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3 years ago

HELP PLEASE!! I DONT KNOW IF I GOT IT RIGHT

Mathematics

1 answer:

lutik1710 [3]3 years ago

4 0

Sin E= 0. 198.

E= 11.42°

Step-by-step explanation:

using Pythagoras theorem.

Hyp²= base² + height ².

101²= 99²+x²

x²= 101²-99²

x²=10201-9801.

x²=400

x=√400

x=20

sin E = opposite

________

adjacent.

sin E= 20

101.

sin E=0.198.

E= sin^-1(0.198)

E= 11.42°

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Vic is standing on the ground at a point directly south of the base of the CN Tower and can see the top when looking at an angle

mel-nik [20]

Vic and Dan are 2, 897m apart.

<h3>How to determine the distance</h3>

It is important to note that the distance between Vic and Dan is the base of CN

Let's say the distance to Dan is x

The distance to Vic is y

Using cosine ratio, we have

cos α = opposite / adjacent

α = 72°

opposite = 553. 3cm

Adjacent = x

cos 72° = $\frac{553. 3}{x}$

Cross multiply

$cos 72$ × $x$ = $553. 3$

$0. 3090x= 553. 3$

$x = \frac{553. 3}{0. 3090}$

$x = 1, 790. 61$ m

The distance to Vic is y

Using the cosine ratio, we have

$cos 60 = \frac{553. 3}{y}$

Cross multiply

$0. 5y = 553. 3$

$y = \frac{553. 3}{0. 5}$

$y = 1,106. 6$ m

To determine how far apart Vic and Dan, we use = x + y

= 1790. 61 + 1106. 6

= 2, 897. 21m

= 2, 897m

Thus, Vic and Dan are 2, 897m apart.

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7 0

2 years ago

Help me please!!

Mice21 [21]

Answer:

1.) - 1/2

2.) the slope is 0 which means it is a vertical line

Step-by-step explanation:

In order to solve this you take the formula

y-y / x-x

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7 0

3 years ago

No Link or I’m blocking you :)

DaniilM [7]

Answer:

- 55

Step-by-step explanation:

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8 0

3 years ago

Find two vectors in R2 with Euclidian Norm 1<br> whoseEuclidian inner product with (3,1) is zero.

alina1380 [7]

Answer:

$v_1=(\frac{1}{10},-\frac{3}{10})$

$v_2=(-\frac{1}{10},\frac{3}{10})$

Step-by-step explanation:

First we define two generic vectors in our $\mathbb{R}^2$ space:

$v_1 = (x_1,y_1)$
$v_2 = (x_2,y_2)$

By definition we know that Euclidean norm on an 2-dimensional Euclidean space $\mathbb{R}^2$ is:

$\left \| v \right \|= \sqrt{x^2+y^2}$

Also we know that the inner product in $\mathbb{R}^2$ space is defined as:

$v_1 \bullet v_2 = (x_1,y_1) \bullet(x_2,y_2)= x_1x_2+y_1y_2$

So as first condition we have that both two vectors have Euclidian Norm 1, that is:

$\left \| v_1 \right \|= \sqrt{x^2+y^2}=1$

and

$\left \| v_2 \right \|= \sqrt{x^2+y^2}=1$

As second condition we have that:

$v_1 \bullet (3,1) = (x_1,y_1) \bullet(3,1)= 3x_1+y_1=0$

$v_2 \bullet (3,1) = (x_2,y_2) \bullet(3,1)= 3x_2+y_2=0$

Which is the same:

$y_1=-3x_1\\y_2=-3x_2$

Replacing the second condition on the first condition we have:

$\sqrt{x_1^2+y_1^2}=1 \\\left | x_1^2+y_1^2 \right |=1 \\\left | x_1^2+(-3x_1)^2 \right |=1 \\\left | x_1^2+9x_1^2 \right |=1 \\\left | 10x_1^2 \right |=1 \\x_1^2= \frac{1}{10}$

Since $x_1^2= \frac{1}{10}$ we have two posible solutions, $x_1=\frac{1}{10}$ or $x_1=-\frac{1}{10}$ . If we choose $x_1=\frac{1}{10}$ , we can choose next the other solution for $x_2$ .