All categories

4 years ago

Solve for x. round your answer to the nearest tenth

Mathematics

1 answer:

Yakvenalex [24]4 years ago

8 0

Answer:

11.9

Step-by-step explanation:

Use sin

Sin ratio is opposite over hypotenuse

Sin $57^{o}$ = $\frac{10.8}{x}$

x = $\frac{10.8}{sin57^{o} }$

x = 11.9

You might be interested in

Which term describes a line segment that connects a vertex of a triangle to the midpoint of the opposite side? A. Angle bisector

Anettt [7]

The term describes a line segment that connects a vertex of a triangle to the midpoint of the opposite side is :

C. Median

7 0

3 years ago

I need help will mark you the brainliest

guapka [62]

Answer:

i thinnk A

Step-by-step explanation:

5 0

3 years ago

Help im behind my math class. pls

maks197457 [2]

Answer:

Please check the explanation.

Step-by-step explanation:

From the given diagram, we can observe that r and s are the two parallel lines intersected by the two transversal lines.

Therefore, the pairs of alternating interior angles formed by the two transversal lines are:

∠1 & ∠5

∠2 & ∠4

∠2 & ∠7

∠3 & ∠6

We know that alternating interior angles formed by a transversal line are congruent.

Thus,

∠1 = ∠5

∠2 = ∠4

∠2 = ∠7

∠3 = ∠6

Part a)

Given

m∠1 = 3x + 42

m∠5 = 8x - 8

∠1 = ∠5

3x + 42 = 8x - 8

flipe the equation

8x - 8 = 3x + 42

subtract 3x from both sides

8x - 8 - 3x = 3x + 42 - 3x

5x - 8 = 42

add 8 to both sides

5x - 8 + 8 = 42 + 8

5x = 50

divide both sides by 5

5x/5 = 50/5

x = 10.0000 (Rounded to four decimal places)

Thus, the value of x = 10.0000 (Rounded to four decimal places)

Part b)

Given

m∠6 = 12°

It is clear that angles ∠6 and angle ∠7 lie on a straight line.

Thus,

The sum of ∠6 and ∠7 is 180°.

∠6 + ∠7 = 180°

substituting m ∠ 6 = 12° in ∠6 + ∠7 = 180°

12° + ∠7 = 180°

subtract 12 from both sides

12° + ∠7 - 12° = 180° - 12°

∠7 = 168°

Thus, measure of angle ∠7 = 168°.

We already know that alternating interior angles formed by a transversal line are congruent.

∠7 and ∠2 are alternating interior angles.

Thus,

∠2 = ∠7

As ∠7 = 168°.

Therefore,

∠2 = 168°

Hence, we conclude that

∠2 = 168°

7 0

3 years ago

Core Math 6 2014 - MA3106 IC

Ratling [72]

Answer:

yes

Step-by-step explanation:

when you add 6 and 5 you get 11

7 times 11 is 77

42 plus 35 is 77

they equal eachother

4 0

3 years ago

Evaluate the integral, show all steps please!

Aloiza [94]

Answer:

$\displaystyle \int \dfrac{1}{(9-x^2)^{\frac{3}{2}}}\:\:\text{d}x=\dfrac{x}{9\sqrt{9-x^2}} +\text{C}$

Step-by-step explanation:

Fundamental Theorem of Calculus

$\displaystyle \int \text{f}(x)\:\text{d}x=\text{F}(x)+\text{C} \iff \text{f}(x)=\dfrac{\text{d}}{\text{d}x}(\text{F}(x))$

If differentiating takes you from one function to another, then integrating the second function will take you back to the first with a constant of integration.

Given indefinite integral:

$\displaystyle \int \dfrac{1}{(9-x^2)^{\frac{3}{2}}}\:\:\text{d}x$

Rewrite 9 as 3² and rewrite the 3/2 exponent as square root to the power of 3:

$\implies \displaystyle \int \dfrac{1}{\left(\sqrt{3^2-x^2}\right)^3}\:\:\text{d}x$

Integration by substitution

$\boxed{\textsf{For }\sqrt{a^2-x^2} \textsf{ use the substitution }x=a \sin \theta}$

$\textsf{Let }x=3 \sin \theta$

$\begin{aligned}\implies \sqrt{3^2-x^2} & =\sqrt{3^2-(3 \sin \theta)^2}\\ & = \sqrt{9-9 \sin^2 \theta}\\ & = \sqrt{9(1-\sin^2 \theta)}\\ & = \sqrt{9 \cos^2 \theta}\\ & = 3 \cos \theta\end{aligned}$

Find the derivative of x and rewrite it so that dx is on its own:

$\implies \dfrac{\text{d}x}{\text{d}\theta}=3 \cos \theta$

$\implies \text{d}x=3 \cos \theta\:\:\text{d}\theta$

Substitute everything into the original integral:

$\begin{aligned}\displaystyle \int \dfrac{1}{(9-x^2)^{\frac{3}{2}}}\:\:\text{d}x & = \int \dfrac{1}{\left(\sqrt{3^2-x^2}\right)^3}\:\:\text{d}x\\\\& = \int \dfrac{1}{\left(3 \cos \theta\right)^3}\:\:3 \cos \theta\:\:\text{d}\theta \\\\ & = \int \dfrac{1}{\left(3 \cos \theta\right)^2}\:\:\text{d}\theta \\\\ & = \int \dfrac{1}{9 \cos^2 \theta} \:\: \text{d}\theta\end{aligned}$

Take out the constant:

$\implies \displaystyle \dfrac{1}{9} \int \dfrac{1}{\cos^2 \theta}\:\:\text{d}\theta$

$\textsf{Use the trigonometric identity}: \quad\sec^2 \theta=\dfrac{1}{\cos^2 \theta}$

$\implies \displaystyle \dfrac{1}{9} \int \sec^2 \theta\:\:\text{d}\theta$

$\boxed{\begin{minipage}{5 cm}\underline{Integrating $\sec^2 kx$}\\\\$\displaystyle \int \sec^2 kx\:\text{d}x=\dfrac{1}{k} \tan kx\:\:(+\text{C})$\end{minipage}}$