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2 years ago

14

Draw a diagram using positive an negative tiles for this equation: -5-(-7)=2

1 answer:

Wewaii [24]2 years ago

5 0

Sadly there is no way to graph properly on this website.

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The horizontal viewing angle is the angle subtended by a straight line from

SSSSS [86.1K]

The measure of the arc is given as π/2. See the explanation below.

<h3>What is an arc?</h3>

An "arc" is a curve that connects two points in mathematics.

It can also be depicted as a section of a circle. It is essentially a portion of a circle's circumference. An arc is a kind of curve.

<h3>What is the calculation for the above solution?</h3>

Note that the viewing angle is 45°.

Thus, the center angle is:
45 X 2 = 90°

Measure of the arc therefore is:
= (π/180°) x 90

= π/2

Learn more about arcs at:
brainly.com/question/2005046
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7 0

1 year ago

The Spice Girls latest single sells 40% fewer singles this week than last week. Last week it sold 120,000 singles. How many did

valentina_108 [34]

Answer:

72000

Step-by-step explanation:

Multiply 120,000 by 0.6 OR multiply 120,000 by 0.4 and subtract 120,000 by the product.

Both equal 72,000

Hope this helps my fellow mike winsowski memer

-Scorpio

8 0

3 years ago

An environment engineer measures the amount ( by weight) of particulate pollution in air samples ( of a certain volume ) collect

Serggg [28]

Answer:

$k = 1$

$P(x > 3y) = \frac{2}{3}$

Step-by-step explanation:

Given

$f \left(x,y \right) = \left{ \begin{array} { l l } { k , } & { 0 \leq x} \leq 2,0 \leq y \leq 1,2 y \leq x } & { \text 0, { elsewhere. } } \end{array} \right.$

Solving (a):

Find k

To solve for k, we use the definition of joint probability function:

$\int\limits^a_b \int\limits^a_b {f(x,y)} \, = 1$

Where

${ 0 \leq x} \leq 2,0 \leq y \leq 1,2 y \leq x }$

Substitute values for the interval of x and y respectively

So, we have:

$\int\limits^2_{0} \int\limits^{x/2}_{0} {k\ dy\ dx} \, = 1$

Isolate k

$k \int\limits^2_{0} \int\limits^{x/2}_{0} {dy\ dx} \, = 1$

Integrate y, leave x:

$k \int\limits^2_{0} y {dx} \, [0,x/2]= 1$

Substitute 0 and x/2 for y

$k \int\limits^2_{0} (x/2 - 0) {dx} \,= 1$

$k \int\limits^2_{0} \frac{x}{2} {dx} \,= 1$

Integrate x

$k * \frac{x^2}{2*2} [0,2]= 1$

$k * \frac{x^2}{4} [0,2]= 1$

Substitute 0 and 2 for x

$k *[ \frac{2^2}{4} - \frac{0^2}{4} ]= 1$

$k *[ \frac{4}{4} - \frac{0}{4} ]= 1$

$k *[ 1-0 ]= 1$

$k *[ 1]= 1$

$k = 1$

Solving (b): $P(x > 3y)$

We have:

$f(x,y) = k$

Where $k = 1$

$f(x,y) = 1$

To find $P(x > 3y)$ , we use:

$\int\limits^a_b \int\limits^a_b {f(x,y)}$

So, we have:

$P(x > 3y) = \int\limits^2_0 \int\limits^{y/3}_0 {f(x,y)} dxdy$

$P(x > 3y) = \int\limits^2_0 \int\limits^{y/3}_0 {1} dxdy$

$P(x > 3y) = \int\limits^2_0 \int\limits^{y/3}_0 dxdy$

Integrate x leave y

$P(x > 3y) = \int\limits^2_0 x [0,y/3]dy$

Substitute 0 and y/3 for x

$P(x > 3y) = \int\limits^2_0 [y/3 - 0]dy$

$P(x > 3y) = \int\limits^2_0 y/3\ dy$

Integrate

$P(x > 3y) = \frac{y^2}{2*3} [0,2]$

$P(x > 3y) = \frac{y^2}{6} [0,2]\\$

Substitute 0 and 2 for y

$P(x > 3y) = \frac{2^2}{6} -\frac{0^2}{6}$

$P(x > 3y) = \frac{4}{6} -\frac{0}{6}$

$P(x > 3y) = \frac{4}{6}$

$P(x > 3y) = \frac{2}{3}$

8 0

3 years ago

2. When do national party conventions meet?

kogti [31]

Answer:

Step-by-step In presidential election years. (Every 4 years)

7 0

3 years ago

Read 2 more answers

2.5 meters cloth is $28.30the cost of 18 meters?

Butoxors [25]

$495.28 I think I'm not sure if I'm right.

8 0

2 years ago

Read 2 more answers