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2 years ago

Need help from a math wiz

Mathematics

2 answers:

ruslelena [56]2 years ago

4 0

1) A vertex is a point. A point cannot be bisected. The problem is poorly worded.
If it means that each diagonal bisects two opposite angles, then the quadrilateral must be a rhombus.

2) In a trapezoid, one pair of opposite sides must be parallel.

3) A rhombus is a parallelogram with 4 congruent sides.

leva [86]2 years ago

3 0

I know the 3rd question the answer is rectangle

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Q11.

Kamila [148]

Answer:

Step-by-step explanation:

The area of a trapezium is given as;

Area = $\frac{1}{2}$ (a + b)h

where: a is the length of the upper side, b is the length of its base and h is its height.

This implies that; a = x - 4, b = x + 5, h = 2x and area = 351

Thus,

351 = $\frac{1}{2}$ ((x - 4) + (x + 50)) 2x

= (x - 4 + x + 5) x

531 = (2x + 1) x

= 2 $x^{2}$ + x

So that;

2 $x^{2}$ + x - 531 = 0

This gives a quadratic equation.

5 0

2 years ago

Determine the number of degrees of freedom for the two-sample t test or CI in each of the following situations. (Round your answ

gulaghasi [49]

Answer:

Part a ) The degrees of freedom for the given two sample non-pooled t test is 24

Part b ) The degrees of freedom for the given two sample non-pooled t test is 30

Part c ) The degrees of freedom for the given two sample non-pooled t test is 30

Part d ) The degrees of freedom for the given two sample non-pooled t test is 25

Step-by-step explanation:

Degrees of freedom for a non-pooled two sample t-test is given by;

Δf = {[ s₁²/m + s₂²/n ]²} / {[( s₁²/m)²/m-1] + [(s₂²/n)²/n-1]}

Now given the information;

a) :- m = 12, n = 15, s₁ = 4.0, s₂ = 6.0

we substitute

Δf = {[ 4²/12 + 6²/15 ]²} / {[( 4²/12)²/12-1] + [(6²/15)²/15-1]}

Δf = 30184 / 1241

Δf = 24.3223 ≈ 24 (down to the nearest whole number)

b) :- m = 12, n = 21, s₁ = 4.0, s₂ = 6.0

we substitute using same formula

Δf = {[ s₁²/m + s₂²/n ]²} / {[( s₁²/m)²/m-1] + [(s₂²/n)²/n-1]}

Δf = {[ 4²/12 + 6²/21 ]²} / {[( 4²/12)²/12-1] + [(6²/21)²/21-1]}

Δf = 56320 / 1871

Δf = 30.1015 ≈ 30 (down to the nearest whole number)

c) :- m = 12, n = 21, s₁ = 3.0, s₂ = 6.0

we substitute using same formula

Δf = {[ s₁²/m + s₂²/n ]²} / {[( s₁²/m)²/m-1] + [(s₂²/n)²/n-1]}

Δf = {[ 3²/12 + 6²/21 ]²} / {[( 3²/12)²/12-1] + [(6²/21)²/21-1]}

Δf = 29095 / 949

Δf = 30.6585 ≈ 30 (down to the nearest whole number)

d) :- m = 10, n = 24, s₁ = 4.0, s₂ = 6.0

we substitute using same formula

Δf = {[ s₁²/m + s₂²/n ]²} / {[( s₁²/m)²/m-1] + [(s₂²/n)²/n-1]}

Δf = {[ 4²/10 + 6²/24 ]²} / {[( 4²/10)²/10-1] + [(6²/24)²/24-1]}

Δf = 1044 / 41

Δf = 25.4634 ≈ 25 (down to the nearest whole number).

6 0

2 years ago

the length of a rectangle is 5 times as long as its breadth and its area is 1620 m² find area of square whose perimeter is equal

Rus_ich [418]

Answer:

Area of the square is 2916 $m^{2}$ .

Step-by-step explanation:

Let the breadth of the rectangle be represented by w. So that;

length of rectangle = 5w

Area of rectangle = length x breadth

= 5w x w

1620 = 5 $w^{2}$

divide through by 5 to have,

$w^{2}$ = 324

w = $\sqrt{324}$

= 18

Thus, the breadth of the rectangle is 18 m.

length = 5 w = 5 x 18

= 90 m

The length of the rectangle is 90 m.

Perimeter of the rectangle = 2(l + w)

= 2( 90 + 18)

= 216 m

Perimeter of a square = 4l

where l is the length of its side.

Given that the perimeters of the rectangle and square are equal, then;

4l = 216

l = $\frac{216}{4}$

= 54

length of the side of square is 54 m.

Therefore,

Area of square = $l^{2}$

= $54^{2}$

= 2916

Area of the square whose perimeter is equal to that of the rectangle is 2916 $m^{2}$ .

6 0

2 years ago

3x+4y=-20<br> 6x+5y=-7<br> whats the solution to the systum exasons

Ganezh [65]

Answer 5x6+

Step-by-step explanation:

5 0

2 years ago

Geometry hw 2.6 big ideas math

Mandarinka [93]

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6 0

3 years ago

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