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3 years ago

Please help me answer this question

Mathematics

1 answer:

MAXImum [283]3 years ago

7 0

Step-by-step explanation:

BC is the bisector of $\angle ABQ$

$\therefore \angle ABC \cong \angle CBQ..(1)\\\angle CBQ \cong \angle CAB..(2) \\(\angle 's\: in \: alternate\: segments) \\$

From equations (1) & (2)

$\angle ABC \cong \angle CAB\\\therefore AC = BC$

(Because sides opposite to equal angles are equal)

Hence Proved

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3 years ago

Please can I have an explanation also, I am terrible at these kinds of questions!

wlad13 [49]

Answer:

<em>The fraction of the beads that are red is</em>

Step-by-step explanation:

<u>Algebraic Expressions</u>

A bag contains red (r), yellow (y), and blue (b) beads. We are given the following ratios:

r:y = 2:3

y:b = 5:4

We are required to find r:s, where s is the total of beads in the bag, or

s = r + y + b

Thus, we need to calculate:

$\displaystyle \frac{r}{r+y+b} \qquad\qquad [1]$

Knowing that:

$\displaystyle \frac{r}{y}=\frac{2}{3} \qquad\qquad [2]$

$\displaystyle \frac{y}{b}=\frac{5}{4}$

Multiplying the equations above:

$\displaystyle \frac{r}{y}\frac{y}{b}=\frac{2}{3}\frac{5}{4}$

Simplifying:

$\displaystyle \frac{r}{b}=\frac{5}{6} \qquad\qquad [3]$

Dividing [1] by r:

$\displaystyle \frac{r}{r+y+b}=\displaystyle \frac{1}{1+y/r+b/r}$

Substituting from [2] and [3]:

$\displaystyle \frac{r}{r+y+b}=\displaystyle \frac{1}{1+3/2+6/5}$

Operating:

$\displaystyle \frac{r}{r+y+b}=\displaystyle \frac{1}{\frac{10+3*5+6*2}{10}}$

$\displaystyle \frac{r}{r+y+b}=\displaystyle \frac{10}{10+15+12}$

$\displaystyle \frac{r}{r+y+b}=\displaystyle \frac{10}{37}$

The fraction of the beads that are red is $\mathbf{\frac{10}{37}}$

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2 years ago

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3 years ago