All categories

3 years ago

Someone help please!

Mathematics

1 answer:

saveliy_v [14]3 years ago

5 0

Answer:

BC ≈ 14.7 m

Step-by-step explanation:

Using the Sine rule in Δ ABC

$\frac{AB}{sinC}$ = $\frac{BC}{sinA}$

To find ∠ A subtract the 2 given angles from 180°

∠ A = 180° - (90 + 28)° = 180° - 118° = 62°

Then

$\frac{7.8}{sin28}$ = $\frac{BC}{sin62}$ ( cross- multiply )

BC × sin28° = 7.8 × sin62° ( divide both sides by sin28° )

BC = $\frac{7.8sin62}{sin28}$ ≈ 14.7 m ( to 3 significant figures )

You might be interested in

QUESTION 5

kenny6666 [7]

Answer:

c, d

Step-by-step explanation:

26*10= 260 total paid

260-219.98 = 40.02

8 0

3 years ago

Read 2 more answers

Please help me quickly :)

Otrada [13]

Answer: The answer is B.

Step-by-step explanation:

Angle ABC is a straight angle, which is 180 degrees, so you can rule out answers C and D.

For angle EBG, simply count the measures of the angle on the protractor. From the right, you start at 55. You end at 100. Subtract 55 from 100 to get a measure of 45 degrees.

4 0

3 years ago

Just wanted to say this..

slavikrds [6]

Thank you so much, is it an app or on the internet?

7 0

3 years ago

HURRY I NEED HELP: Write an equation in point-slope form of the line that passes through the point (3, 5) and has a slope of m=−

skelet666 [1.2K]

Answer:

$\huge y - 5 = - 1(x - 3) \\$

Step-by-step explanation:

To find an equation of a line in point slope form when given the slope and a point we use the formula

$y - y_1 = m(x - x_1)$

From the question we have the final answer as

$y - 5 = - 1(x - 3)$

Hope this helps you

6 0

3 years ago

A collection of coins consists of nickels, dimes, and quarters. There are three fewer quarters than nickels and six more dimes t

weeeeeb [17]

Answer:

You have 15 dimes. You have 9 quarters. You have 12 nickels

Step-by-step explanation:

lets set some variables:

let "n" = the number of nickels

let "d" = the number of dimes

let "q" = the number of quarters

So, the total amount of money you have should be: $4.35 = 0.25q + 0.10d + 0.05n

Now let's look at the relationships between the coins:

"There are three fewer quarters than nickels": n - 3 = q

"six more dimes than quarters": q + 6 = d

So now you have three equations with three variables, all you need to do is solve.

$\left \{ {{4.35 = 0.25q + 0.10d + 0.05n} \atop {n - 3 = q}} \atop {q + 6 = d}}\right.$

first, you can substitute "n-3" for "q" (according to the 2nd equation) in the 1st and 3rd equation, you get:

$\left \{ {{4.35=0.25(n-3)+0.10d+0.05n} \atop {(n-3)+6=d}} \right.$

You now only have two equations and two variables.

Simplify:

$\left \{ {{4.35=0.25n-0.75+0.10d+0.05n} \atop {n+3=d}} \right.$

$\left \{ {{4.35=0.30n-0.75+0.10d} \atop {n+3=d}} \right.$

Now substitute "n+3" for "d" (according to the 2nd equation) in the 1st equation:

4.35=0.30n-0.75+0.10(n+3)

simplify:

4.35=0.30n-0.75+0.10n+0.30

4.35=0.40n-0.45

4.35+0.45=0.40n

4.80=0.40n

n=12

You have 12 nickels. Now sub "n" back into your equations to find the number of dimes and quarters:

n - 3 = q

12 - 3 = q

q = 9

You have 9 quarters.

q + 6 = d

9 + 6 = d

d = 15

You have 15 dimes.

3 0

3 years ago