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2 years ago

Find the error sienna used base-ten blocks to find 45÷3 .explain her error

Mathematics

1 answer:

Harrizon [31]2 years ago

5 0

The answer to your question is 15

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Three of the angles of a polygon are

Liono4ka [1.6K]

Answer:

three angles of a polygon are each 105.5,the sum of the angles in the polygon is 2520. find each of the other angles if they are equal to each other. For a polygon with n sides and n angles, the sum of the measures of all the interior angles equals (n-2)180 degrees. 16-3=13

4 0

2 years ago

Randy presses RAND on his calculator twice to obtain two random numbers between 0 and 1. Let $p$ be the probability that these t

anygoal [31]

Answer:

$\frac{\pi}{4}$

Step-by-step explanation:

Lets call x,y the numbers we obtain from the calculator. x and y are independent random variables of uniform[0,1] distribution.

Lets note that, since both x and y are between 0 and 1, then 1 is the biggest side of the triangle.

Lets first make a geometric interpretation. If the triangle were to be rectangle, then the side of lenght 1 should be its hypotenuse, and therefore x and y should satisfy this property:

x²+y² = 1

Remember that in this case we are supposing the triangle to be rectangle. But the exercise asks us to obtain an obtuse triangle. For that we will need to increase the angle obtained by the sides of lenght x and y. We can do that by 'expanding' the triangle, but if we do that preserving the values of x and y, then the side of lenght 1 should increase its lenght, which we dont want to. Thus, if we expand the triangle then we should also reduce the value of x and/or y so that the side of lenght 1 could preserve its lenght. With this intuition we could deduce that

x²+y² < 1

Now lets do a more mathematical approach. According to the Cosine theorem, a triangle of three sides of lenght a,b,c satisfies

a² = b²+c² - 2bc* cos(α), where α is the angle between the sides of lenght b and c.

Aplying this formula to our triangle, we have that

$1^2 = x^2 + y^2 - 2bc* cos(\alpha)$ , where $\alpha$ is the angle between the sides of lenght x and y.

Since the triangle is obtuse, then $\pi/2 < \alpha < \pi$ , and for those values $cos(\alpha)$ is negative , hence we also obtain

1 > x² + y²

Thus, we need to calculate P(x²+y² <1). This probability can be calculated throught integration. We need to use polar coordinates.

(x, y) = (r*cosФ,r*senФ)

Where r is between 0 and 1, and Ф is between 0 and $\pi /2$ (that way, the numbers are positive).

The jacobian matrix has determinant r, therefore,

${\int\int}\limits_{x^2+y^2 < 1} \, dxdy = \int\limits^1_0\int\limits^{\frac{\pi}{2}}_0 {r} \, d\phi dr = \frac{\pi}{2} * \int\limits^1_0 {r} \, dr = \frac{\pi}{2} * (\frac{r^2}{2} |^1_0) = \frac{\pi}{4}$

As a conclusion, the probability of obtaining an obtuse triangle is $\frac{\pi}{4}$ .

6 0

2 years ago

Read 2 more answers

Rita is planting saplings along her garden fence. When she started, she had x packages of saplings with 5 saplings per package.

Arturiano [62]

Answer:

A. When Rita started, she had 7 packages of saplings.

B. If we represent x (number of packages of saplings) on a number line graph, it will start on number 4 (number of packages Rita still has) then moves to the right to number 7, that is the number of packages when Rita started to plant.

Step-by-step explanation:

1. Let's review the information given to us to answer the question correctly:

Number of packages Rita started = x

Number of saplings per package = 5

Number of saplings planted by Rita = 15

Number of saplings left = 20

2. How many packages of saplings did she start with?

Number of packages Rita started = (Number of saplings planted by Rita + Number of saplings left)/Number of saplings per package

Replacing with the real values:

x = (15 + 20)/5

x = 35/5 = 7

When Rita started, she had 7 packages of saplings.

3. What would the solution look like on a number line graph?

If we represent x (number of packages of saplings) on a number line graph, it will start on number 4 (number of packages Rita still has) then moves to the right to number 7, that is the number of packages when Rita started to plant.

5 0

2 years ago

Given that tangent theta = negative 1, what is the value of secant theta, for StartFraction 3 pi Over 2 EndFraction less-than th

Rashid [163]

Answer:

$sec(\theta)=\frac{2}{\sqrt{2} }=\sqrt{2}$

Step-by-step explanation:

Start by noticing that the angle $\theta$ is on the 4th quadrant (between $\frac{3\pi}{2}$ and $2\pi$ . Recall then that in this quadrant the functions tangent and cosine are positive, while the function sine is negative in value. This is important to remember given the fact that tangent of an angle is defined as the quotient of the sine function at that angle divided by the cosine of the same angle:

$tan(\theta)=\frac{sin(\theta)}{cos(\theta)}$

Now, let's use the information that the tangent of the angle in question equals "-1", and understand what that angle could be:

$tan(\theta)=\frac{sin(\theta)}{cos(\theta)}\\-1=\frac{sin(\theta)}{cos(\theta)}\\-cos(\theta)=sin(\theta)$

The particular special angle that satisfies this (the magnitude of sine and cosine the same) in the 4th quadrant, is the angle $\frac{7\pi}{4}$

which renders for the cosine function the value $\frac{\sqrt{2} }{2}$ .

Now, since we are asked to find the value of the secant of this angle, we need to remember the expression for the secant function in terms of other trig functions: $sec(\theta)=\frac{1}{cos(\theta)}$

Therefore the value of the secant of this angle would be the reciprocal of the cosine of the angle, that is: $sec(\theta)=\frac{2}{\sqrt{2} }=\sqrt{2}$

6 0

1 year ago

Read 2 more answers

What is the following simplified product? Assume x greater-than-or-equal-to 0 (StartRoot 10 x Superscript 4 Baseline EndRoot min

9966 [12]

Answer:

$\bold{10x^4\sqrt{6}+x^3\sqrt{30x}-10x^4\sqrt{3}-x^3\sqrt{15x}}$

Step-by-step explanation:

To find:

Simplified product of:

$(\sqrt{10x^4}-x\sqrt{5x^2})(2\sqrt{15x^4}+\sqrt{3x^3})$

Solution:

First of all, let us have a look at some of the formula:

1. $(a+b) (c+d) = ac+bc+ad+bd$

2. $a^b\times a^c =a^{b+c }$

3. $\sqrt{a^{2b}} = \sqrt{a^b.a^b}=a^b$

4. $\sqrt a \times \sqrt b = \sqrt{a\times b}$

Now, let us apply the above formula to solve the given expression.

$(\sqrt{10x^4}-x\sqrt{5x^2})(2\sqrt{15x^4}+\sqrt{3x^3})\\\\\Rightarrow(\sqrt{10x^4})(2\sqrt{15x^4})+(\sqrt{10x^4})(\sqrt{3x^3})-(x\sqrt{5x^2})(2\sqrt{15x^4})-(x\sqrt{5x^2})(\sqrt{3x^3})\\\\\Rightarrow2\sqrt{150x^8}+\sqrt{30x^7}-2x\sqrt{75x^6}-x\sqrt{15x^5}\\\\\Rightarrow\bold{10x^4\sqrt{6}+x^3\sqrt{30x}-10x^4\sqrt{3}-x^3\sqrt{15x}}$

The answer is:

$\bold{10x^4\sqrt{6}+x^3\sqrt{30x}-10x^4\sqrt{3}-x^3\sqrt{15x}}$

8 0

2 years ago