All categories

3 years ago

16.2=2.7(h+1) solve for h

Mathematics

2 answers:

Vedmedyk [2.9K]3 years ago

6 0

Answer:

$\boxed{ \bold{ \huge{ \boxed{ \sf{h = 5}}}}}$

Step-by-step explanation:

$\bigstar{ \sf{ \: \: 16. 2 = 2.7( \: h \: + \: 1)}}$

$\tt{Step \: 1 \: }$ : Swap the sides of the equation

$\longmapsto{ \sf{2.7(h + 1) = 16.2}}$

$\tt{Step \: 2 \: }$ : Distribute 2.7 through the parentheses

$\longmapsto{ \sf{2.7 \times h + 2.7 \times 1 = 16.2}}$

$\longmapsto{ \sf{2.7h + 2.7 = 16.2}}$

$\tt{Step \: 3 \: }$ : Move 2.7 to right hand side and change it's sign

$\longmapsto{ \sf{2.7h = 16.2 - 2.7}}$

$\tt{Step \: 4 \: }$ : Subtract 2.7 from 16.2

$\longmapsto{ \sf{2.7h = 13.5}}$

$\tt{Step \: 5 \: }$ : Divide both sides by 2.7

$\longmapsto{ \sf{ \frac{2.7h}{2.7} = \frac{13.5}{2.7}}}$

$\tt{Step \: 6 \: }$ : Calculate

$\longmapsto{ \sf{h = 5}}$

The value of h is 5

Hope I helped!

Best regards! :D

~TheAnimeGirl

Rom4ik [11]3 years ago

5 0

Step-by-step explanation:

16.2 = 2.7(h+1)

2.7(h+1) = 16.2

(h+1) = 16.2 ÷ 2.7

h+1 = 6

h = 5

You might be interested in

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 $\theta$ :