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2 years ago

A rectangular drawing is enlarged by 30%. The original dimensions of this drawing are 16cm x 24cm.

Mathematics

1 answer:

Phoenix [80]2 years ago

4 0

Answer:

Step-by-step explanation: Scale $\frac{130}{100} = \frac{13}{10}$

New dimensions $16 * 1.3 --- 24*1.3 =20.8 cm * 31.2 cm$

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in exercises 15-20 find the vector component of u along a and the vecomponent of u orthogonal to a u=(2,1,1,2) a=(4,-4,2,-2)

Nimfa-mama [501]

Answer:

The component of $\vec u$ orthogonal to $\vec a$ is $\vec u_{\perp\,\vec a} = \left(\frac{9}{5},\frac{6}{5},\frac{9}{10},\frac{21}{10}\right)$ .

Step-by-step explanation:

Let $\vec u$ and $\vec a$ , from Linear Algebra we get that component of $\vec u$ parallel to $\vec a$ by using this formula:

$\vec u_{\parallel \,\vec a} = \frac{\vec u \bullet\vec a}{\|\vec a\|^{2}} \cdot \vec a$ (Eq. 1)

Where $\|\vec a\|$ is the norm of $\vec a$ , which is equal to $\|\vec a\| = \sqrt{\vec a\bullet \vec a}$ . (Eq. 2)

If we know that $\vec u =(2,1,1,2)$ and $\vec a=(4,-4,2,-2)$ , then we get that vector component of $\vec u$ parallel to $\vec a$ is:

$\vec u_{\parallel\,\vec a} = \left[\frac{(2)\cdot (4)+(1)\cdot (-4)+(1)\cdot (2)+(2)\cdot (-2)}{4^{2}+(-4)^{2}+2^{2}+(-2)^{2}} \right]\cdot (4,-4,2,-2)$

$\vec u_{\parallel\,\vec a} =\frac{1}{20}\cdot (4,-4,2,-2)$

$\vec u_{\parallel\,\vec a} =\left(\frac{1}{5},-\frac{1}{5},\frac{1}{10},-\frac{1}{10} \right)$

Lastly, we find the vector component of $\vec u$ orthogonal to $\vec a$ by applying this vector sum identity:

$\vec u_{\perp\,\vec a} = \vec u - \vec u_{\parallel\,\vec a}$ (Eq. 3)

If we get that $\vec u =(2,1,1,2)$ and $\vec u_{\parallel\,\vec a} =\left(\frac{1}{5},-\frac{1}{5},\frac{1}{10},-\frac{1}{10} \right)$ , the vector component of $\vec u$ is:

$\vec u_{\perp\,\vec a} = (2,1,1,2)-\left(\frac{1}{5},-\frac{1}{5},\frac{1}{10},-\frac{1}{10} \right)$

$\vec u_{\perp\,\vec a} = \left(\frac{9}{5},\frac{6}{5},\frac{9}{10},\frac{21}{10}\right)$

The component of $\vec u$ orthogonal to $\vec a$ is $\vec u_{\perp\,\vec a} = \left(\frac{9}{5},\frac{6}{5},\frac{9}{10},\frac{21}{10}\right)$ .

4 0

2 years ago

The length of a rectangle is 4 less than twice the width. The perimeter of the rectangle is 106 cm. Determine the length of the

horsena [70]

Answer:

Length = 34

Step-by-step explanation:

Perimeter of rectangle = a+a+b+b = 2(a+b)

Let a = length = 2b - 4

Let b = width

Therefore

(2b-4) + (2b-4) + b + b = 106 cm

Bracket off

2b-4+2b-4+b+b = 106cm

Collect like terms

2b+2b+b+b-4-4 = 106cm

6b-8 = 106

6b = 106+8

6b = 114

b = 114÷6

b = 19

To get a

2b - 4

2(19) - 4

38-4

= 34

4 0

2 years ago

In the adjoining figure, PQRS is a square and TQRS is a trapezium. If the area of the square PQRS is 25sq.cm and TP=12cm, find t

Wewaii [24]

Answer:

area of square=l²

25sq.cm²=l²

l=5sq.cm

area of trapezium =(1/2)(a+b)h

a=5,b=(12+5)=17,h=5

area of trapezium=(1/2)(22)(5)

area=55cm²

Pls mark as brainliest

6 0

2 years ago

Write the ratio 6ft to 4 yd as a fraction in simplest form

vlabodo [156]

6:4 or 3:2 is the answer

5 0

2 years ago

Read 2 more answers

CAN SOMEONE HELP ME!!! With 13 and 14! Plzzzz

lyudmila [28]

Answer:

Step-by-step explanation:

14x13

= 48

if wrong then tell me are you dividing or multiplying

7 0

2 years ago