , -2 , -1 ) = (-3 -t , 3 +6t , 4 -3t)

For EC to be perpendicular to AB then dot product of AB . EC = 0

Hence,

EC . AB = -1 * (-3-t) -2*(3+6t) -3*(4-3t) = 0

3 + t -6 -12t -12+9t = 0

-2t-3 = 0

t = -3/2

Vector EC = (-3 -t , 3 +6t , 4 -3t) = (-1.5 , -6 , 8.5)

Area = magnitude (AB) * magnitude (EC)

Area = sqrt ((-1)^2 + 6^2 + (-3)^2) * sqrt ((-1.5)^2 + 6^2 + (8.5)^2)

Area = sqrt (46) * sqrt (442) / 2

Area = 71.3 unit^2

6 0

3 years ago

35 POINTS AVAILABLE

aliina [53]

Answer:

Part 1) The length of each side of square AQUA is $3.54\ cm$

Part 2) The area of the shaded region is $(486\pi-648)\ units^{2}$

Step-by-step explanation:

Part 1)

step 1

Find the radius of the circle S

The area of the circle is equal to

$A=\pi r^{2}$

we have

$A=25\pi\ cm^{2}$

substitute in the formula and solve for r

$25\pi=\pi r^{2}$

simplify

$25=r^{2}$

$r=5\ cm$

step 2

Find the length of each side of square SQUA

In the square SQUA

we have that

SQ=QU=UA=AS

$SU=r=5\ cm$

Let

x------> the length side of the square

Applying the Pythagoras Theorem

$5^{2}=x^{2} +x^{2}$

$5^{2}=2x^{2}$

$x^{2}=\frac{25}{2}\\ \\x=\sqrt{\frac{25}{2}}\ cm\\ \\ x=3.54\ cm$

Part 2) we know that

The area of the shaded region is equal to the area of the larger circle minus the area of the square plus the area of the smaller circle

Find the area of the larger circle

The area of the circle is equal to

$A=\pi r^{2}$

we have

$r=AB=18\ units$

substitute in the formula

$A=\pi (18)^{2}=324\pi\ units^{2}$

step 2

Find the length of each side of square BCDE

we have that

$AB=18\ units$

The diagonal DB is equal to

$DB=(2)18=36\ units$

Let

x------> the length side of the square BCDE

Applying the Pythagoras Theorem

$36^{2}=x^{2} +x^{2}$

$1,296=2x^{2}$

$648=x^{2}$

$x=\sqrt{648}\ units$

step 3

Find the area of the square BCDE

The area of the square is

$A=(\sqrt{648})^{2}=648\ units^{2}$

step 4

Find the area of the smaller circle

The area of the circle is equal to

$A=\pi r^{2}$

we have

$r=(\sqrt{648})/2\ units$

substitute in the formula

$A=\pi ((\sqrt{648})/2)^{2}=162\pi\ units^{2}$

step 5

Find the area of the shaded region

$324\pi\ units^{2}-648\ units^{2}+162\pi\ units^{2}=(486\pi-648)\ units^{2}$

7 0

3 years ago

Need help , due tomorrow. Plz give all correct answers for brainliest answer

velikii [3]

1, 2, 3, 5, and 7 are all functions. 4, 6, 8, and 9 are not functions.

4 0

3 years ago