All categories

2 years ago

Find the missing side or angle.

Mathematics

1 answer:

Novay_Z [31]2 years ago

8 0

Answer:

2.69 approximately 3

Step-by-step explanation:

53⁰+80⁰+A=180⁰(sum of angles on a triangle)

133⁰+A=180

A=180-133

A=47⁰

using sine rule

2/sin 47 = b/sin 80

b sin47 = 2 sin 80

b= 1.9696/0.7314

b=2.69 approximately 3

You might be interested in

I need help/ revision of my answer please!

Tresset [83]

Answer:

2 and 3 are correct, not sure if you plan on doing 1, in which case I'd help you

8 0

2 years ago

What is the perimeter of ABC with vertices at A (-11,4) B(-7,8) and C (-4,4)

Alborosie

Answer:

Step-by-step explanation:

Using the distance formula,

$AB=\sqrt{(-11-(-7))^{2}+(4-8)^{2}}=\sqrt{16+16}=4\sqrt{2}\\AC=\sqrt{(-11-(-4))^{2}+(4-4)^{2}}=7\\BC=\sqrt{(-7-(-5))^{2}+(8-4)^{2}}=\sqrt{4+16}=2\sqrt{5}\\\\$

So the perimeter is $4\sqrt{2}+7+2\sqrt{5}$

3 0

2 years ago

D(n) = -5 (1/2)n-1<br> What is the 3rd term in the sequence?

jonny [76]

Answer:

-5

Step-by-step explanation:

dₙ = -5 (1/2)n-1

the 3rd term in the sequence means n = 3

d₃ = -5(1/2)(3-1) = -5(1/2)(2) = -5

3 0

3 years ago

Read 2 more answers

A sphere is inscribed in a cube with a volume of 64 cubic inches. What is the volume of the sphere? Round your answer to the nea

erma4kov [3.2K]

Answer:

the volume of the sphere is

$33.51 in^{3}$

Step-by-step explanation:

This problem bothers on the mensuration of solid shapes, sphere and cube.

Given data

Volume of cube v = 64 cubic inches

since we are dealing with a cube the height and the radius of the sphere is same as the sides of the cube,

we know that volume of cube is expressed as

$v= l*b*h$

$v=l^{3}$

$64= l^{3}$

$l= \sqrt[3]{64}$

$l= 4 in$

also diameter d=length l

Diameter d= $4in$

Radius r = $\frac{d}{2}$ = $\frac{4}{2}$ = ${2 in}$

Height h= $4in$

we know that the volume of a sphere is given by

$v= \frac{4}{3} \pi r^{3}$

substituting into the formula we have

$v= \frac{4}{3} \ *3.142*2^{3} \\v=\frac{4*3.142*8}3} \\v= \frac{100.54}{3} \\v= 33.52in^{3}$

8 0

2 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

1 year ago