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3 years ago

How to solve it and understand it

Mathematics

1 answer:

algol133 years ago

8 0

Volume of rectangle prism = w l h

Volume of rectangle pyramid = 1/3(w l h)

so Volume of rectangle pyramid = 1/3 (Volume of rectangle prism)

= 1/3 (240)

= 80

Answer:

80 cm^3

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How do you solve this triangle by using the law of sine where

mars1129 [50]

Actually, when you know 2 sides and an included angle, you use the Law of Cosines. (and we don't know if theta is an included angle).
Solving for side c
c^2 = a^2 + b^2 -2ab * cos(C)
c^2 = 36 + 16 - 2*6*4 * cos(60)
c^2 = 52 -48*.5
c^2 = 28
c = 5.2915

Using the Law of Sines
side c / sin(C) = side b / sin (B)
5.2915 / sin(60) = 4 / sin (B)
sin(B) = sin(60) * 4 / 5.2915
sin(B) = 0.86603 * 4 / 5.2915
sin(B) = 3.46412 / 5.2915
 sin(B) = 0.6546571451 
Angle B = 40.894 Degrees

sin (A) / side a = sin (B) / side b
sin (A) = 6 * sin (40.894) / 4
sin (A) = 6 * 0.65466 / 4
sin (A) = .98199
angle A = 79.109 Degrees

angle C = 60 Degrees

7 0

3 years ago

HELP TIMED BRAINLY POINTS

kozerog [31]

Answer:

dude you need to show the whole answer

Step-by-step explanation:

sorry I couldn't help

6 0

3 years ago

Find the missing term in the geometric sequence. 180,___,5

Elan Coil [88]

Answer:

30.

Step-by-step explanation:

first term = a

3r term = ar^2

So here ar^2/ a = r^2

r^2 = 5/180 = 1/36

r = 1/6

So missing term = 180 * 1/6 = 30.

7 0

3 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

PERSEVERE Find the values of x and y,

Vlad [161]

y² = 8y - 15 (alternate angles are equal)

y² - 8y + 15 = 0

(y - 5)(y - 3) = 0

y = 5 or 3

x + 8y - 15 = 180 (angles in a straight line add up to 180)

when y = 5

x + 40 - 15 = 180

x = 155°

when y = 3

x + 24 - 15 = 180

x = 171°

8 0

3 years ago