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3 years ago

What is the length of the y-component of the vector shown below? A. 77.3 B. 15.0 C. 20.7 D. 21.4

Mathematics

1 answer:

Black_prince [1.1K]3 years ago

6 0

Answer:

$y = 20.7$

Step-by-step explanation:

Given

The figure above

Required

Calculate the y component

Represent the y component with y, the resultant vector with s and the angle with $\theta$

The y component of a vector (as it is, in this case) is calculated as thus;

$sin\theta = \frac{y}{s}$

$Where\ \theta = 15\\ s = 80$

Substituting these values in the equation

$sin\theta = \frac{y}{s}$ becomes

$sin\ 15= \frac{y}{80}$

Multiply both sides by 80

$80 * sin\ 15= \frac{y}{80} * 80$

$80 * sin\ 15= y$

$80 * 0.2588190451= y$

$20.705523608 = y$

$y = 20.705523608$

$y = 20.7$ (Approximated)

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Let V=ℝ2 and let H be the subset of V of all points on the line 4x+3y=12. Is H a subspace of the vector space V?

sergejj [24]

Answer:

No, it isn't.

Step-by-step explanation:

We have $V=IR^{2}$ and let $H$ be the subset of $V$ of all points on the line

$4x+3y=12$

We need to find if $H$ is a subspace of the vector space $V$ .

In $IR^{2}$ all the possibilities for own subspace of the vector space $IR^{2}$ are :

$IR^{2}$ itself.

The vector $0_{IR^{2}}=\left[\begin{array}{c}0&0\end{array}\right]$

All lines in $IR^{2}$ that passes through the origin ( $0_{IR^{2}}=\left[\begin{array}{c}0&0\end{array}\right]$ )

We know that $H$ is the subset of $IR^{2}$ of all points on the line $4x+3y=12$

If we look at the equation, the point $\left[\begin{array}{c}0&0\end{array}\right]$ doesn't verify it because :

$4x+3y=12\\4(0)+3(0)=12\\0=12$

Which is an absurd. Therefore, $H$ doesn't contain the origin (and $H$ is a line in $IR^{2}$ ). Finally, it can't be a vector space of $V=IR^{2}$

8 0

3 years ago

In 2002, there were 972 students enrolled at Oakview High School. Since then, the number of students has increased by 1.5% each

Rudiy27

Answer:

$N(t) = 972(1.015)^{t}$

Growth function.

The number of students enrolled in 2014 is 1162.

Step-by-step explanation:

The number of students in the school in t years after 2002 can be modeled by the following function:

$N(t) = N(0)(1+r)^{t}$

In which N(0) is the number of students in 2002 and r is the rate of change.

If 1+r>1, the function is a growth function.

If 1-r<1, the function is a decay function.

In 2002, there were 972 students enrolled at Oakview High School.

This means that $N(0) = 972$

Since then, the number of students has increased by 1.5% each year.

Increase, so r is positive. This means that $r = 0.015$

Then

$N(t) = N(0)(1+r)^{t}$

$N(t) = 972(1+0.015)^{t}$

$N(t) = 972(1.015)^{t}$

Growth function.

Find the number of students enrolled in 2014.

2014 is 2014-2002 = 12 years after 2002, so this is N(12).

$N(t) = 972(1.015)^{t}$

$N(12) = 972(1.015)^{12}$

$N(12) = 1162$

The number of students enrolled in 2014 is 1162.

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3 years ago

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Anvisha [2.4K]

Answer:3204.
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3 years ago

What is right less a number is less than or equal to seventeen

USPshnik [31]

Answer:

x≤17

Step-by-step explanation:

5 0

3 years ago

3) What is an equation for the line that passes through the coordinates (-1.2) and (7,6) ?

OLEGan [10]

Answer:

$y=\frac{1}{2} x+\frac{5}{2}$

Step-by-step explanation:

First find the slope using the slope using the given points. Remember that the slope is the change in y over the change in x.

$\frac{6-2}{7- (-1)}=\frac{4}{8}=\frac{1}{2}$

So now we have the equation $y=\frac{1}{2} x+b$ , and we need to find out what b is. We can do this by pointing a point (either one, but I'll use -1,2) into the equation

$2=\frac{1}{2}(-1)+b$

Rearrange the equation so it equals b

$\frac{5}{2}=b$

Put it together and that's the final equation!

$y=\frac{1}{2} x+\frac{5}{2}$

7 0

2 years ago