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2 years ago

The product of three and a squared number is twice the sum of the number and four

2 answers:

7 0

From the text of the task we can write the equation:

$3 x^{2} =2(x+4)\\
\\
3x^2=2x+8\\
\\
3x^2-2x-8=0\\
\\
\Delta=(-2)^2-4.3.(-8)=4+96=100\\
\\
x=\frac{2 \pm\sqrt{100}}{2.3}=\frac{2 \pm10}{6}\\
\\
x_1=\frac{2-10}{6}=-\frac{8}{6}=-\frac{4}{3}\\
\\
x_2=\frac{2+10}{2.3}=\frac{12}{6}=2$

iris [78.8K]2 years ago

3 0

$3x^2=2(x+4)\\
3x^2=2x+8\\
3x^2-2x-8=0\\
3x^2-6x+4x-8=0\\
3x(x-2)+4(x-2)=0\\
(3x+4)(x-2)=0\\
x=-\frac{4}{3} \vee x=2$

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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

The equation (x-6)^2/16 + (y+7)^2/4 = 1 represents an ellipse. What are the vertices of the ellipse?

topjm [15]

Answer:

The correct option is C.

Step-by-step explanation:

The given equation is

$\frac{(x-6)^2}{16}+\frac{(y+7)^2}{4}= 1$

It can be rewritten as

$\frac{(x-6)^2}{4^2}+\frac{(y+7)^2}{2^2}= 1$ .....(1)

The standard form of an ellipse is

$\frac{(x-h)^2}{a^2}+\frac{(y-k)^2}{b^2}= 1$ ....(2)

Where (h,k) is center of the ellipse.

If a>b, then the vertices of the ellipse are $(h\pm a, k)$ .

From (1) and (2) we get

$h=6,k=-7,a=4,b=2$

Since a>b, therefore the vertices of the ellipse are

$(h+a, k)=(6+4,-7)\Rightarrow (10,-7)$

$(h-a, k)=(6-4,-7)\Rightarrow (2,-7)$

The vertices of the given ellipse are (10, –7) and (2, –7). Therefore the correct option is C.

5 0

2 years ago

Read 2 more answers

Help it says to make a triangle and I don't know wut I'm doing wrong. I already got (6,4) like a lot of ppl said but it still sa

Soloha48 [4]

Answer:

this is pythagorus theorem

the height is 7,2

the distance is d

d^2 = 5^2 + 4^2

hope this helps

3 0

2 years ago

Travis has the following problem for homework:

Elina [12.6K]

Answer:

The correct answer is -4

Step-by-step explanation:

First you have to solve what is in the parenthesis. So do 36/6 which is 6 and then subtract 2 to get 4. Then you have to do the multiplication so 16 x 4 and you get 64. then you can go left to right. 82-64+18+4 and when you do that you get -4

5 0

2 years ago

Which of the following graphs shows the solution for the inequality<br> y-4> 2(x+2)?

timofeeve [1]

Answer:

Answer:

Graph C shows the solution for the inequality

Step-by-step explanation:

y - 4 > 2(x + 2)

y > 2x + 4 + 4

y > 2x + 8

Let y = 2x + 8, Then

X-intercept (0, 8)

Y-intercept (-4, 0)

Since the inequality sign is > we use broken line.

Put, (0, 0)

y > 2x + 8

0 > 0+ 8

0 > 8 Which is false

so, the answer will be above the broken line

Hence , Graph C shows inequality

#SPJ1

5 0

2 years ago