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3 years ago

The three inside angles (A,B and C) of a right angled triangle are in the ratio 7:18:11

1 answer:

6 0

This is overdetermined. Without looking at the ratio (except for the order of the sides) the actute angles in a right triangle are complementary so the answer must be:

Answer: A=35°, C=90°, B=55°

We can do it the way they like.

7+18+11 =36

so to get 180 degrees we need to scale the ratio elements by a factor of 5.

7:18:11=35:90:55

A is the smallest. We see C is indeed a right angle and B is complementary to A.

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I need this answered asap please and thankyou :D

suter [353]

Answer:

Step-by-step explanation:

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3 years ago

What's the missing sequence of 37 40 123 ? 372 1116 <br>

erica [24]

Answer:

369

Step-by-step explanation:

37 + 3 = 40

40 × 3 = 120

120 + 3 = 123

123 × 3 = 369

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3 years ago

Determine the perimeter of quadrilateral CGHI

Murljashka [212]

Answer:

if the sides are 2, 3, 4.2, 2.6 then the perimeter is 11.8

Step-by-step explanation:

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2 years ago

Solve the following proportion for v. <br>8/5 = v/6 <br>Round your answer to the nearest tenth.

bekas [8.4K]

Answer:

$v=\frac{48}{5}=9\frac{3}{5} \\$

Step-by-step explanation:

$\frac{8}{5}=\frac{v}{6}$

Swap sides:

$\frac{v}{6}=\frac{8}{5}$

Multiply to both sides by 6:

$\frac{v}{6}\cdot6=\frac{8}{5}\cdot6$

Group like terms:

$\frac{1}{6}\cdot6v=\frac{8}{5}\cdot6$

Simplify the fraction:

$v=\frac{8}{5}\cdot6$

Multiply the fractions

$v=\frac{8\cdot6}{5}$

Simplify the arithmetic:

$v=\frac{48}{5}$

4 0

2 years ago

A computer system uses passwords that are six characters, and each character is one of the 26 letters (a–z) or 10 integers (0–9)

Blababa [14]

First of all, since we have 36 characters available per spot (26 letters and 10 digits), and we have 6 spots, we have a total of

$36^6$

possible passwords.

Event A happens if the password starts with either a, e, i, o or u. If we fix the first character, we're left with 36 characters available for each of the remaining 5 spots, leading to a total of

$5\cdot 36^5$

possible passwords.

So, the probability of event A, computed as the ratio between "good" cases and all possible cases, is

$\dfrac{5\cdot 36^5}{36^6}=\dfrac{5}{36}$

Event B works exactly the same, since we're fixing the last spot, leaving 36 characters available for each of the first 5 spots. So, we have

$P(A)=P(B)=\dfrac{5}{36}$

As for the intersection, we want the first character to be a vowel, and the last character to be an even digits. There are 25 passwords satisfying this request:

$axxxx0,\ axxxx2,\ axxxx4,\ axxxx6,\ axxxx8$