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3 years ago

PLEASE HELP QUICKLY,15 POINTS AND WILL

Mathematics

1 answer:

zepelin [54]3 years ago

7 0

Answer:

x=26

Step-by-step explanation:

The sum of the angles of a triangle add to 180 degrees

(2x-5)° +(4x-1)° + 30° = 180

Combine like terms

6x +24 =180

Subtract 24 from each side

6x+24-24 =180-24

6x = 156

Divide by 6

6x/6 =156/6

x =26

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What is the answer (×+3) (×+7)

kap26 [50]

$x^{2} +10x+21$ . Use FOIL to do this.

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6 0

4 years ago

Read 2 more answers

When a cylindrical tank is filled with water at a rate of 22 cubic meters per hour, the level of water in the tank rises at a ra

Sergeu [11.5K]

Answer:

$r=\sqrt{10}\text{ m}$

Step-by-step explanation:

We have been given that when a cylindrical tank is filled with water at a rate of 22 cubic meters per hour, the level of water in the tank rises at a rate of 0.7 meters per hour. We are asked to find the approximate radius of tank in meters.

We will use volume of cylinder formula to solve our given problem as:

$V=\pi r^2h$ , where,

r = Radius,

h = Height of cylinder.

Since the level of water in the tank rises at a rate of 0.7 meters per hour, so height of cylinder would be $h = 0.7$ meters at $V=22\text{ m}^3$ .

Upon substituting these values in above formula, we will get:

$22\text{ m}^3=\frac{22}{7}\cdot r^2(0.7\text{ m})$

$22\text{ m}^3=\frac{22}{10}\cdot r^2\text{ m}$

$\frac{10}{22}\cdot\frac{22\text{ m}^3}{\text{m}}=r^2$

$r^2=\frac{10}{22}\cdot\frac{22\text{ m}^3}{\text{m}}$

$r^2=10\text{ m}^2$

Now, we will take positive square root of both sides as radius cannot be negative.

$\sqrt{r^2}=\sqrt{10\text{ m}^2}$

$r=\sqrt{10}\text{ m}$

Therefore, radius of tank would be approximately square root of 10 m.

5 0

3 years ago

PLEASE HELP ME AND MY FRIEND NEED THIS QUESTION ANSWERED PLEASE HELP

Bas_tet [7]

Answer:

$V_{cone} =65.94\: cm^3$

Step-by-step explanation:

Height of cone (h) = 7 cm

Base radius (r) = 3 cm

Formula for Volume of cone is given as:

$V_{cone} = \frac{1}{3}\pi r^2h$

Plugging the values of h and r in the above formula, we find:

$V_{cone} = \frac{1}{3}(3.14) (3)^2(7)$

$\implies V_{cone} = \frac{1}{\cancel 3}(3.14) (\cancel 9)(7)$

$\implies V_{cone} = (3.14) (3)(7)$

$\implies V_{cone} =65.94\: cm^3$

7 0

2 years ago

In circle H with m \angle GHJ= 104m∠GHJ=104 and GH=15GH=15 units, find the length of arc GJ. Round to the nearest hundredth.

AlladinOne [14]

Answer: GJ= 27.23

i dont know how to explain this lol sorry but that answer is right

6 0

3 years ago

Given two points P(sinθ+2, tanθ-2) and Q(4sin²θ+4sinθcosθ+2acosθ, 3sinθ-2cosθ+a). Find constant "a" and the corresponding value

vodomira [7]

Answer:

$\rm\displaystyle \displaystyle \displaystyle θ= {60}^{ \circ} , {300}^{ \circ}$

$\rm \displaystyle a = - \frac{ \sqrt{3} }{2} - 1, \frac{\sqrt{3}}{2} - 1$

Step-by-step explanation:

we are given two coincident points

$\displaystyle P( \sin(θ)+2, \tan(θ)-2) \: \text{and } \\ \displaystyle Q(4 \sin ^{2} (θ)+4 \sin(θ) \cos(θ)+2a \cos(θ), 3 \sin(θ)-2 \cos(θ)+a)$

since they are coincident points

$\rm \displaystyle P( \sin(θ)+2, \tan(θ)-2) = \displaystyle Q(4 \sin ^{2} (θ)+4 \sin(θ )\cos(θ)+2a \cos(θ), 3 \sin(θ)-2 \cos(θ)+a)$

By order pair we obtain:

$\begin{cases} \rm\displaystyle \displaystyle 4 \sin ^{2} (θ)+4 \sin(θ) \cos(θ)+2a \cos(θ) = \sin( \theta) + 2 \\ \\ \displaystyle 3 \sin( \theta) - 2 \cos( \theta) + a = \tan( \theta) - 2\end{cases}$

now we end up with a simultaneous equation as we have two variables

to figure out the simultaneous equation we can consider using substitution method

to do so, make a the subject of the equation.therefore from the second equation we acquire:

$\begin{cases} \rm\displaystyle \displaystyle 4 \sin ^{2} (θ)+4 \sinθ \cos(θ)+2a \cos(θ )= \sin( \theta) + 2 \\ \\ \boxed{\displaystyle a = \tan( \theta) - 2 - 3 \sin( \theta) + 2 \cos( \theta) } \end{cases}$

now substitute:

$\rm\displaystyle \displaystyle 4 \sin ^{2} (θ)+4 \sin(θ) \cos(θ)+2 \cos(θ) \{\tan( \theta) - 2 - 3 \sin( \theta) + 2 \cos( \theta) \}= \sin( \theta) + 2$

distribute:

$\rm\displaystyle \displaystyle 4 \sin ^{2}( θ)+4 \sin(θ) \cos(θ)+2 \sin(θ ) - 4\cos( \theta) - 6 \sin( \theta) \cos( \theta) + 4 \cos ^{2} ( \theta) = \sin( \theta) + 2$

collect like terms:

$\rm\displaystyle \displaystyle 4 \sin ^{2}( θ) - 2\sin(θ) \cos(θ)+2 \sin(θ ) - 4\cos( \theta) + 4 \cos ^{2} ( \theta) = \sin( \theta) + 2$

rearrange:

$\rm\displaystyle \displaystyle 4 \sin ^{2}( θ) + 4 \cos ^{2} ( \theta) - 2\sin(θ) \cos(θ)+2 \sin(θ ) - 4\cos( \theta) + = \sin( \theta) + 2$

by Pythagorean theorem we obtain:

$\rm\displaystyle \displaystyle 4 - 2\sin(θ) \cos(θ)+2 \sin(θ ) - 4\cos( \theta) = \sin( \theta) + 2$

cancel 4 from both sides:

$\rm\displaystyle \displaystyle - 2\sin(θ) \cos(θ)+2 \sin(θ ) - 4\cos( \theta) = \sin( \theta) - 2$

move right hand side expression to left hand side and change its sign:

$\rm\displaystyle \displaystyle - 2\sin(θ) \cos(θ)+\sin(θ ) - 4\cos( \theta) + 2 = 0$

factor out sin:

$\rm\displaystyle \displaystyle \sin (θ) (- 2 \cos(θ)+1) - 4\cos( \theta) + 2 = 0$

factor out 2:

$\rm\displaystyle \displaystyle \sin (θ) (- 2 \cos(θ)+1) + 2(- 2\cos( \theta) + 1 ) = 0$

group:

$\rm\displaystyle \displaystyle ( \sin (θ) + 2)(- 2 \cos(θ)+1) = 0$

factor out -1:

$\rm\displaystyle \displaystyle - ( \sin (θ) + 2)(2 \cos(θ) - 1) = 0$

divide both sides by -1:

$\rm\displaystyle \displaystyle ( \sin (θ) + 2)(2 \cos(θ) - 1) = 0$

by Zero product property we acquire:

$\begin{cases}\rm\displaystyle \displaystyle \sin (θ) + 2 = 0 \\ \displaystyle2 \cos(θ) - 1= 0 \end{cases}$

cancel 2 from the first equation and add 1 to the second equation since -1≤sinθ≤1 the first equation is false for any value of theta

$\begin{cases}\rm\displaystyle \displaystyle \sin (θ) \neq - 2 \\ \displaystyle2 \cos(θ) = 1\end{cases}$

divide both sides by 2:

$\rm\displaystyle \displaystyle \displaystyle \cos(θ) = \frac{1}{2}$

by unit circle we get:

$\rm\displaystyle \displaystyle \displaystyle θ= {60}^{ \circ} , {300}^{ \circ}$

so when θ is 60° a is:

$\rm \displaystyle a = \tan( {60}^{ \circ} ) - 2 - 3 \sin( {60}^{ \circ} ) + 2 \cos( {60}^{ \circ} )$

recall unit circle:

$\rm \displaystyle a = \sqrt{3} - 2 - \frac{ 3\sqrt{3} }{2} + 2 \cdot \frac{1}{2}$

simplify which yields:

$\rm \displaystyle a = - \frac{ \sqrt{3} }{2} - 1$

when θ is 300°

$\rm \displaystyle a = \tan( {300}^{ \circ} ) - 2 - 3 \sin( {300}^{ \circ} ) + 2 \cos( {300}^{ \circ} )$

remember unit circle:

$\rm \displaystyle a = - \sqrt{3} - 2 + \frac{3\sqrt{ 3} }{2} + 2 \cdot \frac{1}{2}$

simplify which yields:

$\rm \displaystyle a = \frac{ \sqrt{3} }{2} - 1$

and we are done!

disclaimer: also refer the attachment I did it first before answering the question

5 0

3 years ago