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3 years ago

Number 3 please need help

Mathematics

1 answer:

kirza4 [7]3 years ago

4 0

Im pretty sure it would be 4

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Draw a line from each figure to an area

Georgia [21]

Answer:

A=27 cm^2

B=24 cm^2

C=26 cm^2

D=28 cm^2

Step-by-step explanation:

Break each problem down into individual shapes.

For instance, A can be split into a 3 by 3 square and a 6 by 3 square.

Get the area by multiplying the length & height: A = L * H

For the triangles the area is the same equation divided by 2 A=LH/2

Shapes with unclear dimensions like C can be skipped and have their area revealed through process of elimination.

3 0

3 years ago

Find the volume and surface area of the prism below

Nata [24]

Answer:

a) The volume of the prism is 660 mm³

b) The surface area of the prism is 514 mm²

Step-by-step explanation:

The given prism for which we are to find the volume and the surface area shows the dimensions of the sides

a) To find the volume, we can consider the prism as a composite figure as follows;

The topmost cuboid with dimensions (15 - 2×3) mm, 3 mm, and 9 mm

Therefore, the volume of the topmost cuboid, V₁, is given as follows;

V₁ = 5 mm × 3 mm × 9 mm = 135 mm³

The volume of the cuboid on which the top cuboid rest, V₂, is given as follows;

V₂ = 15 mm × 5 mm × 7 mm = 525 mm³

The volume of the prism, V = V₁ + V₂

Therefore, we have;

V =135 mm³ + 525 mm³ = 660 mm³

The volume of the prism, V = 660 mm³

b) The surface area of the prism is given as follows;

The surface area of the top cuboid, SA₁, is given as follows;

SA₁ = 9 mm × 5 mm + 2 × 3 mm × 9 mm + 2 × 3 mm × 5 mm = 129 mm²

The surface area of the larger cuboid, SA₂, is given as follows;

SA₂ = 2 × 15 mm × 7 mm + 2 × 5 mm × 7 mm + 2 × 3 mm × 5 mm + 15 mm × 5 mm = 385 mm²

The surface area of the prism, SA = SA₁ + SA₂

∴ The surface area of the prism, SA = 129 mm² + 385 mm² = 514 mm²

4 0

2 years ago

in exercises 15-20 find the vector component of u along a and the vecomponent of u orthogonal to a u=(2,1,1,2) a=(4,-4,2,-2)

Nimfa-mama [501]

Answer:

The component of $\vec u$ orthogonal to $\vec a$ is $\vec u_{\perp\,\vec a} = \left(\frac{9}{5},\frac{6}{5},\frac{9}{10},\frac{21}{10}\right)$ .

Step-by-step explanation:

Let $\vec u$ and $\vec a$ , from Linear Algebra we get that component of $\vec u$ parallel to $\vec a$ by using this formula:

$\vec u_{\parallel \,\vec a} = \frac{\vec u \bullet\vec a}{\|\vec a\|^{2}} \cdot \vec a$ (Eq. 1)

Where $\|\vec a\|$ is the norm of $\vec a$ , which is equal to $\|\vec a\| = \sqrt{\vec a\bullet \vec a}$ . (Eq. 2)

If we know that $\vec u =(2,1,1,2)$ and $\vec a=(4,-4,2,-2)$ , then we get that vector component of $\vec u$ parallel to $\vec a$ is:

$\vec u_{\parallel\,\vec a} = \left[\frac{(2)\cdot (4)+(1)\cdot (-4)+(1)\cdot (2)+(2)\cdot (-2)}{4^{2}+(-4)^{2}+2^{2}+(-2)^{2}} \right]\cdot (4,-4,2,-2)$

$\vec u_{\parallel\,\vec a} =\frac{1}{20}\cdot (4,-4,2,-2)$

$\vec u_{\parallel\,\vec a} =\left(\frac{1}{5},-\frac{1}{5},\frac{1}{10},-\frac{1}{10} \right)$

Lastly, we find the vector component of $\vec u$ orthogonal to $\vec a$ by applying this vector sum identity:

$\vec u_{\perp\,\vec a} = \vec u - \vec u_{\parallel\,\vec a}$ (Eq. 3)

If we get that $\vec u =(2,1,1,2)$ and $\vec u_{\parallel\,\vec a} =\left(\frac{1}{5},-\frac{1}{5},\frac{1}{10},-\frac{1}{10} \right)$ , the vector component of $\vec u$ is:

$\vec u_{\perp\,\vec a} = (2,1,1,2)-\left(\frac{1}{5},-\frac{1}{5},\frac{1}{10},-\frac{1}{10} \right)$

$\vec u_{\perp\,\vec a} = \left(\frac{9}{5},\frac{6}{5},\frac{9}{10},\frac{21}{10}\right)$

The component of $\vec u$ orthogonal to $\vec a$ is $\vec u_{\perp\,\vec a} = \left(\frac{9}{5},\frac{6}{5},\frac{9}{10},\frac{21}{10}\right)$ .

4 0

3 years ago

You pick a card at random from an ordinary deck of 52 cards. If the card is an ace, you get 9 points; if not, you lose 1 point.

Andreyy89

Answer:

$a = 9\\b = 48\\c = -1$

Step-by-step explanation:

We know that:

In a deck of 52 cards there are 4 aces.

Therefore the probability of obtaining an ace is:

P (x) = 4/52

The probability of not getting an ace is:

P ('x) = 1-4 / 52

P ('x) = 48/52

In this problem the number of aces obtained when extracting cards from the deck is a discrete random variable.

For a discrete random variable V, the expected value is defined as:

$E(V) = VP(V)$

Where V is the value that the random variable can take and P (V) is the probability that it takes that value.

We have the following equation for the expected value:

$E(V) = \frac{4}{52}(a) + \frac{b}{52}(c)$

In this problem the variable V can take the value V = 9 if an ace of the deck is obtained, with probability of 4/52, and can take the value V = -1 if an ace of the deck is not obtained, with a probability of 48 / 52

Therefore, expected value for V, the number of points obtained in the game is:

$E(V) = \frac{4}{52}(9) + \frac{48}{52}(-1)$