All categories

2 years ago

Each pair of figures is similar. Find the missing side.

Mathematics

1 answer:

Kaylis [27]2 years ago

3 0

Step-by-step explanation:

Similar figures have same ratio of corresponding sides.

Set up ratios and solve for x.

x / 54 = 6 / 36
x / 54 = 1/6
x = 54/6
x = 9

x / 4 = 99 / 11
x / 4 = 9
x = 4*9
x = 36

x / 10 = 6 / 60
x / 10 = 1/6
x = 10/6
x = 5/3

x / 4 = 28 / 7
x / 4 = 4
x = 4*4
x = 16

You might be interested in

Will give brainliest

Anna [14]

Answer: 23.5

Step-by-step explanation: Divide 47.00 by 2 and you get 23.5

hope this helps

7 0

2 years ago

Help please thank you

allsm [11]

Answer:

yes

0.92 cents

4.72

9.26

Step-by-step explanation:

6 0

2 years ago

Given: AD = BC and AD || BC

bearhunter [10]

Answer:

ABCD is a parallelogram.

Step-by-step explanation:

A parallelogram is a quadrilateral that has two parallel and equal pairs of opposite sides.

From the given diagram,

Given: AD = BC and AD || BC, then:

i. AB = DC (both pairs of opposite sides of a parallelogram are congruent)

ii. <ADC = < BCD and < DAB = < CBA

thus, AD || BC and AB || DC (both pairs of opposite sides of a parallelogram are parallel)

iii. < BAC = < DCA (alternate angle property)

iv. Join BD, line AC and BC are the diagonals of the quadrilateral which bisect each other. The two diagonals are at a right angle to each other.

v. <ADC + < BCD + < DAB + < CBA = $360^{0}$ (sum of angles in a quadrilateral equals 4 right angles)

Therefore, ABCD is a parallelogram.

4 0

3 years ago

Express in simplified exponential notation. x3 • x5 =

Usimov [2.4K]

Answer:

x^8

Step-by-step explanation:

The applicable rule of exponents is ...

(x^a)(x^b) = x^(a+b)

Here you have ...

(x^3)(x^5) = x^(3+5) = x^8

8 0

2 years ago

How to solve this trig

n200080 [17]

Hi there!

To find the Trigonometric Equation, we have to isolate sin, cos, tan, etc. We are also given the interval [0,2π).

First Question

What we have to do is to isolate cos first.

$\displaystyle \large{ cos \theta = - \frac{1}{2} }$

Then find the reference angle. As we know cos(π/3) equals 1/2. Therefore π/3 is our reference angle.

Since we know that cos is negative in Q2 and Q3. We will be using π + (ref. angle) for Q3. and π - (ref. angle) for Q2.

Find Q2

$\displaystyle \large{ \pi - \frac{ \pi}{3} = \frac{3 \pi}{3} - \frac{ \pi}{3} } \\ \displaystyle \large \boxed{ \frac{2 \pi}{3} }$

Find Q3

$\displaystyle \large{ \pi + \frac{ \pi}{3} = \frac{3 \pi}{3} + \frac{ \pi}{3} } \\ \displaystyle \large \boxed{ \frac{4 \pi}{3} }$

Both values are apart of the interval. Hence,

$\displaystyle \large \boxed{ \theta = \frac{2 \pi}{3} , \frac{4 \pi}{3} }$

Second Question

Isolate sin(4 theta).

$\displaystyle \large{sin 4 \theta = - \frac{1}{ \sqrt{2} } }$

Rationalize the denominator.

$\displaystyle \large{sin4 \theta = - \frac{ \sqrt{2} }{2} }$

The problem here is 4 beside theta. What we are going to do is to expand the interval.

$\displaystyle \large{0 \leqslant \theta < 2 \pi}$

Multiply whole by 4.

$\displaystyle \large{0 \times 4 \leqslant \theta \times 4 < 2 \pi \times 4} \\ \displaystyle \large \boxed{0 \leqslant 4 \theta < 8 \pi}$

Then find the reference angle.

We know that sin(π/4) = √2/2. Hence π/4 is our reference angle.

sin is negative in Q3 and Q4. We use π + (ref. angle) for Q3 and 2π - (ref. angle for Q4.)

Find Q3

$\displaystyle \large{ \pi + \frac{ \pi}{4} = \frac{ 4 \pi}{4} + \frac{ \pi}{4} } \\ \displaystyle \large \boxed{ \frac{5 \pi}{4} }$

Find Q4

$\displaystyle \large{2 \pi - \frac{ \pi}{4} = \frac{8 \pi}{4} - \frac{ \pi}{4} } \\ \displaystyle \large \boxed{ \frac{7 \pi}{4} }$

Both values are in [0,2π). However, we exceed our interval to < 8π.

We will be using these following:-

$\displaystyle \large{ \theta + 2 \pi k = \theta \: \: \: \: \: \sf{(k \: \: is \: \: integer)}}$

Hence:-

For Q3

$\displaystyle \large{ \frac{5 \pi}{4} + 2 \pi = \frac{13 \pi}{4} } \\ \displaystyle \large{ \frac{5 \pi}{4} + 4\pi = \frac{21 \pi}{4} } \\ \displaystyle \large{ \frac{5 \pi}{4} + 6\pi = \frac{29 \pi}{4} }$

We cannot use any further k-values (or k cannot be 4 or higher) because it'd be +8π and not in the interval.

For Q4

$\displaystyle \large{ \frac{ 7 \pi}{4} + 2 \pi = \frac{15 \pi}{4} } \\ \displaystyle \large{ \frac{ 7 \pi}{4} + 4 \pi = \frac{23\pi}{4} } \\ \displaystyle \large{ \frac{ 7 \pi}{4} + 6 \pi = \frac{31 \pi}{4} }$

Therefore:-

$\displaystyle \large{4 \theta = \frac{5 \pi}{4} , \frac{7 \pi}{4} , \frac{13\pi}{4} , \frac{21\pi}{4} , \frac{29\pi}{4}, \frac{15 \pi}{4} , \frac{23\pi}{4} , \frac{31\pi}{4} }$

Then we divide all these values by 4.

$\displaystyle \large \boxed{\theta = \frac{5 \pi}{16} , \frac{7 \pi}{16} , \frac{13\pi}{16} , \frac{21\pi}{16} , \frac{29\pi}{16}, \frac{15 \pi}{16} , \frac{23\pi}{16} , \frac{31\pi}{16} }$

Let me know if you have any questions!

3 0

2 years ago