Ask question

Ask question

All categories

2 years ago

9

Find the values of x and y in the figure below:

2 answers:

LenKa [72]2 years ago

8 0

B to get my phone taken away I

kenny6666 [7]2 years ago

5 0

Answer:

B

Step-by-step explanation:

You might be interested in

Combine like terms to simplify the expression:<br> 1 + t + 3 − 5t + 5 = ______

lidiya [134]

Answer:

9-4t

Step-by-step explanation:

7 0

2 years ago

Read 2 more answers

Explain how would you find the square root of a number

nirvana33 [79]

Answer:

multiply

Step-by-step explanation:

i know this from 3rd grade

7 0

2 years ago

Read 2 more answers

Help with trigonometry

diamong [38]

Answer:

cos ∠CBD = - 4√41 / 41

Step-by-step explanation:

AB = 4 ΔABC = (4 x AC) / 2 = 10

AC = 5

BC = √5² + 4² = √41

cos ∠CBD = cos (180° - ∠CBA) = cos 180° cos ∠CBA + sin 180° sin ∠CBA

(cos 180° = - 1 sin 180° = 0 cos ∠CBA = 4 / √41 )

cos ∠CBD = (-1) x cos ∠CBA = - 4 / √41 = - 4√41 / 41

4 0

3 years ago

If 3:y=18:24 find y.

VashaNatasha [74]

y = 4

Explanation: $\begin{gathered} 3\text{ : y = 18 : 24} \\ 3\colon y\text{ = }\frac{3}{y} \\ 18\colon24\text{ = }\frac{18}{24} \end{gathered}$ $\begin{gathered} \frac{3}{y}\text{ = }\frac{18}{24} \\ \text{cross mult}iply\colon \\ 3(24)\text{ = y(18)} \\ \text{divide both sides by 18:} \\ \frac{3(24)}{18}\text{ = }\frac{18y}{18} \end{gathered}$ $\begin{gathered} y\text{ = }\frac{24}{6} \\ y\text{ = 4} \end{gathered}$

7 0

1 year ago

In the △ABC, the height AN = 24 in, BN = 18 in, AC = 40 in. Find AB and BC. Consider all possible cases. Case 1 : N∈BC, AB = __,

aivan3 [116]

Answer:

Case 1:

$AB = 30$

$BC = 50$

Case 2:

$AB = 15.9$

$BC = 36.7$

Case 3: Not possible

Step-by-step explanation:

Given

See attachment for illustration of each case

Required

Find AB and BC

Case 1:

Using Pythagoras theorem in ANB, we have:

$AB^2 = AN^2 + BN^2$

This gives:

$AB^2 = 24^2 + 18^2$

$AB^2 = 576 + 324$

$AB^2 = 900$

Take square roots of both sides

$AB = \sqrt{900$

$AB = 30$

To calculate BC, we consider ANC, where:

$AC^2 = AN^2 + NC^2$

$40^2 = 24^2 + NC^2$

$1600 = 576 + NC^2$

Collect like terms

$NC^2 = 1600 - 576$

$NC^2 = 1024$

Take square roots

$NC = \sqrt{1024$

$NC = 32$

So:

$BC = NC + BN$

$BC = 32 + 18$

$BC = 50$

Case 2:

Using Pythagoras theorem in ANB, we have:

$AN^2 = AB^2 + BN^2$

This gives:

$24^2 = AB^2 + 18^2$

$576 = AB^2 + 324$

Collect like terms

$AB^2 = 576 - 324$

$AB^2 = 252$

Take square roots of both sides

$AB = \sqrt{252$

$AB = 15.9$

To calculate BC, we consider ABC, where:

$AC^2 = AB^2 + BC^2$

$40^2 = 252 + BC^2$

$1600 = 252 + BC^2$

Collect like terms

$BC^2 = 1600 - 252$

$BC^2 = 1348$

Take square roots

$BC = \sqrt{1348$

$BC = 36.7$

Case 3:

This is not possible because in ANC

The hypotenuse AN (24) is less than AC (40)

3 0

2 years ago