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3 years ago

Use the side-splitter theorem to solve for x in the triangle below.

Mathematics

1 answer:

fiasKO [112]3 years ago

8 0

Answer:

$x = 7$

Step-by-step explanation:

According to the side-splitter theorem, the two sides of the triangle is divided proportionally if the line that intersects both sides is parallel to the third side of the ∆.

Therefore, based on this theorem, we would have the following:

$\frac{x}{21} = \frac{8}{24}$

Solve for x

$\frac{x}{21} = \frac{1}{3}$

Multiply both sides by 21

$\frac{x}{21}*21 = \frac{1}{3}*21$

$x = \frac{21}{3}$

$x = 7$

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A right cylinder has a radius of 3 and a height of 12. What is its surface area?

shepuryov [24]

Answer:

Surface Area = 283

Step-by-step explanation:

Radius = 3

Height = 12

Formula for suface area is SA= 2πrh+2πr^2

let "r" be for radius

let "h" be for height

SA=2*π*3*12+2*π*3^2

SA=282.74

round to the nearest ones spot and that will make the surface area = 283

8 0

4 years ago

Jakob is asking to simplify the expression -3a + 4b + 5a + (-7b).

garik1379 [7]

Answer:

B. Commutative

Step-by-step explanation:

He can do that because it's a sum

3 0

3 years ago

Read 2 more answers

Write an algebraic expression for the following word phrase.<br> The quotient of 62 and x

madreJ [45]

Answer:

62/x

Step-by-step explanation:

5 0

3 years ago

Prove the following statement.

jolli1 [7]

<h2>Step-by-step explanation:</h2>

As per the question,

Let a be any positive integer and b = 4.

According to Euclid division lemma , a = 4q + r

where 0 ≤ r < b.

Thus,

r = 0, 1, 2, 3

Since, a is an odd integer, and

The only valid value of r = 1 and 3

So a = 4q + 1 or 4q + 3

<u>Case 1 :-</u> When a = 4q + 1

On squaring both sides, we get

a² = (4q + 1)²

= 16q² + 8q + 1

= 8(2q² + q) + 1

= 8m + 1 , where m = 2q² + q

<u>Case 2 :-</u> when a = 4q + 3

On squaring both sides, we get

a² = (4q + 3)²

= 16q² + 24q + 9

= 8 (2q² + 3q + 1) + 1

= 8m +1, where m = 2q² + 3q +1

Now,

<u>We can see that at every odd values of r, square of a is in the form of 8m +1.</u>

Also we know, a = 4q +1 and 4q +3 are not divisible by 2 means these all numbers are odd numbers.

Hence , it is clear that square of an odd positive is in form of 8m +1

3 0

3 years ago

Evaluate the interval (Calculus 2)

Darya [45]

Answer:

$2 \tan (6x)+2 \sec (6x)+\text{C}$

Step-by-step explanation:

<u>Fundamental Theorem of Calculus</u>

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

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:

$\displaystyle \int \dfrac{12}{1-\sin (6x)}\:\:\text{d}x$

$\boxed{\begin{minipage}{5 cm}\underline{Terms multiplied by constants}\\\\$\displaystyle \int a\:\text{f}(x)\:\text{d}x=a \int \text{f}(x) \:\text{d}x$\end{minipage}}$

If the terms are multiplied by constants, take them outside the integral:

$\implies 12\displaystyle \int \dfrac{1}{1-\sin (6x)}\:\:\text{d}x$

Multiply by the conjugate of 1 - sin(6x) :

$\implies 12\displaystyle \int \dfrac{1}{1-\sin (6x)} \cdot \dfrac{1+\sin(6x)}{1+\sin(6x)}\:\:\text{d}x$

$\implies 12\displaystyle \int \dfrac{1+\sin(6x)}{1-\sin^2(6x)} \:\:\text{d}x$

$\textsf{Use the identity} \quad \sin^2 x+ \cos^2 x=1:$

$\implies \sin^2 (6x) + \cos^2 (6x)=1$

$\implies \cos^2 (6x)=1- \sin^2 (6x)$

$\implies 12\displaystyle \int \dfrac{1+\sin(6x)}{\cos^2(6x)} \:\:\text{d}x$

Expand:

$\implies 12\displaystyle \int \dfrac{1}{\cos^2(6x)}+\dfrac{\sin(6x)}{\cos^2(6x)} \:\:\text{d}x$

$\textsf{Use the identities }\:\: \sec \theta=\dfrac{1}{\cos \theta} \textsf{ and } \tan\theta=\dfrac{\sin \theta}{\cos \theta}:$

$\implies 12\displaystyle \int \sec^2(6x)+\dfrac{\tan(6x)}{\cos(6x)} \:\:\text{d}x$

$\implies 12\displaystyle \int \sec^2(6x)+\tan(6x)\sec(6x) \:\:\text{d}x$

$\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})$\end{minipage}}$

$\boxed{\begin{minipage}{6 cm}\underline{Integrating $ \sec kx \tan kx$}\\\\$\displaystyle \int \sec kx \tan kx\:\text{d}x= \dfrac{1}{k}\sec kx\:\:(+\text{C})$\end{minipage}}$

$\implies 12 \left[\dfrac{1}{6} \tan (6x)+\dfrac{1}{6} \sec (6x) \right]+\text{C}$

Simplify:

$\implies \dfrac{12}{6} \tan (6x)+\dfrac{12}{6} \sec (6x)+\text{C}$

$\implies 2 \tan (6x)+2 \sec (6x)+\text{C}$

Learn more about indefinite integration here:

brainly.com/question/27805589

brainly.com/question/28155016

3 0

2 years ago