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3 years ago

Evaluate 4-2f when f=1

Mathematics

2 answers:

iren2701 [21]3 years ago

7 0

The answer is 2 ....
..

Sladkaya [172]3 years ago

6 0

First substitute f for 1 a number and a variable together means multiplication. 4-2*1
Multiplication is first so (2*1=2) which brings us to 4-2.
Subtract 4-2=2
Answer: 2

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I'd go with Checking account B because its cheaper .

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3 years ago

Find the direction cosines and direction angles of the vector. (Give the direction angles correct to the nearest degree.) 5, 1,

Dahasolnce [82]

Answer:

The direction cosines are:

$\frac{5}{\sqrt{42} }$ , $\frac{1}{\sqrt{42} }$ and $\frac{4}{\sqrt{42} }$ with respect to the x, y and z axes respectively.

The direction angles are:

40°, 81° and 52° with respect to the x, y and z axes respectively.

Step-by-step explanation:

For a given vector a = ai + aj + ak, its direction cosines are the cosines of the angles which it makes with the x, y and z axes.

If a makes angles α, β, and γ (which are the direction angles) with the x, y and z axes respectively, then its direction cosines are: cos α, cos β and cos γ in the x, y and z axes respectively.

Where;

cos α = $\frac{a . i}{|a| . |i|}$ ---------------------(i)

cos β = $\frac{a.j}{|a||j|}$ ---------------------(ii)

cos γ = $\frac{a.k}{|a|.|k|}$ ----------------------(iii)

And from these we can get the direction angles as follows;

α = cos⁻¹ ( $\frac{a . i}{|a| . |i|}$ )

β = cos⁻¹ ( $\frac{a.j}{|a||j|}$ )

γ = cos⁻¹ ( $\frac{a.k}{|a|.|k|}$ )

Now to the question:

Let the given vector be

a = 5i + j + 4k

a . i = (5i + j + 4k) . (i)

a . i = 5 [a.i is just the x component of the vector]

a . j = 1 [the y component of the vector]

a . k = 4 [the z component of the vector]

Also

|a|. |i| = |a|. |j| = |a|. |k| = |a| [since |i| = |j| = |k| = 1]

|a| = $\sqrt{5^2 + 1^2 + 4^2}$

|a| = $\sqrt{25 + 1 + 16}$

|a| = $\sqrt{42}$

Now substitute these values into equations (i) - (iii) to get the direction cosines. i.e

cos α = $\frac{5}{\sqrt{42} }$

cos β = $\frac{1}{\sqrt{42} }$

cos γ = $\frac{4}{\sqrt{42} }$

From the value, now find the direction angles as follows;

α = cos⁻¹ ( $\frac{a . i}{|a| . |i|}$ )

α = cos⁻¹ ( $\frac{5}{\sqrt{42} }$ )

α = cos⁻¹ ( $\frac{5}{6.481}$ )

α = cos⁻¹ (0.7715)

α = 39.51

α = 40°

β = cos⁻¹ ( $\frac{a.j}{|a||j|}$ )

β = cos⁻¹ ( $\frac{1}{\sqrt{42} }$ )

β = cos⁻¹ ( $\frac{1}{6.481 }$ )

β = cos⁻¹ ( 0.1543 )

β = 81.12

β = 81°

γ = cos⁻¹ ( $\frac{a.k}{|a|.|k|}$ )

γ = cos⁻¹ ( $\frac{4}{\sqrt{42} }$ )

γ = cos⁻¹ ( $\frac{4}{6.481}$ )

γ = cos⁻¹ (0.6172)

γ = 51.89

γ = 52°

Conclusion:

The direction cosines are:

$\frac{5}{\sqrt{42} }$ , $\frac{1}{\sqrt{42} }$ and $\frac{4}{\sqrt{42} }$ with respect to the x, y and z axes respectively.

The direction angles are:

40°, 81° and 52° with respect to the x, y and z axes respectively.

3 0

3 years ago

¿Cómo las razones de seno y cos eno son semejantes?

Sauron [17]

Las razones de los lados de un triángulo rectángulo se llaman razones trigonométricas.

Espero te ayude :)

6 0

3 years ago

Solve the inequality w/2<-42

exis [7]

Answer:

w<-84

Step-by-step explanation:

solved the equation

4 0

3 years ago

Find all solutions of the equation: 2cos^2x-cosx=1

Art [367]

If you're using the app, try seeing this answer through your browser: brainly.com/question/3166243

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Solve the trigonometric equation:

$\mathsf{2\,cos^2\,x-cos\,x=1}\\\\ \mathsf{2\,cos^2\,x-cos\,x-1=0}$

Make a substitution:

$\mathsf{cos\,x=t\qquad (-1\le t\le 1)}$

and the equation becomes

$\mathsf{2t^2-t-1=0}$

Rewrite conveniently – t as + t – 2t, and then factor the left-hand side by grouping:

$\mathsf{2t^2+t-2t-1=0}\\\\ \mathsf{t\cdot (2t+1)-1\cdot (2t+1)=0}$

Factor out 2t + 1:

$\mathsf{(2t+1)\cdot (t-1)=0}\\\\ \begin{array}{rcl} \mathsf{2t+1=0}&~\textsf{ or }~&\mathsf{t-1=0}\\\\ \mathsf{2t=1}&~\textsf{ or }~&\mathsf{t=1}\\\\ \mathsf{t=\dfrac{\,1\,}{2}}&~\textsf{ or }~&\mathsf{t=1} \end{array}$

Substitute back for t = cos x:

$\begin{array}{rcl}\mathsf{cos\,x=\dfrac{\,1\,}{2}}&~\textsf{ or }~&\mathsf{cos\,x=1}\\\\ \mathsf{cos\,x=cos\,60^\circ}&~\textsf{ or }~&\mathsf{cos\,x=cos\,0} \end{array}$

Therefore,

$\begin{array}{rcl} \mathsf{x=\pm\,60^\circ+k\cdot 360^\circ}&~\textsf{ or }~&\mathsf{cos\,x=0+k\cdot 360^\circ} \end{array}$

where k is an integer.

Solution set:

$\mathsf{S=\left\{x\in\mathbb{R}:~~x=-\,60^\circ+k\cdot 360^\circ~~or~~x=60^\circ+k\cdot 360^\circ~~or~~x=k\cdot 360^\circ,~~k\in\mathbb{Z}\right\}}$

I hope this helps. =)

3 0

3 years ago