PLS HELP MEE

1st equation: a^2 + b^2 = c^2
2nd equation: a^2 + b^2 = n^2

Since the equations for both triangles have a^2 + b^2, you can think of the two equations for c^2 and n^2 as a system of equations. Substitute what a^2 + b^2 equals in the first equation for a^2 + b^2 in the second equation. After you substitute, what equation do
you get?

Answer:

4 cm

Step-by-step explanation:

V = πr²h

or, 36π = π(3)²×h

or, 36 = 9h

or, 4 = h

So the height is 4 cm

Answered by GAUTHMATH

9514 1404 393

Answer:

  u = 7√2; v = 7

Step-by-step explanation:

The ratios of side lengths in an isosceles right triangle are ...

  1 : 1 : √2 = 7 : v : u

Then v = 7 and u = 7√2.

Answer:

B

Step-by-step explanation:

option B is not similar.

the ratio of each side isn't same

Answer:

See Below.

Step-by-step explanation:

We want to verify the equation:

[tex]\displaystyle \frac{1}{1+\sin\theta} = \sec^2\theta - \sec\theta \tan\theta[/tex]

To start, we can multiply the fraction by (1 - sin(θ)). This yields:

[tex]\displaystyle \frac{1}{1+\sin\theta}\left(\frac{1-\sin\theta}{1-\sin\theta}\right) = \sec^2\theta - \sec\theta \tan\theta[/tex]

Simplify. The denominator uses the difference of two squares pattern:

[tex]\displaystyle \frac{1-\sin\theta}{\underbrace{1-\sin^2\theta}_{(a+b)(a-b)=a^2-b^2}} = \sec^2\theta - \sec\theta \tan\theta[/tex]

Recall that sin²(θ) + cos²(θ) = 1. Hence, cos²(θ) = 1 - sin²(θ). Substitute:

[tex]\displaystyle \displaystyle \frac{1-\sin\theta}{\cos^2\theta} = \sec^2\theta - \sec\theta \tan\theta[/tex]

Split into two separate fractions:

[tex]\displaystyle \frac{1}{\cos^2\theta} -\frac{\sin\theta}{\cos^2\theta} = \sec^2\theta - \sec\theta\tan\theta[/tex]

Rewrite the two fractions:

[tex]\displaystyle \left(\frac{1}{\cos\theta}\right)^2-\frac{\sin\theta}{\cos\theta}\cdot \frac{1}{\cos\theta}=\sec^2\theta - \sec\theta \tan\theta[/tex]

By definition, 1 / cos(θ) = sec(θ) and sin(θ)/cos(θ) = tan(θ). Hence:

[tex]\displaystyle \sec^2\theta - \sec\theta\tan\theta \stackrel{\checkmark}{=} \sec^2\theta - \sec\theta\tan\theta[/tex]

Hence verified.

Answer:

Slope: 0

x-intercept: there isnt one

y-intercept: 2

Step-by-step explanation:

the easiest way to look at this is to put it in the form y=mx+b where m is the slope and b is the y intercept.  when we just think about what y=2 would look like, we can imagine a straight horizontal line at y=2.  No matter what x value you choose, y will always equal 2.  We know the slope (m) of any horizontal line is zero because there is no rise and zero divided by anything is going to be zero.  we also know if y is always equal to 2 the y intercept will be 2.  this would give us y=0x+2.  to find the x intercept we just need to set y equal to zero in this equation.  this gives us 0=0x+2 or 0=2 which can never be true, therefore there will be no x intercept.

*insert shrek script*

...................................

Answer: 27

Step-by-step explanation:

[tex]Mean=\frac{16 + 25 + 23 + 12 + 17 + x}{6}=20 \\16 + 25 + 23 + 12 + 17 + x=20(6)\\93+x=120\\x=120-93=27[/tex]

Answer:

27 years old

Step-by-step explanation:

There are originally 5 people in the room.

Their ages are: 16, 25, 23, 12, 17

The sum of their ages is: 93

One person enters the room. That person's age is unknown, so we call it x.

The sum of the ages of the 6 people is:

x + 93

There are 6 people in the room now, so the mean age is (total age divided by number of people):

(x + 93)/6

We are told the mean age is 20.

(x + 93)/6 = 20

x + 93 = 120

x = 27

Answer: 27 years old

Answer:

75 - 35y / 3 + 2y = -4

( 75 - 35y + 6y ) / 3 = -4 [ Taking LCM ]

( 75 - 29y ) / 3 = -4

75 - 29y = -12

29y = 75+12

29y = 87

y = 3

hope that helps ✌

Answer:

x = 6

Step-by-step explanation:

Since GH bisects LM at K , then

LK = [tex]\frac{1}{2}[/tex] LM , that is

5x + 2 = [tex]\frac{1}{2}[/tex] × 64 = 32 ( subtract 2 from both sides )

5x = 30 ( divide both sides by 5 )

x = 6

The value of x is 6.

Here,

GH bisects LM at K.

LK = 5x+2 and LM = 64

We have to find the value of x.

What is Bisection in line?

Bisecting a line is cutting a line exactly in half. It may also be referred to as constructing a perpendicular bisector as the line you are drawing will be at a right angle to the original line.

Now,

GH bisects LM at K.

LK + KM = LM

2 LK = LM

2 ( 5x + 2 ) = 64

5x + 2 = 32

5x = 30

x = 6

Hence, The value of x is 6.

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Answer:

B

Step-by-step explanation:

Answer:23

Step-by-step explanation:

Answer: 0 and 4

Step-by-step explanation:

Answer:

number of pennies = 54

Step-by-step explanation:

(4,60) ; (6, 65)

Find the slope

Slope = [tex]\frac{y_{2}-y_{1}}{x_{2}-x_{1}}[/tex]

          [tex]=\frac{65-60}{6-4}\\\\=\frac{5}{2}[/tex]

m = 5/2 and (4,60)

Equation of the line: y - y1 = m (x -x1)

[tex]y - 60 = \frac{5}{2}(x -4)\\\\y - 60 = \frac{5}{2}x-\frac{5}{2}*4\\\\y- 60 = \frac{5}{2}x-10\\\\y =\frac{5}{2}x-10+60\\\\y=\frac{5}{2}x + 50[/tex]

x-----> determine the number of pennies and y determines the mass of pennies.

Here mass (y -value) is given as 185. We have find the corresponding x-value. So plugin  y  = 185 in the above equation,

[tex]185 = \frac{5}{2}x+50\\\\185-50=\frac{5}{2}x\\\\135=\frac{5}{2}x\\\\135*\frac{2}{5}=x\\\\27*2=x[/tex]

x = 54

Answer:

-21

Step-by-step explanation:

comments

Answer:

x=-1

Step-by-step explanation:

f (x)= -x + 8

Let f(x) = 9

9 = -x+8

Subtract 8 from each side

9-8 = -x+8-8

1 = -x

Multiply each side by -1

-1 = x

Answer: $420

Step-by-step explanation:

40% of 1st year commission

1st year commission is $1050

40% x 1050 = .40 x 1050 = 420

$420

Hi there!

[tex]\large\boxed{(2, 8)}[/tex]

Solve by setting the two equations equal to each other:

-2x² + 4x + 8 = -4x + 16

Move all terms to one side:

-2x² + 4x + 4x + 8 - 16 = 0

Simplify:

-2x² + 8x - 8 = 0

Factor out a negative 2 from each term:

-2(x² - 4x + 4) = 0

Factor:

-2(x - 2)² = 0

Set the factor equal to 0 to solve:

(x - 2)² = 0

x - 2 = 0

x = 2

Substitute this value of x into an equation to find the shared y value:

y = -4(2) + 16 = 8

I'm assuming the integral is

[tex]\displaystyle \int \frac{\mathrm dx}{x+\sqrt{x^2-1}}[/tex]

Rationalize the denominator:

[tex]\dfrac1{x+\sqrt{x^2-1}} \times \dfrac{x-\sqrt{x^2-1}}{x-\sqrt{x^2-1}} = \dfrac{x-\sqrt{x^2-1}}{x^2-\left(\sqrt{x^2-1}\right)^2} = x-\sqrt{x^2-1}[/tex]

Then the integral is

[tex]\displaystyle \int\left(x-\sqrt{x^2-1}\right)\,\mathrm dx = \dfrac12x^2 - \int\sqrt{x^2-1}\,\mathrm dx[/tex]

For the remaining integral, substitute x = sec(t ) and dx = sec(t ) tan(t ) dt. Then over an appropriate domain, we have

[tex]\displaystyle\int\sqrt{x^2-1}\,\mathrm dx = \int\sec(t)\tan(t)\sqrt{\sec^2(t)-1}\,\mathrm dt = \int\sec(t)\tan^2(t)\,\mathrm dt[/tex]

Integrate by parts, taking

u = tan(t )   ==>   du = sec²(t ) dt

dv = sec(t ) tan(t ) dt   ==>   v = sec(t )

Then

[tex]\displaystyle\int\sec(t)\tan^2(t)\,\mathrm dt = \sec(t)\tan(t) - \int\sec^3(t)\,\mathrm dt[/tex]

Now for *this* remaining integral, integrate by parts again, taking

u = sec(t )   ==>   du = sec(t ) tan(t ) dt

dv = sec²(t ) dt   ==>   v = tan(t )

so that

[tex]\displaystyle\int\sec^3(t)\,\mathrm dt = \sec(t)\tan(t) - \int\sec(t)\tan^2(t)\,\mathrm dt \\\\ \int\sec^3(t)\,\mathrm dt = \sec(t)\tan(t) - \int\sec(t)(\sec^2(t)-1)\,\mathrm dt \\\\ \int\sec^3(t)\,\mathrm dt = \sec(t)\tan(t) - \int\sec^3(t)\,\mathrm dt + \int\sec(t)\,\mathrm dt \\\\ \int\sec^3(t)\,\mathrm dt = \frac12\sec(t)\tan(t)+\frac12\int\sec(t)\,\mathrm dt \\\\ \int\sec^3(t)\,\mathrm dt = \frac12\sec(t)\tan(t) + \frac12\ln\left|\sec(t)+\tan(t)\right| + C[/tex]

To summarize, if I denotes the original integral, we have

[tex]\displaystyle I = \frac12x^2 - \int\sqrt{x^2-1}\,\mathrm dx \\\\ I = \frac12x^2 - \int\sec(t)\tan^2(t)\,\mathrm dt \\\\ I = \frac12x^2 - \sec(t)\tan(t) + \int\sec^3(t)\,\mathrm dt \\\\ I = \frac12x^2 - \sec(t)\tan(t) + \frac12\sec(t)\tan(t) + \frac12\ln\left|\sec(t)+\tan(t)\right| + C \\\\ I = \frac12x^2 - \frac12\sec(t)\tan(t) + \frac12\ln\left|\sec(t)+\tan(t)\right| + C[/tex]

Putting everything back in terms of x, we have

sec(t ) = x

tan(t ) = √(x ² - 1)

so that

[tex]\displaystyle I = \boxed{\frac12x^2 - \frac12x\sqrt{x^2-1}+\frac12\ln\left|x+\sqrt{x^2-1}\right|+C}[/tex]

Hi there!  

»»————-　★　————-««

I believe your answer is:  

D) 8

»»————-　★　————-««  

Here’s why:  

⸻⸻⸻⸻

[tex]\boxed{\text{Solving for 'b'...}}\\\\\sqrt[3]{b} =2\\--------------\\\rightarrow ( \sqrt[3]{b} )^3= 2^3\\\\\boxed{b = 8}[/tex]

⸻⸻⸻⸻

»»————-　★　————-««  

Hope this helps you. I apologize if it’s incorrect.  

Answer:

18.

Domain: X E \{-3,2}

Y-Intercept: 1/2

X-Intercept: 3

19.

Domain: (I think it's none)

Y-Intercept: -2

X-Intercept: 2

20.

Domain: X E R\ {-3,1}

y-INTERCEPT: 2/3

x-INTERCEPT: -1

Step-by-step explanation:

Answer:

let,

x=ky

y=39 when x=7 so,

7=39k

or, k=7/39

now, when x=22

22=7y/39

or, y=39×22/7

or, y=858/7

or, y=122.57 (rounded to the nearest hundredth)

Answer:

D) 4x +3y = -6

Step-by-step explanation:

paralell lines so m1 and m2 are equal

m = (3 +1 )/ (0 - 3 )

m = -4/ 3

y -2 = -4/3 (x +3)

y =-4x/3 -2

3y = -4x -6

4x +3y = -6

Answer:

a) The horizontal asymptote is y = 0

The y-intercept is (0, 9)

b) The horizontal asymptote is y = 0

The y-intercept is (0, 5)

c) The horizontal asymptote is y = 3

The y-intercept is (0, 4)

d) The horizontal asymptote is y = 3

The y-intercept is (0, 4)

e) The horizontal asymptote is y = -1

The y-intercept is (0, 7)

The x-intercept is (-3, 0)

f) The asymptote is y = 2

The y-intercept is (0, 6)

Step-by-step explanation:

a) f(x) = [tex]3^{x + 2}[/tex]

The asymptote is given as x → -∞, f(x) = [tex]3^{x + 2}[/tex] → 0

∴ The horizontal asymptote is f(x) = y = 0

The y-intercept is given when x = 0, we get;

f(x) = [tex]3^{0 + 2}[/tex] = 9

The y-intercept is f(x) = (0, 9)

b) f(x) = [tex]5^{1 - x}[/tex]

The asymptote is fx) = 0 as x → ∞

The asymptote is y = 0

Similar to question (1) above, the y-intercept is f(x) = [tex]5^{1 - 0}[/tex] = 5

The y-intercept is (0, 5)

c) f(x) = 3ˣ + 3

The asymptote is 3ˣ → 0 and f(x) → 3 as x → ∞

The asymptote is y = 3

The y-intercept is f(x) = 3⁰ + 3= 4

The y-intercept is (0, 4)

d) f(x) = 6⁻ˣ + 3

The asymptote is 6⁻ˣ → 0 and f(x) → 3 as x → ∞

The horizontal asymptote is y = 3

The y-intercept is f(x) = 6⁻⁰ + 3 = 4

The y-intercept is (0, 4)

e) f(x) = [tex]2^{x + 3}[/tex] - 1

The asymptote is [tex]2^{x + 3}[/tex]  → 0 and f(x) → -1 as x → -∞

The horizontal asymptote is y = -1

The y-intercept is f(x) =  [tex]2^{0 + 3}[/tex] - 1 = 7

The y-intercept is (0, 7)

When f(x) = 0, [tex]2^{x + 3}[/tex] - 1 = 0

[tex]2^{x + 3}[/tex] = 1

x + 3 = 0, x = -3

The x-intercept is (-3, 0)

f) [tex]f(x) = \left (\dfrac{1}{2} \right)^{x - 2} + 2[/tex]

The asymptote is [tex]\left (\dfrac{1}{2} \right)^{x - 2}[/tex] → 0 and f(x) → 2 as x → ∞

The asymptote is y = 2

The y-intercept is f(x) = [tex]f(0) = \left (\dfrac{1}{2} \right)^{0 - 2} + 2 = 6[/tex]

The y-intercept is (0, 6)

Answer:yes

Step-by-step explanation:

Yes

Answer:

i dont know the total sum of marbles

Step-by-step explanation:

A fraction is written in the form of a numerator and a denominator

where the denominator is greater than the numerator.

1/4 of 80 is 20.

The number of white marbles is 20.

A fraction is written in the form of a numerator and a denominator where the denominator is greater than the numerator.

Example: 1/2, 1/3 is a fraction.

We have,

Total marbles = x

The number of white marbles = 1/4x

The number of blue marbles = 3/8x

The number of yellow marbles = 30

Total marbles.

x = (1/4)x + (3/8)x + 30

x - (1/4)x - (3/8)x = 30

(8x - 2x - 3x) / 8 = 30

3x/8 = 30

3x = 30 x 8

3x = 240

x = 80

The number of white marbles.

= (1/4)x

= (1/4) x 80

= 20 marbles.

Thus,

The number of white marbles is 20.

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Answer:

x=1

Step-by-step explanation:

3х - 6 = 6x - 9

Subtract 3x from each side

3х-3x - 6 = 6x-3x - 9

-6 = 3x-9

Add 9 to each side

-6+9 = 3x-9+9

3 =3x

Divide by 3

3/3 = 3x/3

1 =x

Answer:

3x-6=6x-9

3x-6x=-9+6

-3x=-3

-3x÷-3=-3÷-3

x=1

Answer:

With the formula of lwh/3 the answer is 305.37

Step-by-step explanation:

l×w×h /3 =305.37

Answer:

916.11 ft³

Step-by-step explanation:

(9×6)×7.83/2×13/3

= 211.41×13/3

= 916.11 ft³

Answered by GAUTHMATH

Answer:

∠8 and ∠16.

Step-by-step explanation:

Given the diagram we have to find the angles which are corresponding angles with ∠12. Those angles which are on the same side of traversal called corresponding angles. By definition of corresponding angles the angles which are corresponding to ∠12 are ∠8 and ∠16.