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2 years ago

Seventh grade

Mathematics

1 answer:

nirvana33 [79]2 years ago

4 0

Answer:

Step-by-step explanation

47,400-36,498 = 10,902

10,902/ 47,400 = 0.23

0.23 = 23%

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Area of a trapezium is 31.5cm. If the parallel sides are of lengths 7.3cm and 5.3cm. Calculate the perpendicular distance betwee

Anna11 [10]

Answer:

5cm

Step-by-step explanation:

Given area of trapezium = 31.5 sq. cm

Length of parallel sides = 7.3cm and 5.3cm

Formula to calculate area of trapezium is

1/2*(sum of parallel sides)*perpendicular distance between parallel sides

sum of parallel sides = (7.3 + 5.3) = 12.6cm

substituting value of area given and sum of parallel sides we have

31.5 = 1/2* 12.6 * perpendicular distance between parallel sides

(31.5 * 2)/12.6 = perpendicular distance between parallel sides

perpendicular distance between parallel sides = 63/12.6 = 5

Therefore perpendicular distance between parallel sides for given trapezium is 5cm.

5 0

3 years ago

Really need this todayyyyyy

Bond [772]

Answer:

Q3: x1=2+i, x2=2-i Q4: x1=-3/2+i, x2=-3/2-i

Step-by-step explanation:

$x^{2} - 4x + 5 =0$

( $4$ ± $\sqrt{16 - 4*1*5}$ ) / $2$ = (

x1= 2+i

x2= 2-i

$2x^{2} +6x + 5 =0$

( $-6$ ± $\sqrt{36 - 4*2*5}$ ) / $4$ = (-6 ± $\sqrt{-4}$ )/4= (-3 ± i)/2

x1 = -3/2 +i

x2 = -3/2 -i

7 0

3 years ago

Read 2 more answers

Evaluate the surface integral S F · dS for the given vector field F and the oriented surface S.

sweet-ann [11.9K]

Answer:

2.794

Step-by-step explanation:

Recall that if G(x,y) is a parametrization of the surface S and F and G are smooth enough then

$\bf \displaystyle\iint_{S}FdS=\displaystyle\iint_{R}F(G(x,y))\cdot(\displaystyle\frac{\partial G}{\partial x}\times\displaystyle\frac{\partial G}{\partial y})dxdy$

F can be written as

F(x,y,z) = (xy, yz, zx)

and S has a straightforward parametrization as

$\bf G(x,y) = (x, y, 3-x^2-y^2)$

with 0≤ x≤1 and 0≤ y≤1

$\bf \displaystyle\frac{\partial G}{\partial x}= (1,0,-2x)\\\\\displaystyle\frac{\partial G}{\partial y}= (0,1,-2y)\\\\\displaystyle\frac{\partial G}{\partial x}\times\displaystyle\frac{\partial G}{\partial y}=(2x,2y,1)$

we also have

$\bf F(G(x,y))=F(x, y, 3-x^2-y^2)=(xy,y(3-x^2-y^2),x(3-x^2-y^2))=\\\\=(xy,3y-x^2y-y^3,3x-x^3-xy^2)$

and so

$\bf F(G(x,y))\cdot(\displaystyle\frac{\partial G}{\partial x}\times\displaystyle\frac{\partial G}{\partial y})=(xy,3y-x^2y-y^3,3x-x^3-xy^2)\cdot(2x,2y,1)=\\\\=2x^2y+6y^2-2x^2y^2-2y^4+3x-x^3-xy^2$

we just have then to compute a double integral of a polynomial on the unit square 0≤ x≤1 and 0≤ y≤1

$\bf \displaystyle\int_{0}^{1}\displaystyle\int_{0}^{1}(2x^2y+6y^2-2x^2y^2-2y^4+3x-x^3-xy^2)dxdy=\\\\=2\displaystyle\int_{0}^{1}x^2dx\displaystyle\int_{0}^{1}ydy+6\displaystyle\int_{0}^{1}dx\displaystyle\int_{0}^{1}y^2dy-2\displaystyle\int_{0}^{1}x^2dx\displaystyle\int_{0}^{1}y^2dy-2\displaystyle\int_{0}^{1}dx\displaystyle\int_{0}^{1}y^4dy+\\\\+3\displaystyle\int_{0}^{1}xdx\displaystyle\int_{0}^{1}dy-\displaystyle\int_{0}^{1}x^3dx\displaystyle\int_{0}^{1}dy-\displaystyle\int_{0}^{1}xdx\displaystyle\int_{0}^{1}y^2dy$

=1/3+2-2/9-2/5+3/2-1/4-1/6 = 2.794

5 0

3 years ago

Kira drew PQR and STU so that P S, Q T, PR = 12, and SU = 3. Are PQR and STU similar? If so, identify the similarity postulate o

Debora [2.8K]

B. Similar- Angle Angle Similarity Theorem
This theorem states that if 2 angles of a triangle are congruent to two angles of another triangle, then the two triangles are similar.
Hope this helps :)

8 0

3 years ago

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The function h(x) is given below.

Elina [12.6K]

Answer: Option (2)

Step-by-step explanation:

The inverse of a function swaps the domain and range.

5 0

1 year ago