All categories

3 years ago

Quadrilateral ABCD is similar to Quadrilateral EFGH. The scale factor is 3:2. If AB = 18, find EF.

Mathematics

2 answers:

lord [1]3 years ago

8 0

Answer: EF = 12 units

Step-by-step explanation:

Given : Quadrilateral ABCD is similar to Quadrilateral EFGH.

The scale factor is 3:2.

We know that the scale factor is the ratio of the corresponding sides of two similar figures.

So if Quadrilateral ABCD is similar to Quadrilateral EFGH, the side AB corresponds to side EF.

If AB = 18, then we have the following equation:-

$\dfrac{AB}{EF}=\dfrac{3}{2}\\\\\Rightarrow\dfrac{18}{EF}=\dfrac{3}{2}\\\\\Rightarrow\ EF=\dfrac{2\times18}{3}\\\\\Rightarrow\ EF=12$

Olenka [21]3 years ago

6 0

EF =12

do you want to know how I got my answer...

You might be interested in

If the squared difference of the zeroes of the quadratic polynomial x2+kx+30 is equal to 169 find the value of k and the zeroes

anyanavicka [17]

ANSWER

$x = 2 \: \: or \: \: x = 15$

$x = - 2 \: \: or \: \: x = - 15$

EXPLANATION

The given polynomial is

$f(x) = {x}^{2} + kx + 30$

where a=1,b=k, c=30

Let the zeroes of this polynomial be m and n.

Then the sum of roots is

$m + n = - \frac{b}{a} = -k$

and the product of roots is

$mn = \frac{c}{a} = 30$

The square difference of the zeroes is given by the expression.

$( {m - n})^{2} = {(m + n)}^{2} - 4mn$

From the question, this difference is 169.

This implies that:

$( { - k)}^{2} - 4(30) = 169$

${ k}^{2} -120= 169$

$k^{2} = 289$

$k= \pm \sqrt{289}$

$k= \pm17$

We substitute the values of k into the equation and solve for x.

$f(x) = {x}^{2} \pm17x + 30$

$f(x) = (x \pm2)(x \pm 15)$

The zeroes are given by;

$(x \pm2)(x \pm 15) = 0$

$x = \pm2 \: \: or \: \: x = \pm 15$

5 0

3 years ago

Simplify each expression using the laws of exponents. No negative exponents should be in your final answer. (On the image attach

Cloud [144]

Answer:

Answer is given below with explanations.

Step-by-step explanation:

$by \: simplifying \: the \: given \: expression \: using \: \\ law \: of \: exponents \: we \: get \\ \frac{15 {x}^{15} {y}^{ - 3} {z}^{4} }{18 {x}^{18} { {y}^{4} {z}^{ - 4} }^{} } \\ = \frac{5 \: {x}^{ - 3} {y}^{ - 7} {z}^{8} }{6} \\ we \: may \: also \: write \: as \\ = \frac{5 {z}^{8} }{6 {x}^{3} {y}^{7} }$