help me pls pls help

Answer:

It's D:5.3

Step-by-step explanation:

√28 =5.29

Round off therefore is 5.3

3/5kg + 760g

3/5kg = 3/5×1000

= 600g

Now

600g + 760g = 1360g

Or 1.36kg

Answered by Gauthmath must click thanks and mark brainliest

Answer:

Step-by-step explanation:

xa+yb=c

xb+yc=a

xc+ya=b

add

x(a+b+c)+y(a+b+c)=a+b+c

x+y=1 ... (1)

xac+ybc=c²

xab+yac=a²

xbc+yab=b²

add

x(ab+bc+ca)+y(ab+bc+ca)=a²+b²+c²

[tex]x+y=\frac{a^2+b^2+c^2}{ab+bc+ca} \\\frac{a^2+b^2+c^2}{ab+bc+ca} =1\\a^2+b^2+c^2=ab+bc+ca\\a^2+b^2+c^2-ab-bc-ca=0\\a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ca)=(a+b+c)(0)=0\\a^3+b^3+c^3=3abc\\\frac{a^3}{abc} +\frac{b^3}{abc} +\frac{c^3}{abc} =3\\\frac{a^2}{bc} +\frac{b^2}{ca} +\frac{c^2}{ab} =3[/tex]

Answer:

Let be a rational complex number of the form [tex]z = \frac{a + i\,b}{c + i\,d}[/tex], we proceed to show the procedure of resolution by algebraic means:

1) [tex]\frac{a + i\,b}{c + i\,d}[/tex]   Given.

2) [tex]\frac{a + i\,b}{c + i\,d} \cdot 1[/tex] Modulative property.

3) [tex]\left(\frac{a+i\,b}{c + i\,d} \right)\cdot \left(\frac{c-i\,d}{c-i\,d} \right)[/tex]   Existence of additive inverse/Definition of division.

4) [tex]\frac{(a+i\,b)\cdot (c - i\,d)}{(c+i\,d)\cdot (c - i\,d)}[/tex]   [tex]\frac{x}{y}\cdot \frac{w}{z} = \frac{x\cdot w}{y\cdot z}[/tex]  

5) [tex]\frac{a\cdot (c-i\,d) + (i\,b)\cdot (c-i\,d)}{c\cdot (c-i\,d)+(i\,d)\cdot (c-i\,d)}[/tex]  Distributive and commutative properties.

6) [tex]\frac{a\cdot c + a\cdot (-i\,d) + (i\,b)\cdot c +(i\,b) \cdot (-i\,d)}{c^{2}-c\cdot (i\,d)+(i\,d)\cdot c+(i\,d)\cdot (-i\,d)}[/tex] Distributive property.

7) [tex]\frac{a\cdot c +i\,(-a\cdot d) + i\,(b\cdot c) +(-i^{2})\cdot (b\cdot d)}{c^{2}+i\,(c\cdot d)+[-i\,(c\cdot d)] +(-i^{2})\cdot d^{2}}[/tex] Definition of power/Associative and commutative properties/[tex]x\cdot (-y) = -x\cdot y[/tex]/Definition of subtraction.

8) [tex]\frac{(a\cdot c + b\cdot d) +i\cdot (b\cdot c -a\cdot d)}{c^{2}+d^{2}}[/tex] Definition of imaginary number/[tex]x\cdot (-y) = -x\cdot y[/tex]/Definition of subtraction/Distributive, commutative, modulative and associative properties/Existence of additive inverse/Result.

Step-by-step explanation:

Let be a rational complex number of the form [tex]z = \frac{a + i\,b}{c + i\,d}[/tex], we proceed to show the procedure of resolution by algebraic means:

1) [tex]\frac{a + i\,b}{c + i\,d}[/tex]   Given.

2) [tex]\frac{a + i\,b}{c + i\,d} \cdot 1[/tex] Modulative property.

3) [tex]\left(\frac{a+i\,b}{c + i\,d} \right)\cdot \left(\frac{c-i\,d}{c-i\,d} \right)[/tex]   Existence of additive inverse/Definition of division.

4) [tex]\frac{(a+i\,b)\cdot (c - i\,d)}{(c+i\,d)\cdot (c - i\,d)}[/tex]   [tex]\frac{x}{y}\cdot \frac{w}{z} = \frac{x\cdot w}{y\cdot z}[/tex]  

5) [tex]\frac{a\cdot (c-i\,d) + (i\,b)\cdot (c-i\,d)}{c\cdot (c-i\,d)+(i\,d)\cdot (c-i\,d)}[/tex]  Distributive and commutative properties.

6) [tex]\frac{a\cdot c + a\cdot (-i\,d) + (i\,b)\cdot c +(i\,b) \cdot (-i\,d)}{c^{2}-c\cdot (i\,d)+(i\,d)\cdot c+(i\,d)\cdot (-i\,d)}[/tex] Distributive property.

7) [tex]\frac{a\cdot c +i\,(-a\cdot d) + i\,(b\cdot c) +(-i^{2})\cdot (b\cdot d)}{c^{2}+i\,(c\cdot d)+[-i\,(c\cdot d)] +(-i^{2})\cdot d^{2}}[/tex] Definition of power/Associative and commutative properties/[tex]x\cdot (-y) = -x\cdot y[/tex]/Definition of subtraction.

8) [tex]\frac{(a\cdot c + b\cdot d) +i\cdot (b\cdot c -a\cdot d)}{c^{2}+d^{2}}[/tex] Definition of imaginary number/[tex]x\cdot (-y) = -x\cdot y[/tex]/Definition of subtraction/Distributive, commutative, modulative and associative properties/Existence of additive inverse/Result.

Answer:

(2,5)

Step-by-step explanation:

4x - 2y = -2

6x + 3y = 27

Divide the first equation by 2 and the second equation by 3

2x - y = -1

2x + y = 9

Add the equations together

2x - y = -1

2x + y = 9

-------------------

4x = 8

Divide by 4

4x/4 = 8/2

x =2

2x+y = 9

2(2) +y = 9

4+u = 9

y = 9-4

y=5

(2,5)

Answer:

the number is 2.45

Step-by-step explanation:

let the original number = n

[tex]\frac{n^4}{n/2} = \frac{2n^4}{n} = 2n^3\\\\2n^3 + \frac{141}{2} = 100\\\\4n^3 + 141= 200\\\\4n^3 = 200 - 141\\\\4n^3 = 59\\\\n^3 = \frac{59}{4} \\\\n^3 = 14.75\\\\n = \sqrt[3]{14.75} \\\\n = 2.45[/tex]

Therefore, the number is 2.45

Answer:

[tex]sin {45}^{ \circ} = \frac{x}{2} \\ = > x = 2 \: sin {45}^{ \circ} \\ = > x = 2 \times \frac{1}{ \sqrt{2} } \\ = > \green{x = \sqrt{2} }[/tex]

[tex]tan {45}^{ \circ} = \frac{x}{y} = \frac{ \sqrt{2} }{y} \\ = > y = \frac{ \sqrt{2} }{tan {45}^{ \circ} } \\ = > y = \frac{ \sqrt{2} }{1} \\ = > \pink{ y = \sqrt{ 2 } }[/tex]

Answer:

$17000

Step-by-step explanation:

Given

[tex]Total = 18275[/tex]

[tex]Tax = 7.5\%[/tex]

Required

The original price

This is calculated using:

[tex]Price(1 + Tax) = Total[/tex]

Make Price the subject

[tex]Price = \frac{Total}{(1 + Tax)}[/tex]

So, we have:

[tex]Price = \frac{18275}{(1 + 7.5\%)}[/tex]

[tex]Price = \frac{18275}{1.075}[/tex]

[tex]Price = 17000[/tex]

The degree 3 polynomial with the zeros {1, 4, 2} and a leading coefficient equal to 1 is:

p(x) = x^3 -7x^2 + 14x - 8

We know that for a polynomial of degree n, with a leading coefficient "a" and the zeros {x₁, x₂, ..., xₙ} can be written as:

p(x) = a*(x - x₁)*(x - x₂)*...*(x - xₙ)

Knowing that here we have a polynomial of degree n = 3, with a leading coefficient a = 1, and the zeros {1, 4, 2}

Replacing these in the above form, we get:

p(x) = 1*(x - 1)*(x - 4)*(x - 2)

Now we can expand that to get:

p(x) = (x^2 - x - 4x + 4)*(x - 2) = (x^2 - 5x + 4)*(x - 2)

p(x) = x^3 - 5x^2 + 4x - 2x^2 + 10x - 8

p(x) = x^3 -7x^2 + 14x - 8

If you want to read more about polynomials, you can read:

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The answer:

[tex]\sqrt{9}1 /10[/tex]

Explanation to your question:

Since the sin of theta is 0.3, we can reasonably deduct that the opposite side to theta has a ration of 3 to 10 to that of the hypotenuse. Thus, the adjacent side to theta, using the pythagorean theorem, will be root91. Therefore, since the cosine of theta is the adjacent/hypotenuse, we get root 91/10

Answer:

See explanation.

General Formulas and Concepts:

Pre-Algebra

Order of Operations: BPEMDAS

Algebra I

Functions

Calculus

Differentiation

Derivative Property [Multiplied Constant]:                                                           [tex]\displaystyle \frac{d}{dx} [cf(x)] = c \cdot f'(x)[/tex]

Derivative Property [Addition/Subtraction]:                                                         [tex]\displaystyle \frac{d}{dx}[f(x) + g(x)] = \frac{d}{dx}[f(x)] + \frac{d}{dx}[g(x)][/tex]

Basic Power Rule:

Derivative Rule [Chain Rule]:                                                                                 [tex]\displaystyle \frac{d}{dx}[f(g(x))] =f'(g(x)) \cdot g'(x)[/tex]

Step-by-step explanation:

We are given the following and are trying to find the second derivative at x = 2:

[tex]\displaystyle f(2) = 2[/tex]

[tex]\displaystyle \frac{dy}{dx} = 6\sqrt{x^2 + 3y^2}[/tex]

We can differentiate the 1st derivative to obtain the 2nd derivative. Let's start by rewriting the 1st derivative:

[tex]\displaystyle \frac{dy}{dx} = 6(x^2 + 3y^2)^\big{\frac{1}{2}}[/tex]

When we differentiate this, we must follow the Chain Rule:                             [tex]\displaystyle \frac{d^2y}{dx^2} = \frac{d}{dx} \Big[ 6(x^2 + 3y^2)^\big{\frac{1}{2}} \Big] \cdot \frac{d}{dx} \Big[ (x^2 + 3y^2) \Big][/tex]

Use the Basic Power Rule:

[tex]\displaystyle \frac{d^2y}{dx^2} = 3(x^2 + 3y^2)^\big{\frac{-1}{2}} (2x + 6yy')[/tex]

We know that y' is the notation for the 1st derivative. Substitute in the 1st derivative equation:

[tex]\displaystyle \frac{d^2y}{dx^2} = 3(x^2 + 3y^2)^\big{\frac{-1}{2}} \big[ 2x + 6y(6\sqrt{x^2 + 3y^2}) \big][/tex]

Simplifying it, we have:

[tex]\displaystyle \frac{d^2y}{dx^2} = 3(x^2 + 3y^2)^\big{\frac{-1}{2}} \big[ 2x + 36y\sqrt{x^2 + 3y^2} \big][/tex]

We can rewrite the 2nd derivative using exponential rules:

[tex]\displaystyle \frac{d^2y}{dx^2} = \frac{3\big[ 2x + 36y\sqrt{x^2 + 3y^2} \big]}{\sqrt{x^2 + 3y^2}}[/tex]

To evaluate the 2nd derivative at x = 2, simply substitute in x = 2 and the value f(2) = 2 into it:

[tex]\displaystyle \frac{d^2y}{dx^2} \bigg| \limits_{x = 2} = \frac{3\big[ 2(2) + 36(2)\sqrt{2^2 + 3(2)^2} \big]}{\sqrt{2^2 + 3(2)^2}}[/tex]

When we evaluate this using order of operations, we should obtain our answer:

[tex]\displaystyle \frac{d^2y}{dx^2} \bigg| \limits_{x = 2} = 219[/tex]

Topic: AP Calculus AB/BC (Calculus I/I + II)

Unit: Differentiation

Answer:

it should be $30 so letter a

Answer:

if the diameter is 11, them the answer is 47.52275cm

Answer:

=17x²-x

Step-by-step explanation:

=x.(9x²+18x-x-2)-x.(9x²-1)

=x.(9x²+17x-2-9x²+1)

=x.(17x-1)

=17x²-x

I'm sorry but there's not enough info

Step-by-step explanation:

Answer:

The triangle with vertices A (2 , 0) , B (3 , 2) , C (5 , 1) is isosceles right Δ

The triangle with vertices A (-3 , 1) , B (-3 , 4) , C (-1 , 1) is right Δ

The triangle with vertices A (-5 , 2) , B (-4 , 4) , C (-2 , 2) is acute scalene Δ

The triangle with vertices A (-4 , 2) , B (-2 , 4) , C (-1 , 4) is obtuse scalene Δ

Answer:

Answer:

(-4,17)

Step-by-step explanation:

y = f(x)

f(-4) = (-4)^2+1 = 17

y-coordinate = 17

Answer:

D). (-4, 17)

Step-by-step explanation:

Plug in -4 for x.

[tex]f(-4)=(-4)^2}+1[/tex]

Solve.

[tex]f(-4)=16+1[/tex]

[tex]f(-4)=17[/tex]

We already know that the x-coordinate is -4. (-4, y)

f(x) stands for y, so y=17.

(-4, 17)

I hope this helps!

Standard form of a quadratic equation: ax^2 + bx + c = 0

3x - 4 = -x^2

x^2 + 3x - 4 = 0

Hope this helps!

The measure of the angle AFE or m∠AFE is 173 degrees option (B) 173 is correct if the angle AFB = 25 degrees, Angle BFC = 57 degrees, Angle CFD = 34 degrees, and Angle DFE = 57 degrees.

When two lines or rays converge at the same point, the measurement between them is called a "Angle."

We have angles shown in the picture.

Angle AFB = 25 degrees

Angle BFC = 57 degrees

Angle CFD = 34 degrees

Angle DFE = 57 degrees

Angle AFE is the sum of the angle AFB, Angle BFC, Angle CFD, and Angle DFE.

Angle AFE = Angle AFB + Angle BFC + Angle CFD + Angle DFE

Angle AFE = 25 + 57 + 34 + 57

Angle AFE = 173 degrees

Thus, the measure of the angle AFE or m∠AFE is 173 degrees option (B) 173 is correct if the angle AFB = 25 degrees, Angle BFC = 57 degrees, Angle CFD = 34 degrees, and Angle DFE = 57 degrees.

Learn more about the angle here:

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Answer:

a) .287

b) .293

Step-by-step explanation:

The answers are boxed in red in the picture.

First I found how many people only golfed. Then I did the same for the people that only bowled. Next I found how many members didn't golf or bowl.

From there I found the probabilities by dividing

a.) # of members that only golf / total # of members

b.) # of members that don't bowl or golf / total # of members

Answer:

x = .5

Step-by-step explanation:

Since we have a right triangle, we can use trig functions

tan theta = opp / adj

tan 63 = 1/x

x tan 63 = 1

x = 1/ tan 63

x=0.50952

Rounding to the nearest tenth

x = .5

Answer:

x = 14.5/a

Step-by-step explanation:

ax – 2 = 12.5

Add 2 to each side

ax – 2+2 = 12.5+2

ax = 14.5

Divide by a

ax/a = 14.5/a

x = 14.5/a

Answer:

Step-by-step explanation:

Equation of circle: (x - h)² + (y - k)² = r²   where (h,k) is the center.

Center( -4 , 4) and r = 5

(x -[-4])² + (y - 4)²= 5²

(x + 4)² + (y-4)² = 25

x²  + 2*4*x +4²  + y²  - 2*y*4 + 4²  = 25

x²  +8x + 16 + y²  - 8y + 16 = 25

x²  + 8x + y²  - 8y + 16 + 16 -25 = 0

x²  + 8x + y²  - 8y +7 = 0

We have that the an equation of a circle given the center (-4,4) and radius r=5  is mathematically given as

(x-4)^2+(y-4)^2=5^2

Question Parameters:

Given the center (-4,4) and radius r=5

Generally the equation for the Equation of a circle   is mathematically given as

(x-x')^2+(y-y')^2=r^2

Therefore, The resultant equation will be

(x-x')^2+(y-y')^2=r^2

(x-4)^2+(y-4)^2=5^2

Hence,an equation of a circle given the center (-4,4) and radius r=5 is

(x-4)^2+(y-4)^2=5^2

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Answer: y= x^2/16-6x/16+41/16

Step-by-step explanation:

The equation of a parabola will be; y = x^2/16 - 6x/16 + 41/16

If a quadratic equation is written in the form

y=a(x-h)^2 + k

then it is called to be in vertex form. It is called so because when you plot this equation's graph, you will see vertex point(peak point) is on (h,k)

Otherwise, we had to use calculus to get critical points, then second derivative of functions to find the character of critical points as minima or maxima or saddle etc to get the location of vertex point.

This point (h,k) is called the vertex of the parabola that quadratic equation represents.

WE need to find the equation of a parabola that has a vertical axis, passes through the point (–1, 3), and has its vertex at (3, 2)

Thus, the equation of a parabola will be;

y = x^2/16 - 6x/16 + 41/16

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Convert to base 10:

10 0110₂ = 2⁵ + 2² + 2¹ = 38

Convert to base 3:

38 = 27 + 11 = 27 + 9 + 2 = 3³ + 3² + 2×3⁰ = 1102₃

Answer: No solutions

Step-by-step explanation:

[tex]\large \bf \boldsymbol{ \boxed{\sqrt{a}\cdot \sqrt{b}=\sqrt{a\cdot b} }} \\\\\\ \sqrt{25x+75} +3\sqrt{x-2} =2+4\sqrt{x-3} +\sqrt{9x-18} \\\\ \sqrt{25} \cdot \sqrt{x+3}+3\sqrt{x-2} =2+4\sqrt{x-3} +\sqrt{9}\cdot \sqrt{x-2} \\\\5\sqrt{x+3} +3\sqrt{x-2} \!\!\!\!\!\!\!\!\!\!\bigg{/} \ \ =2 +4\sqrt{x-3} +3\sqrt{x-2} \!\!\!\!\!\!\!\!\!\!\bigg{/} \\\\(5\sqrt{x+3})^2 =(2+4\sqrt{x-3} )^2 \\\\ \ \ \ let \ \ t=x+3 \ \ ; \ \ \ t-6=x-3 \\\\ \big(5\sqrt{t} \ \big)^2=(2+\sqrt{t-6} )^2 \\\\[/tex]                                   [tex]\large \boldsymbol{} \bf 25t=4+16\sqrt{t-6} +16(t-6) \\\\(9t+92)^2=(16\sqrt{t-6} )^2 \\\\81t^2+1656t+8464=256(t-6)\\\\81t^2+1400t+10000=0 \\\\ D=1400^2-324000=-128000=> \\\\D<0 \ \ no \ \ solutions[/tex]