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Marta_Voda [28]

2 years ago

15

A wire is on the shape of a square of side 10 cm. If the wire is rebent into rectangle of length 12 cm, find its breadth. Which

encloses more area, the square or the rectangle?

1 answer:

frutty [35]2 years ago

4 0

Answer:

The square kase ahhh! Tag 12cm ang sides so same sila ng rectangle

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Zane needs 20 yards of fencing to go around his rectangular yard. What are possible dimensions for Zane's yard? Select all that

stiv31 [10]

Answer:

1 yard by 9 yards

2 yards by 8 yards

3 yards by 7 yards

4 yards by 6 yards

Step-by-step explanation:

the perimeter of the rectangular yard needs to be 20 yards

Perimeter of a rectangle = 2 x ( length + width)

20 = 2 x ( length + width)

( length + width) = 20 /2

( length + width) = 10

The sum of the length and width should equal 10

Possible dimensions that would equal 10 are :

1 yard by 9 yards

2 yards by 8 yards

3 yards by 7 yards

4 yards by 6 yards

7 0

3 years ago

Can somebody plz help give the correct equations for both of the word problems (only if u done this before) thx :D

fgiga [73]

Answer:

A. 12 cm

B. brent scored 9

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4 0

2 years ago

Read 2 more answers

All vectors are in Rn. Check the true statements below:

Oduvanchick [21]

Answer:

A), B) and D) are true

Step-by-step explanation:

A) We can prove it as follows:

$Proy_{cv}y=\frac{(y\cdot cv)}{||cv||^2}cv=\frac{c(y\cdot v)}{c^2||v||^2}cv=\frac{(y\cdot v)}{||v||^2}v=Proy_{v}y$

B) When you compute the product Ax, the i-th component is the matrix of the i-th column of A with x, denote this by Ai x. Then, we have that $||Ax||=\sqrt{(A_1 x)^2+\cdots (A_n x)^2}$ . Now, the colums of A are orthonormal so we have that (Ai x)^2=x_i^2. Then $||Ax||=\sqrt{(x_1)^2+\cdots (x_n)^2}=||x||$ .

C) Consider $S=\{(0,2),(2,0)\}\subseteq \mathbb{R}^2$ . This set is orthogonal because $(0,2)\cdot(2,0)=0(2)+2(0)=0$ , but S is not orthonormal because the norm of (0,2) is 2≠1.

D) Let A be an orthogonal matrix in $\mathbb{R}^n$ . Then the columns of A form an orthonormal set. We have that $A^{-1}=A^t$ . To see this, note than the component $b_{ij}$ of the product $A^t A$ is the dot product of the i-th row of $A^t$ and the jth row of $A$ . But the i-th row of $A^t$ is equal to the i-th column of $A$ . If i≠j, this product is equal to 0 (orthogonality) and if i=j this product is equal to 1 (the columns are unit vectors), then $A^t A=I$

E) Consider S={e_1,0}. S is orthogonal but is not linearly independent, because 0∈S.

In fact, every orthogonal set in R^n without zero vectors is linearly independent. Take a orthogonal set $\{u_1,u_2\cdots u_p\}$ and suppose that there are coefficients a_i such that $a_1u_1+a_2u_2\cdots a_nu_n=0$ . For any i, take the dot product with u_i in both sides of the equation. All product are zero except u_i·u_i=||u_i||. Then $a_i||u_i||=0$ then $a_i=0$ .

5 0

3 years ago

∆ABC has vertices A(–2, 0), B(0, 8), and C(4, 2)

Natali [406]

Answer:

Part 1) The equation of the perpendicular bisector side AB is $y=-\frac{1}{4}x+\frac{15}{4}$

Part 2) The equation of the perpendicular bisector side BC is $y=\frac{2}{3}x+\frac{11}{3}$

Part 3) The equation of the perpendicular bisector side AC is $y=-3x+4$

Part 4) The coordinates of the point P(0.091,3.727)

Step-by-step explanation:

Part 1) Find the equation of the perpendicular bisector side AB

we have

A(–2, 0), B(0, 8)

<em>step 1</em>

Find the slope AB

The formula to calculate the slope between two points is equal to

$m=\frac{y2-y1}{x2-x1}$

substitute the values

$m=\frac{8-0}{0+2}$

$m=4$

<em>step 2</em>

Find the slope of the perpendicular line to side AB

Remember that

If two lines are perpendicular, then their slopes are opposite reciprocal (the product of their slopes is equal to -1)

therefore

The slope is equal to

$m=-\frac{1}{4}$

<em>step 3</em>

Find the midpoint AB

The formula to calculate the midpoint between two points is equal to

$M(\frac{x1+x2}{2},\frac{y1+y2}{2})$

substitute the values

$M(\frac{-2+0}{2},\frac{0+8}{2})$

$M(-1,4)$

<em>step 4</em>

Find the equation of the perpendicular bisectors of AB

the slope is $m=-\frac{1}{4}$

passes through the point $(-1,4)$

The equation in slope intercept form is equal to

$y=mx+b$

substitute

$4=(-\frac{1}{4})(-1)+b$

solve for b

$b=4-\frac{1}{4}$

$b=\frac{15}{4}$

so

$y=-\frac{1}{4}x+\frac{15}{4}$

Part 2) Find the equation of the perpendicular bisector side BC

we have

B(0, 8) and C(4, 2)

<em>step 1</em>

Find the slope BC

The formula to calculate the slope between two points is equal to

$m=\frac{y2-y1}{x2-x1}$

substitute the values

$m=\frac{2-8}{4-0}$

$m=-\frac{3}{2}$

<em>step 2</em>

Find the slope of the perpendicular line to side BC

Remember that

If two lines are perpendicular, then their slopes are opposite reciprocal (the product of their slopes is equal to -1)

therefore

The slope is equal to

$m=\frac{2}{3}$

<em>step 3</em>

Find the midpoint BC

The formula to calculate the midpoint between two points is equal to

$M(\frac{x1+x2}{2},\frac{y1+y2}{2})$

substitute the values

$M(\frac{0+4}{2},\frac{8+2}{2})$

$M(2,5)$

<em>step 4</em>

Find the equation of the perpendicular bisectors of BC

the slope is $m=\frac{2}{3}$

passes through the point $(2,5)$

The equation in slope intercept form is equal to

$y=mx+b$

substitute

$5=(\frac{2}{3})(2)+b$

solve for b

$b=5-\frac{4}{3}$

$b=\frac{11}{3}$

so

$y=\frac{2}{3}x+\frac{11}{3}$

Part 3) Find the equation of the perpendicular bisector side AC

we have

A(–2, 0) and C(4, 2)

<em>step 1</em>

Find the slope AC

The formula to calculate the slope between two points is equal to

$m=\frac{y2-y1}{x2-x1}$

substitute the values

$m=\frac{2-0}{4+2}$

$m=\frac{1}{3}$

<em>step 2</em>

Find the slope of the perpendicular line to side AC

Remember that

If two lines are perpendicular, then their slopes are opposite reciprocal (the product of their slopes is equal to -1)

therefore

The slope is equal to

$m=-3$

<em>step 3</em>

Find the midpoint AC

The formula to calculate the midpoint between two points is equal to

$M(\frac{x1+x2}{2},\frac{y1+y2}{2})$

substitute the values

$M(\frac{-2+4}{2},\frac{0+2}{2})$

$M(1,1)$

<em>step 4</em>

Find the equation of the perpendicular bisectors of AC

the slope is $m=-3$

passes through the point $(1,1)$

The equation in slope intercept form is equal to

$y=mx+b$

substitute

$1=(-3)(1)+b$

solve for b

$b=1+3$

$b=4$

so

$y=-3x+4$

Part 4) Find the coordinates of the point of concurrency of the perpendicular bisectors (P)

we know that

The point of concurrency of the perpendicular bisectors is called the circumcenter.

Solve by graphing

using a graphing tool

the point of concurrency of the perpendicular bisectors is P(0.091,3.727)

see the attached figure

5 0

3 years ago

Find the new cost including Sales Tax:<br> $99 headphones; 5% tax

rjkz [21]

Answer:

103.95

Step-by-step explanation:

99×5/100=4.95

99+4.95=103.95

6 0

3 years ago