Math Help please 15 points

Mathematics

1 answer:

Law Incorporation [45]3 years ago

4 0

1) You have vertical angles here. This means the big and small triangles are similar, meaning the ratio of corresponding sides of the two triangles are equal.

In this case, 32cm corresponds to 4cm and 40cm corresponds to x. To find x, set up the ratio as follows: 4/32 = x/40.

4/32 can be reduce to 1/8. Multiply each side by 40 to isolate x. you get 1/8 * 40

This yields x = 5 cm

2) set up a proportion, similar to the first question.

(x + 4)/8 = (2x + 1)/12
12(x + 4) = 8(2x + 1)
12x + 48 = 16x + 8
40 = 4x
x = 10

the length of BC is (x + 4) + (2x + 1) = 3x + 5

this means BC = 35

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What is the distance between the points (4,5) and (10,13) on a coordinate plane?

Readme [11.4K]

The distacne between the points (x1,y1) and (x2,y2) is
$D=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$
so the distance between (4,5) and (10,13) is
well, first
x1=4
y1=5
x2=10
y2=13
so

$D=\sqrt{(10-4)^2+(13-5)^2}$
$D=\sqrt{(6)^2+(8)^2}$
$D=\sqrt{36+64}$
D=√100
D=10
the distance is 10 units

7 0

3 years ago

Please help me!!!!!

denpristay [2]

Answer: see proof below

<u>Step-by-step explanation:</u>

Given: A + B + C = π → A = π - (B + C)

→ B = π - (A + C)

→ C = π - (A + B)

Use Sum to Product Identity: sin A - sin B = 2 cos [(A + B)/2] · sin [(A - B)/2]

Use the following Cofunction Identity: cos (π/2 - A) = sin A

<u>Proof LHS → RHS:</u>

LHS: sin A - sin B + sin C

= (sin A - sin B) + sin C

$\text{Sum to Product:}\quad 2\cos \bigg(\dfrac{A+B}{2}\bigg)\cdot \sin \bigg(\dfrac{A-B}{2}\bigg)+2\cos \bigg(\dfrac{C}2{}\bigg)\cdot \sin \bigg(\dfrac{C}{2}\bigg)$

$\text{Given:}\qquad 2\cos \bigg(\dfrac{\pi -(B+C)}{2}+\dfrac{B}{2}}\bigg)\cdot \sin \bigg(\dfrac{A-B}{2}\bigg)+2\cos \bigg(\dfrac{C}2{}\bigg)\cdot \sin \bigg(\dfrac{C}{2}\bigg)\\\\\\.\qquad \qquad =2\cos \bigg(\dfrac{\pi -C}{2}\bigg)\cdot \sin \bigg(\dfrac{A-B}{2}\bigg)+2\cos \bigg(\dfrac{C}2{}\bigg)\cdot \sin \bigg(\dfrac{C}{2}\bigg)$

$.\qquad \qquad =2\cos \bigg(\dfrac{\pi}{2} -\dfrac{C}{2}\bigg)\cdot \sin \bigg(\dfrac{A-B}{2}\bigg)+2\cos \bigg(\dfrac{C}2{}\bigg)\cdot \sin \bigg(\dfrac{C}{2}\bigg)$

$\text{Cofunction:} \qquad 2\sin \bigg(\dfrac{C}{2}\bigg)\cdot \sin \bigg(\dfrac{A-B}{2}\bigg)+2\cos \bigg(\dfrac{C}2{}\bigg)\cdot \sin \bigg(\dfrac{C}{2}\bigg)$

$\text{Factor:}\qquad 2\sin \bigg(\dfrac{C}{2}\bigg)\bigg[ \sin \bigg(\dfrac{A-B}{2}\bigg)+\cos \bigg(\dfrac{C}{2}\bigg)\bigg]$

$\text{Given:}\qquad 2\sin \bigg(\dfrac{C}{2}\bigg)\bigg[ \sin \bigg(\dfrac{A-B}{2}\bigg)+\cos \bigg(\dfrac{\pi -(A+B)}{2}\bigg)\bigg]\\\\\\.\qquad \qquad =2\sin \bigg(\dfrac{C}{2}\bigg)\bigg[ \sin \bigg(\dfrac{A-B}{2}\bigg)+\cos \bigg(\dfrac{\pi}{2} -\dfrac{(A+B)}{2}\bigg)\bigg]$

$\text{Cofunction:}\qquad 2\sin \bigg(\dfrac{C}{2}\bigg)\bigg[ \sin \bigg(\dfrac{A-B}{2}\bigg)+\sin \bigg(\dfrac{A+B}{2}\bigg)\bigg]$

$\text{Sum to Product:}\qquad 2\sin \bigg(\dfrac{C}{2}\bigg)\bigg[ 2\sin \bigg(\dfrac{A}{2}\bigg)\cdot \cos \bigg(\dfrac{B}{2}\bigg)\bigg]\\\\\\.\qquad \qquad \qquad \qquad =4\sin \bigg(\dfrac{A}{2}\bigg)\cdot \cos \bigg(\dfrac{B}{2}\bigg)\cdot \sin \bigg(\dfrac{C}{2}\bigg)$

$\text{LHS = RHS:}\quad 4\sin \bigg(\dfrac{A}{2}\bigg)\cdot \cos \bigg(\dfrac{B}{2}\bigg)\cdot \sin \bigg(\dfrac{C}{2}\bigg)=4\sin \bigg(\dfrac{A}{2}\bigg)\cdot \cos \bigg(\dfrac{B}{2}\bigg)\cdot \sin \bigg(\dfrac{C}{2}\bigg)\quad \checkmark$