All categories

2 years ago

Thursday: For the equation y=3x+5 what is the value of x when y=11? * *

Mathematics

1 answer:

poizon [28]2 years ago

7 0

Answer:

$\huge\boxed{\bf\: 2 = x}$

Step-by-step explanation:

Given,

y = 11

We need to find x.

Equation: y = 3x + 5.

On solving,

$y = 3x + 5\\(11) = 3x + 5\\11 - 5 = 3x\\6 = 3x\\6 \div 3 = x\\\boxed{\bf\:2=x}$

$\rule{150pt}{2pt}$

You might be interested in

Help me out pls!!!!!!

klio [65]

It is at (2,4) hope it helps

4 0

3 years ago

Read 2 more answers

What is the y-intercept of the function, and what does this tell you about the car?

scZoUnD [109]

<span>Write an equation in the standard form to represent the total rent (y) that Sam has to ... is the y-intercept of the function, and what does this tell you about the car?</span><span>
</span>

6 0

3 years ago

Log (X) + 5 = 8 el valor de X es *

GaryK [48]

Log (x) = 3
10^3 = x
x = 1,000

4 0

3 years ago

Find the angle between u =the square root of 5i-8j and v =the square root of 5i+j.

fenix001 [56]

Answer:

The angle between vector $\vec{u} = 5\, \vec{i} - 8\, \vec{j}$ and $\vec{v} = 5\, \vec{i} + \, \vec{j}$ is approximately $1.21$ radians, which is equivalent to approximately $69.3^\circ$ .

Step-by-step explanation:

The angle between two vectors can be found from the ratio between:

their dot products, and
the product of their lengths.

To be precise, if $\theta$ denotes the angle between $\vec{u}$ and $\vec{v}$ (assume that $0^\circ \le \theta < 180^\circ$ or equivalently $0 \le \theta < \pi$ ,) then:

$\displaystyle \cos(\theta) = \frac{\vec{u} \cdot \vec{v}}{\| u \| \cdot \| v \|}$ .

<h3>Dot product of the two vectors</h3>

The first component of $\vec{u}$ is $5$ and the first component of $\vec{v}$ is also

The second component of $\vec{u}$ is $(-8)$ while the second component of $\vec{v}$ is $1$ . The product of these two second components is $(-8) \times 1= (-8)$ .

The dot product of $\vec{u}$ and $\vec{v}$ will thus be:

$\begin{aligned} \vec{u} \cdot \vec{v} = 5 \times 5 + (-8) \times1 = 17 \end{aligned}$ .

<h3>Lengths of the two vectors</h3>

Apply the Pythagorean Theorem to both $\vec{u}$ and $\vec{v}$ :

$\| u \| = \sqrt{5^2 + (-8)^2} = \sqrt{89}$ .
$\| v \| = \sqrt{5^2 + 1^2} = \sqrt{26}$ .

<h3>Angle between the two vectors</h3>

Let $\theta$ represent the angle between $\vec{u}$ and $\vec{v}$ . Apply the formula $\displaystyle \cos(\theta) = \frac{\vec{u} \cdot \vec{v}}{\| u \| \cdot \| v \|}$ to find the cosine of this angle:

$\begin{aligned} \cos(\theta)&= \frac{\vec{u} \cdot \vec{v}}{\| u \| \cdot \| v \|} = \frac{17}{\sqrt{89}\cdot \sqrt{26}}\end{aligned}$ .

Since $\theta$ is the angle between two vectors, its value should be between $0\; \rm radians$ and $\pi \; \rm radians$ ( $0^\circ$ and $180^\circ$ .) That is: $0 \le \theta < \pi$ and $0^\circ \le \theta < 180^\circ$ . Apply the arccosine function (the inverse of the cosine function) to find the value of $\theta$ :