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3 years ago

If Justine earns £10920 in year, how much does she earn each month?

Mathematics

1 answer:

Sholpan [36]3 years ago

7 0

Answer:

910.

Step-by-step explanation:

you take your year earning then divide it by twelve months to give you ur answers. 10920÷12=910

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A. The middle value of a set of data arranged in numerical order is the

Effectus [21]

Answer:

Median

Step-by-step explanation:

Median is the number present in the middle when the numbers in a set of data are arranged in ascending or descending order.

6 0

3 years ago

14% of 163<br> What is 14% of 163 estimated

Elina [12.6K]

Answer:

22.82

Step-by-step explanation:

it is about 22.82 when taken 14% away

3 0

3 years ago

Please help- <br> Find the equation of the line with slope 3 that passes through (1, 8).

MakcuM [25]

Answer:

y = 3x + 5

Step-by-step explanation:

The equation of a line in slope- intercept form is

y = mx + c ( m is the slope and c the y- intercept )

Here m = 3 , then

y = 3x + c ← is the partial equation

To find c substitute (1, 8 ) into the partial equation

8 = 3 + c ⇒ c = 8 - 3 = 5

y = 3x + 5 ← equation of line

7 0

3 years ago

List the angles in order from the largest to the smallest and explain

steposvetlana [31]

Answer:

The correct answer is= A
Solution
Angle A is opposite to side BC
Angle B is opposite to side CA
Angla C is opposite to side BC
In triangle, the largest angle will be opposite the largest side. So Angle A is the largest angle.
Angle B is the second largest and Angle C is the smallest.

3 0

3 years ago

Find the volume of the solid bounded by z = 2 - x2 - y2 and z = 1. Express your answer as a decimal rounded to the hundredths pl

butalik [34]

Answer:

The answer is " $(\frac{\pi}{2})$ ".

Step-by-step explanation:

$z = 2 - x^2 - y^2.........(1) \\\\z = 1.............(2)$

Let add equation 1 and 2:

Using formula: $x^2+y^2=1 \\\\$

convert to polar coordinates

$r=2$

$\theta \varepsilon (0=\pi)=z\\\\V=\int^{2\pi}_{\theta=0}\int^{1}_{\pi=0}\int^{z_{2}}_{z_1} r \ dr \d \theta\\\\$

$=\int^{2\pi}_{0}\int^{1}_{0} (Z_2-z_1)r \ dr \d \theta\\\\=\int^{2\pi}_{0}\int^{1}_{0} 1- (-1-x^2-y^2) r \ dr \d \theta\\\\=\int^{2\pi}_{0}\int^{1}_{0} (+1 \pm r^2) r \ dr \d \theta\\\\=\int^{2\pi}_{0}\int^{1}_{0} (-r^3 + r) \ dr \d \theta\\\\=\int^{2\pi}_{0} (-\frac{r^4}{4}+\frac{r^2}{1})^{1}_{0} \d \theta\\\\=\int^{2\pi}_{0} (\frac{1}{4}) \d \theta\\\\=(\frac{2 \pi}{4}) \\\\=(\frac{\pi}{2}) \\\\$

8 0

3 years ago