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3 years ago

What is c divided by 22?

Mathematics

1 answer:

Allushta [10]3 years ago

3 0

Answer:

Step-by-step explanation:

looked it up because I have no Idea lol

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The vector (a) is a multiple of the vector (2i +3j) and (b) is a multiple of (2i+5j) The sum (a+b) is a multiple of the vector (

kow [346]

Answer:

$\|a\| = 5\sqrt{13}$ .

$\|b\| = 3\sqrt{29}$ .

Step-by-step explanation:

Let $m$ , $n$ , and $k$ be scalars such that:

$\displaystyle a = m\, (2\, \vec{i} + 3\, \vec{j}) = m\, \begin{bmatrix}2 \\ 3\end{bmatrix}$ .

$\displaystyle b = n\, (2\, \vec{i} + 5\, \vec{j}) = n\, \begin{bmatrix}2 \\ 5\end{bmatrix}$ .

$\displaystyle (a + b) = k\, (8\, \vec{i} + 15\, \vec{j}) = k\, \begin{bmatrix}8 \\ 15\end{bmatrix}$ .

The question states that $\| a + b \| = 34$ . In other words:

$k\, \sqrt{8^{2} + 15^{2}} = 34$ .

$k^{2} \, (8^{2} + 15^{2}) = 34^{2}$ .

$289\, k^{2} = 34^{2}$ .

Make use of the fact that $289 = 17^{2}$ whereas $34 = 2 \times 17$ .

$\begin{aligned}17^{2}\, k^{2} &= 34^{2}\\ &= (2 \times 17)^{2} \\ &= 2^{2} \times 17^{2} \end{aligned}$ .

$k^{2} = 2^{2}$ .

The question also states that the scalar multiple here is positive. Hence, $k = 2$ .

Therefore:

$\begin{aligned} (a + b) &= k\, (8\, \vec{i} + 15\, \vec{j}) \\ &= 2\, (8\, \vec{i} + 15\, \vec{j}) \\ &= 16\, \vec{i} + 30\, \vec{j}\\ &= \begin{bmatrix}16 \\ 30 \end{bmatrix}\end{aligned}$ .

$(a + b)$ could also be expressed in terms of $m$ and $n$ :

$\begin{aligned} a + b &= m\, (2\, \vec{i} + 3\, \vec{j}) + n\, (2\, \vec{i} + 5\, \vec{j}) \\ &= (2\, m + 2\, n) \, \vec{i} + (3\, m + 5\, n) \, \vec{j} \end{aligned}$ .

$\begin{aligned} a + b &= m\, \begin{bmatrix}2\\ 3 \end{bmatrix} + n\, \begin{bmatrix} 2\\ 5 \end{bmatrix} \\ &= \begin{bmatrix}2\, m + 2\, n \\ 3\, m + 5\, n\end{bmatrix}\end{aligned}$ .

Equate the two expressions and solve for $m$ and $n$ :

$\begin{cases}2\, m + 2\, n = 16 \\ 3\, m + 5\, n = 30\end{cases}$ .

$\begin{cases}m = 5 \\ n = 3\end{cases}$ .

Hence:

$\begin{aligned} \| a \| &= \| m\, (2\, \vec{i} + 3\, \vec{j})\| \\ &= m\, \| (2\, \vec{i} + 3\, \vec{j}) \| \\ &= 5\, \sqrt{2^{2} + 3^{2}} = 5 \sqrt{13}\end{aligned}$ .

$\begin{aligned} \| b \| &= \| n\, (2\, \vec{i} + 5\, \vec{j})\| \\ &= n\, \| (2\, \vec{i} + 5\, \vec{j}) \| \\ &= 3\, \sqrt{2^{2} + 5^{2}} = 3 \sqrt{29}\end{aligned}$ .

6 0

3 years ago

Which number sentence has a difference of 5? 8-(-3) 0 -3-8 -3-(-8) 3-(-8)

valina [46]

Answer:

The 4th one

Step-by-step explanation:

-3-(-8)=5

6 0

3 years ago

Which of the lines is perpendicular to the line shown in the graph

Nitella [24]

Answer:

The line passing through (-8, 10) and (-1, 4).

Step-by-step explanation:

Two lines are perpendicular if the product of their slopes is -1. The slope of the line in the picture is $7\over 6$ , so we should find a line with slope of $-6\over 7$ .

Note that the slope of the line in the last option is $-7\over 6$ .

4 0

3 years ago

Please help!!!!! It's urgent

Evgesh-ka [11]

2b 2a
----------------- + -----------------
(b+a)^2 (b^2 - a^2)

2b 2a
= ----------------- + -------------------
(b+a)(b+a) (b+a)(b-a)

2b(b - a) + 2a(b + a)
= ------------------------------------
   (b+a)(b+a)(b-a)

2b^2 - 2ab + 2ab + 2a^2
= ---------------------------------------
     (b+a)(b+a)(b-a)

2b^2 + 2a^2
= ------------------------
   (b+a)(b+a)(b-a)

2(b^2 + a^2)
= ------------------------
   (b+a)^2 (b-a)

Answer:

Numerator: 2(b^2 + a^2)
Denominator: (b+a)^2 (b-a)

7 0

3 years ago

Can some one answer this plz ill give !100 POINTS!

Lorico [155]

Hello!

First of all let's find the perimeter (circumference) of the semi circles. We can combine them to make one circle with a diameter of 4 (as we can see the side length of one semi circle is 4 cm. We now plug it into the circumference equation ( $\pi$ =3.14).

4(3.14)=12.56

Now we add up the side lengths of the rectangle.

4+6+4+6=20

Now we add up the length of our circle and rectangle.

20+12.56=32.56

Therefore our answer is 32.56 cm.
----------------------------------------------------------

Now to find the area! If we combine the two semicircles, we get a circle with a diameter of four. This means that is has a radius of two. We use the equation below to find the area of the two circles.

A= $\pi$ r²

First we will square our radius.

2(2)=4

Now we multiply by pi.

4(3.14)=12.56

Now we need to find the area of the rectangle.

6(4)=24

Now we add.

24+12.56=36.56.

I hope this helps!

8 0

4 years ago

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