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2 years ago

How do u do this question:(

Mathematics

2 answers:

aksik [14]2 years ago

5 0

C=3.14r^2 that's the formula a

Flura [38]2 years ago

4 0

The formula for the circumference of a circle is...

C=2rπ were r is the radius of the circle

plug in the data

C=2(1)π
C=2π (i used π and not 3.14)
C=6.28

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Step-by-step explanation:

G(6)=3(6)^2-2(6)-1

=108-12-1

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2 years ago

2x + 7 - 3x ⥶ 10<br> show work

gavmur [86]

Answer:

x ⥶ -3

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2 years ago

A clock was reading the time accurately on Friday at noon. On Monday at 6pm the clock was running late by 468 seconds. On averag

Setler [38]

The clock was skipping 3 seconds every 30 minutes from Friday noon to Monday 6 pm.

The clock was still accurate by Friday noon. The clock was late by 468 seconds by Monday, 6 pm.

To solve the problem, we must:

Know how many 30-minutes have passed during the time period.

1 day = 24 hours

1 hour = 60 minutes = 2 × (30 minutes)

1 day = 24 hours × 2 × (30 minutes)

1 day = 48 × (30 minutes)

Thus, there are 48, 30-minutes in a day. On Friday, however, we start counting at noon, which is half of the day. Moreover, on Monday, the mark is only up to 6 pm, which is three-fourths of the day.

Friday = 48 × $\frac{1}{2}$ = 24

Saturday = 48

Sunday = 48

Monday = 48 × $\frac{3}{4}$ = 36

TOTAL = 24 + 48 + 48 + 36 = 156

Therefore, the total number of 30-minutes that have passed is 156. There were 156, 30-minutes that passed during the time period.

Divide the number of total seconds late by the number of 30-minutes passed.

That is, the number of total seconds late= 468 seconds ÷ 156

= 3 seconds

Therefore, the clock was skipping 3 seconds every 30 minutes from Friday noon to Monday 6 pm.

To learn more about clock problems visit:

brainly.com/question/27122093.

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2 years ago

Solve the system of linear equations for x and y.(cos θ)x + (sin θ)y=1(−sin θ)x + (cos θ)y = 0x=y=

lys-0071 [83]

Answer:

$\bold{x =cos\theta}\\\bold{y=sin\theta}$

Step-by-step explanation:

The given system of linear equations is:

$(cos\theta)x+(sin\theta)y=1\\(-sin\theta)x+(cos\theta)y=0$

We have to solve the equations for the values of $x, y$ .

Let us use elimination method in which we eliminate one of the variables from the two variables.

For this, let us multiply the first equation by $sin\theta$ and second equation by $cos\theta$

Now, the equations become:

$(cos\theta.sin\theta)x+(sin\theta.sin\theta)y=sin\theta\\\Rightarrow (cos\theta.sin\theta)x+(sin^2\theta)y=sin\theta ....... (1)\\\\(-sin\theta.cos\theta)x+(cos\theta.cos\theta)y=0\\\Rightarrow (-sin\theta.cos\theta)x+(cos^2\theta)y=0 ..... (2)$