All categories

3 years ago

Please help me I've posted the same questions like 20 times I need help fast!!

Mathematics

2 answers:

Tanzania [10]3 years ago

8 0

Answer:

Step-by-step explanation:

2). Definition of bisector.

4). Reflexive property

5). ASA postulate of congruency.

Inga [223]3 years ago

7 0

Answer:

Step-by-step explanation:

1. given

2. A bisected angle forms 2 congruent angles

3.given

4.reflexive propriety

5. ASA theorem

You might be interested in

Rewrite each expression as a sum or difference of terms. (i) (x+3)(x-3) (ii) (x^2-3)(x^2+3) (iii) (x^15+3)(x^15-3) (iv) (x-3)(x^

klasskru [66]

Answer:

All the answers are given below.

Step-by-step explanation:

(i) (x+3)(x-3) =x² -3² =x² -9 {Since (a+b)(a-b) =a² -b²}

(ii) (x²-3)(x²+3) =(x²)² -3² = $x^{4} -9$ {Since (a+b)(a-b) =a² -b²}

(iii) $(x^{15} +3)(x^{15}-3) =(x^{15} )^{2}-3^{2} =x^{30} -9$

(iv) (x-3)(x²-9)(x+3) =(x²-9)(x²-9)=(x²-9)²= $x^{4}-18x^{2} +81$ {Since, (a-b)² =a²-2ab+b²}

(v) (x²+y²)(x²-y²) = (x²)² -(y²)² = $x^{4}-y^{4}$

(vi) (x²+y²)² = $x^{4}+2x^{2} y^{2} +y^{4}$ {Since (a+b)² =a²+2ab+b²}

(vii) (x-y)²(x+y)² =(x²-y²)² = $x^{4} -2x^{2} y^{2}+y^{4}$ {Since (a-b)² =a²-2ab+b² and (a+b)(a-b) =a²-b²}

(viii) (x-y)²(x²+y²)²(x+y)² =(x²+y²)²(x²-y²)² = $(x^{4} -y^{4} )^{2} =x^{8}-2x^{4} y^{4} +y^{8}$ (Answer)

5 0

4 years ago

Which is the BEST estimate of the average rate of change for the function given in the table, over the interval 2 ≤ x ≤ 4?

ladessa [460]

Answer:

Step-by-step explanation:

Use the slope formula:

$m = \frac{y_2-y_1}{x_2-x_1}$

Picking x = 2 and x = 4 as my two points (since the interval is 2 ≤ x ≤ 4), we get:

$m = \frac{15-3}{4-2} = \frac{12}{2} = 6$

So, the BEST estimate of the average rate of change for the function given in the table is 6.

3 0

3 years ago

Read 2 more answers

It's a question from real and complex numbers which I can't solve. so someone PLZ HeLp

Semmy [17]

Answer:

$\frac{1}{5}$

Step-by-step explanation:

Using the rules of exponents

$a^{m}$ × $a^{n}$ = $a^{(m+n)}$ , $\frac{a^{m} }{a^{n} }$ = $a^{(m-n)}$ , $(a^m)^{n}$ = $a^{mn}$

Simplifying the product of the first 2 terms

$\frac{a^{p^2+pq} }{a^{pq+q^2} }$ × $\frac{a^{q^2+qr} }{a^{qr+r^2} }$

= $a^{p^2-q^2}$ × $a^{q^2-r^2}$

= $a^{p^2-r^2}$

Simplifying the third term

5( $(a^p+r)^{p-r}$

= 5 $a^{(p+r)(p-r)}$ = 5 $a^{(p^2-r^2)}$

Performing the division, that is

$\frac{a^{(p^2-r^2)} }{5a^{(p^2-r^2)} }$ ← cancel $a^{(p^2-r^2)}$ on numerator/ denominator leaves

= $\frac{1}{5}$

4 0

3 years ago

The purchase cost of a TV set is $320. If the dealer wants to earn 15% profits, how much should be added into the purchase cost

____ [38]

Answer:

$48

..........................

7 0

3 years ago

Read 2 more answers

Solve for b. Put your answer in simplest radical form. Show Steps plz!<br><br><br> A = 4πbc + πb^2

sladkih [1.3K]

Answer:

b = -2c ± [√(4π²c² + πA)]/π

Step-by-step explanation:

A = 4πbc + πb^2

A = 4πbc + πb²

πb² + 4πbc - A = 0

Using the quadratic formula to solve this quadratic equation.

The quadratic formula for the quadratic equation, pb² + qb + r = 0, is given as

b = [-q ± √(q² - 4pr)] ÷ 2p

Comparing

πb² + 4πbc - A = 0 with pb² + qb + r = 0,

p = π

q = 4πc

r = -A

b = [-q ± √(q² - 4pr)] ÷ 2p

b = {-4πc ± √[(4πc)² - 4(π)(-A)]} ÷ 2π

b = {-4πc ± √[16π²c² + 4πA]} ÷ 2π

b = (-4πc/2π) ± {√[16π²c² + 4πA] ÷ 2π}

b = -2c ± [√(4π²c² + πA)]/π

Hope this Helps!!!

4 0

3 years ago