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1 year ago

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what steps do you need to do to evaluate the following expression? Select two options. –5(–2) Drag 5 sets of –2 tiles to the win

dow. Drag zero pairs to the window. Remove the 5 groups of –2 tiles. Remove 5 negative tiles from the window.

1 answer:

kvv77 [185]1 year ago

5 0

To solve the expression given , Drag zero pairs to the window and then Remove the 5 groups of –2 tiles , Option b and c is the right answer

<h3>What is an Integer Tile method ?</h3>

It is a teaching method in which materialistic representation is done for concepts like addition and subtraction .

It is used for kids who find it difficult to understand adding a negative integer etc.

For the expression given

-5 (-2)

Drag zero pairs to the window and then Remove the 5 groups of –2 tiles.

Therefore Option B and C is the right answer.

To know more about Integer Tile Method

brainly.com/question/18097183

#SPJ1

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Use the given information to solve the triangle for A = 150°, b = 4.8, a = 9.4

ELEN [110]

Well you can start by drawing a triangle with the information

the size of angle B can be found using the Law of Sine

$\frac{sin(150)}{9.4)} = \frac{sin(B)}{4.8}$

the size of angle C can be found using angle sum of a triangle.

The length of side c can be found using the Law of Cosines

$c^2 = 4.8^2 + 9.4^2 - 2 \times4.8\times9.4\times cos(C)$

hope it helps

7 0

3 years ago

*Please Help!* (Easy question) 20 POINTS

mariarad [96]

Yes - they have to be similar. Since they are using the same line as the hypotenuse, the ratio of the other two sides are in the same ratio (ie, the slope of the line), the 3 inner angles will be the same. Thus, the triangles will be similar.

7 0

2 years ago

Read 2 more answers

Find the mass and center of mass of the lamina that occupies the region D and has the given density function rho. D is the trian

Alla [95]

Answer: mass (m) = 4 kg

center of mass coordinate: (15.75,4.5)

Step-by-step explanation: As a surface, a lamina has 2 dimensions (x,y) and a density function.

The region D is shown in the attachment.

From the image of the triangle, lamina is limited at x-axis: 0≤x≤2

At y-axis, it is limited by the lines formed between (0,0) and (2,1) and (2,1) and (0.3):

<u>Points (0,0) and (2,1):</u>

y = $\frac{1-0}{2-0}(x-0)$

y = $\frac{x}{2}$

<u>Points (2,1) and (0,3):</u>

y = $\frac{3-1}{0-2}(x-0) + 3$

y = -x + 3

Now, find total mass, which is given by the formula:

$m = \int\limits^a_b {\int\limits^a_b {\rho(x,y)} \, dA }$

Calculating for the limits above:

$m = \int\limits^2_0 {\int\limits^a_\frac{x}{2} {2(x+y)} \, dy \, dx }$

where a = -x+3

$m = 2.\int\limits^2_0 {\int\limits^a_\frac{x}{2} {(xy+\frac{y^{2}}{2} )} \, dx }$

$m = 2.\int\limits^2_0 {(-x^{2}-\frac{x^{2}}{2}+3x )} \, dx }$

$m = 2.\int\limits^2_0 {(\frac{-3x^{2}}{2}+3x)} \, dx }$

$m = 2.(\frac{-3.2^{2}}{2}+3.2-0)$

m = 2(-4+6)

m = 4

<u>Mass of the lamina that occupies region D is 4.</u>

<u />

Center of mass is the point of gravity of an object if it is in an uniform gravitational field. For the lamina, or any other 2 dimensional object, center of mass is calculated by:

$M_{x} = \int\limits^a_b {\int\limits^a_b {y.\rho(x,y)} \, dA }$

$M_{y} = \int\limits^a_b {\int\limits^a_b {x.\rho(x,y)} \, dA }$

$M_{x}$ and $M_{y}$ are moments of the lamina about x-axis and y-axis, respectively.

Calculating moments:

For moment about x-axis:

$M_{x} = \int\limits^a_b {\int\limits^a_b {y.\rho(x,y)} \, dA }$

$M_{x} = \int\limits^2_0 {\int\limits^a_\frac{x}{2} {2.y.(x+y)} \, dy\, dx }$

$M_{x} = 2\int\limits^2_0 {\int\limits^a_\frac{x}{2} {y.x+y^{2}} \, dy\, dx }$

$M_{x} = 2\int\limits^2_0 { ({\frac{y^{2}x}{2}+\frac{y^{3}}{3})}\, dx }$

$M_{x} = 2\int\limits^2_0 { ({\frac{x(-x+3)^{2}}{2}+\frac{(-x+3)^{3}}{3} -\frac{x^{3}}{8}-\frac{x^{3}}{24} )}\, dx }$

$M_{x} = 2.(\frac{-9.x^{2}}{4}+9x)$

$M_{x} = 2.(\frac{-9.2^{2}}{4}+9.2)$

$M_{x} = 18$

Now to find the x-coordinate:

x = $\frac{M_{y}}{m}$

x = $\frac{63}{4}$

x = 15.75

For moment about the y-axis:

$M_{y} = \int\limits^2_0 {\int\limits^a_\frac{x}{2} {2x.(x+y))} \, dy\,dx }$

$M_{y} = 2.\int\limits^2_0 {\int\limits^a_\frac{x}{2} {x^{2}+yx} \, dy\,dx }$

$M_{y} = 2.\int\limits^2_0 {y.x^{2}+x.{\frac{y^{2}}{2} } } \,dx }$

$M_{y} = 2.\int\limits^2_0 {x^{2}.(-x+3)+\frac{x.(-x+3)^{2}}{2} - {\frac{x^{3}}{2}-\frac{x^{3}}{8} } } \,dx }$

$M_{y} = 2.\int\limits^2_0 {\frac{-9x^3}{8}+\frac{9x}{2} } \,dx }$

$M_{y} = 2.({\frac{-9x^4}{32}+9x^{2})$

$M_{y} = 2.({\frac{-9.2^4}{32}+9.2^{2}-0)$

$M{y} =$ 63

To find y-coordinate:

y = $\frac{M_{x}}{m}$

y = $\frac{18}{4}$

y = 4.5

<u>Center mass coordinates for the lamina are (15.75,4.5)</u>

3 0

3 years ago

Is a number of television in a household a discrete quantitative or continuous quantitative

mezya [45]

discrete, you cannot have 1.5 televisions

generally the number of X is a discrete variable.

continous = can contain decimal value (float values)

discrete = no decimal (integers)

7 0

3 years ago

A(1) = -11<br> a(n) = a(n − 1). 10<br> What is the 4th term in the sequence?

neonofarm [45]

Answer:

a(n)= a(n-1) .10

Clearly see the equation

you can do it directly Apply logic

Every n th term is 10 times of previous n-1 th term

That is common ratio an/a(n-1)= 10

So, We have to find 4 th term

So, 4 th term is simply

1st term × (common ratio)^(4-1)

-11( 10)^3

-11000

Thanks

3 0

3 years ago