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3 years ago

Pls help me with this questions. Look at the image

Mathematics

1 answer:

Andrei [34K]3 years ago

5 0

21 is 19 get ur facts straight

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If BC = 6.5, CD = 7.8, and AD = 12.2, find AB to the nearest tenth.

Sladkaya [172]

This can be solved using the angle bisector theorem. Note that line segment AD is the angle bisector of triangle DAB from angle DAB. Therefore, we know that the ratios $\frac{\text{AD}}{\text{CD}}$ and $\frac{\text{AB}}{\text{CB}}$ are equal. Solving, we get $\frac{\text{12.2}}{\text{7.8}}=\frac{\text{AB}}{\text{6.5}}$ . Cross multiplying, we get $79.3 = 7.8x$ , and solving this equation we get $x = 10.1666... \approx 10.2$ . 10.2 is your answer.

6 0

3 years ago

What does one half divided by five tenths equal

Masteriza [31]

It would equal 1

Explanation
- keep (keep first fraction the same)
- change (change division to multiplication)
- flip (flip the last fraction)

3 0

2 years ago

Read 2 more answers

8. Let α, β, γ be the angles made by a vector u ∈ R 3 with positive x, y, z−axis respectively. Then, the numbers l = cos α, m =

alex41 [277]

Answer:

The answers are given in the attachment

Step-by-step explanation:

The detailed step by step calculations are shown in the attachment.

8 0

2 years ago

What is the equation of a line that passes

OleMash [197]

Answer:y=9.5x-0.5

Step-by-step explanation:

Ok, so you need to use the equation y2-y1/x2-x1 to find the slope. So (18.5-4.25)/(2-0.5) so 14.25/1.5 , calculate that and you get 9.5 which is your slope, apply that to the slope formula which is y=mx+b where (m) is slope and (b) is y intercept and you get y=9.5x-0.5

4 0

2 years ago

Help I am very bad at math I need answers by tomorrow if you can help please do

d1i1m1o1n [39]

Answer:

Step-by-step explanation:

(a - b)(a +b) = a² - b²

1 - Sin² A = Cos² A

$LHS = \frac{1}{1- Sin A} + \frac{1}{1 + Sin A}\\\\= \frac{1*(1 + Sin A)}{(1- Sin A)(1 + Sin A)} + \frac{1*(1- Sin A)}{(1 + Sin A)(1- Sin A)}\\\\= \frac{1 + Sin A+ 1 - Sin A}{1^{2}- Sin^{2} A}\\\\= \frac{2}{1 - Sin^{2} A}\\\\= \frac{2}{Cos^{2} A}\\\\= 2 Sec^{2} A$

2) Sec² A - Tan² A = 1

$LHS = \frac{1}{Sec A - Tan A}\\\\=\frac{1*(Sec A + Tan A)}{(Sec A - Tan A)(Sec A + Tan A)}\\\\=\frac{Sec A + Tan A}{Sec^{2} A - Tan^{2} A}\\\\=\frac{Sec A + Tan A }{1}\\\\= Sec A + Tan A = RHS\\\\\\$

3) LHS = Cosec² A + Cot² A

= Cosec² A + Cosec² A - 1

= 2Cosec² A - 1 = RHS

$4) LHS = \frac{Sec A}{Cos A}- \frac{Tan A}{Cot A}\\\\ = Sec A*\frac{1}{Cos A}-Tan A*\frac{1}{Cot A}\\\\ = Sec A * Sec A - Tan A * Tan A\\\\= Sec^{2} A - Tan^{2} A \\\\= 1$

3 0

2 years ago