What is the coefficient of x6 y4 in the expansion of (x + y)10?

Answer:

-4

Step-by-step explanation:

6 − 4x − 8 + 2

The variable is x

The coefficient is the number in front of the variable ( it will include the sign)

-4 is the coefficient

Answer:

-4

Step-by-step explanation:

Answer:

2(x + 9)(x + 1)

Step-by-step explanation:

-2(x^2 + 10x + 9)

→ The main quadratic equation is equal to ax^2 + bx + c.

2(x + 9)(x + 1)

The expression cannot be factored any further.

[tex]35pq^2=5\cdot7\cdot p\cdot q^2\\-14pq^2=-1 \cdot 2\cdot 7\cdot p \cdot q^2\\-21p^2q^3=-1\cdot 3\cdot 7\cdot p^2 \cdot q^3\\\\\text{hcf}(35pq^2,-14pq^2,-21p^2q^3)=7\cdot p \cdot q^2=7pq^2[/tex]

Answer:

33

Step-by-step explanation:

For every 5 squares, two are red and three are green.  So 3/5 of the squares are green.

3/5x55=33

Answer:

V(t) = 14,000(0.89)^t  

Step-by-step explanation:

Present value of the Dodge Challenger = $14,000

Present percentage value = 100%

Depreciation value = 11%

Number of years = t

Future value = V(t)

V(t) = Present value of the Dodge Challenger(Present percentage value - Depreciation value)^t

= 14,000(100% - 11%)^t

= 14,000(89%)^t

= 14,000(0.89)^t

V(t) = 14,000(0.89)^t

Answer:

p=5/8

Step-by-step explanation:

you first have to work out the brackets

5p+7=3(4-p)

5p+7=12-3p

then you have to group the like terms

5p+3p=12-7

8p=5

8p/8=5/8

p=5/8

you can also live it as a decimal

I hope this helps

Answer: p = 5/8

Explanation:

⇒5p + 7 = 3(4 -p)

⇒5p + 7 = 12 - 3p

⇒5p + 3p = 12 - 7

⇒8p = 5

⇒p = 5/8

Must click thanks and mark brainliest

Hi, I'm happy to help!

First, we need to solve for x. We know that ABD is 78 degrees. (5x+3), and (5x-5) combined are also 78 degrees. So, we combine the like terms of these to get a new expression.

(5x+3) +(5x-5)

10x-2=78

From here, we want to single out the x, by editing both sides in the same way.

10x-2=78

    +2  +2

10x=80

÷10 ÷10

x=8

Now that we know what x is, we can plug it in to the previous expressions to get the angles of ABC and DBC.

(5x-5)

((5·8)-5)

(40-5)

35

Now, we know that ABC is 35 degrees.

(5x+3)

((5·8)+3)

(40+3)

43

Now, we also know that DBC is 43 degrees.

To verify our answer we can add ABC and DBC together to check if it totals to ABD.

35+43=78

Now, we know that the answer: ABC is 35 degrees and DBC is 43 degrees is correct.

I hope this was helpful, keep learning! :D

Answer:

x=2

Step-by-step explanation:

3(x + 2) = 12

Distribute

3x+6 = 12

Subtract 6 from each sdie

3x+6-6 = 12-6

3x = 6

Divide by 3

3x/3 = 6/2

x=2

━━━━━━━☆☆━━━━━━━

▹ Answer

x = 2

▹ Step-by-Step Explanation

3(x + 2) = 12

Distribute:

3x + 6 = 12

Subtract 6 from each side:

6 - 6 = 0

12 - 6 = 6

Rewrite:

3x = 6

Divide each side by 3:

3/3 = x

6/3 = 2

x = 2

Hope this helps!

CloutAnswers ❁

━━━━━━━☆☆━━━━━━━

Answer:

33 1/3 %

Step-by-step explanation:

120degree is shaded

There are 360 degrees in a circle

120/360

1/3

.33333

33 1/3 %

Answer:

x = 67°

Step-by-step explanation:

The 2 base angles x are congruent , then

x = [tex]\frac{180-46}{2}[/tex] = [tex]\frac{134}{2}[/tex] = 67°

Answer:

X= 67 degrees.

Step-by-step explanation:

All triangles must add up to 180°. This is an isosceles triangle which means once you have one degree you can find the other two.

In order to get the answer start with 180°-46°

180-46=134

The two x's equal each other. So you can divide the remaining degrees by 2 to get the answer.

134÷2=67.

So the answer is 67°.

Answer:

Step-by-step explanation:

The standard from of equation of a line written in slope-intercept format is expressed as y = mx+c where c is the slope of the line and c is the y-intercept.

Given the equation of the line x+3y = 10, to get the slope of the line, we need write he equation in standard from first by making y the subject of the formula as shown;

[tex]x+3y = 10\\\\subtract\ x \ from \ both \ sides\\\\x+3y-x = 10 -x\\\\3y = -x+10\\\\Divide \ through\ by \ 3\\\\\frac{3y}{3} = -\frac{x}{3} +\frac{10}{3} \\\\[/tex]

[tex]y = -\frac{1}{3}(x) +\frac{10}{3} \\[/tex]

Comparing the resulting equation with y = mx+c, the slope 'm' of the equation is -1/3

The slope of the line is -1/3.

the correct option is D.

To find the slope of the line given by the equation x + 3y = 10, we need to rewrite it in slope-intercept form, which is y = mx + b, where m represents the slope.

Let's rearrange the equation to solve for y:

x + 3y = 10

3y = -x + 10

y = (-1/3)x + 10/3

Comparing this equation to the slope-intercept form, we can see that the coefficient of x (-1/3) represents the slope.

Therefore, the slope of the line is -1/3.

Learn more about Slope here:

https://brainly.com/question/17158875

#SPJ6

Hi

2.30 for small and 2.80 for large .  

7 small + 1 large  :   7*2.30 +1*2.80 = 16.10 + 2.80 = 18.90

Answer:

2.30 for small and 2.80 for large .  

7 small + 1 large  :   7*2.30 +1*2.80 = 16.10 + 2.80 = 18.90

Step-by-step explanation:

Answer:

4x^2−2

Step-by-step explanation:

Let's simplify step-by-step.

7x^2−2−3x^2

=7x^2+(−2)+(−3x^2)

Combine Like Terms:

=7x^2(+−2)+(−3x^2)

=(7x^2+(−3x^2))+(−2)

=4x^2+−2

Answer:

=4x2−2

Answer:

hello there...

you know BODMAS for sure

First brackets then of then divi then multi then add and then subtract

going by BODMAS

we need to first of all multiply of -6 and 3

hence ans is multiplying -6 and 3

( option 1 Is wrong by the equation u sent if the equation is correct then ans is which given bur by options I think equation is wrong becoz I don't see divide symbol)

Answer/Step-by-step explanation:

Use the distance formula, [tex] d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} [/tex], to solve each problem.

1. Distance between A(0, 0) and B(-5, 7):

[tex] (0, 0) = (x_1, y_1) [/tex]

[tex] (-5, 7) = (x_2, y_2) [/tex]

[tex] AB = \sqrt{(-5 - 0)^2 + (7 - 0)^2} = \sqrt{(-5)^2 + (7)^2} = \sqrt{25 + 49} [/tex]

[tex] AB = \sqrt{74} [/tex]

2. Distance between A(-3, 1) and B(-11, 1):

[tex] (-3, 1) = (x_1, y_1) [/tex]

[tex] (-11, 1) = (x_2, y_2) [/tex]

[tex] AB = \sqrt{(-11 - (-3))^2 + (1 - 1)^2} = \sqrt{(-8)^2 + (0)^2} = \sqrt{64 + 0} [/tex]

[tex] AB = \sqrt{64} = 8 [/tex]

3. Distance between A(-3, -2) and B(5, 6):

[tex] (-3, -2) = (x_1, y_1) [/tex]

[tex] (5, 6) = (x_2, y_2) [/tex]

[tex] AB = \sqrt{(5 - (-3))^2 + (6 - (-2))^2} = \sqrt{(8)^2 + (8)^2} = \sqrt{64 + 64} [/tex]

[tex] AB = \sqrt{128} = \sqrt{64*2} [/tex]

[tex] AB = 8\sqrt{2} [/tex]

4. Distance between A(-2, -5) and B(-6, 1):

[tex] (-2, -5) = (x_1, y_1) [/tex]

[tex] (-6, 1) = (x_2, y_2) [/tex]

[tex] AB = \sqrt{(-6 - (-2))^2 + (1 - (-5))^2} = \sqrt{(-4)^2 + (6)^2} = \sqrt{16 + 36} [/tex]

[tex] AB = \sqrt{52} [/tex]

None of the choices are correct

Answer:

Step-by-step explanation:

Given that :

the  perpendicular distance of the point from x axis = 2 units

the perpendicular distance from y axis is 3 units

The objective is to the write the coordinates of the points if it lies in

(i) | Quadrant  (ii) || Quadrant (iii) ||| Quadrant (iv) |v Quadrant

The position of a point in a plane is conveniently specified by the distances fro two perpendicular lines.The lines are called the x-axis and y-axis and their [point of intersection is called the origin.

The perpendicular distance from the y-axis is called the x- coordinate or abscissa and the perpendicular distance from x-axis is called the y-coordinate or ordinate,The  coordinate form an ordered pair with the abscissa written as first.

With that be said, So In the given question,

the  perpendicular distance of the point from x axis = 2 units

y coordinate = ±2

the perpendicular distance from y axis is 3 units

x coordinate = ±3

The ordered pair of the coordinates = ( ±3, ±2)

Therefore;

in | quadrant ; we have (3,2)   where  x and y are both on the positive axis

in || quadrant ; we have (-3,2)   x is on the negative axis and y is on the positive axis

In ||| quadrant ; we have (-3. -2) both x and y are on the negative axis

In |v quadrant ; we have ( 3, -2)  x is on the positive axis and y is on the negative axis.

Answer:

66

Step-by-step explanation:

The figure of triangle JKL is attached. JA = 9, AL = 10, and CK = 14.

According to two tangent theorem, the tangent to a circle that meets at the same point is equal to each other. Therefore:

AJ = BJ, AL = CL, CK = BK.

Since AJ = 9, BJ = AJ = 9

AL = 10, CL = AL = 10

CK = 14, BK = CK = 14.

Therefore the lengths of the triangle sides are:

JL = AJ + AL = 9 + 10 = 19

JK = BJ + BK = 9 + 14 = 23

KL = CL + CK = 14 + 10 = 24

The perimeter of the triangle is the sum of all its sides, it is given as:

Perimeter = JL + JK + KL = 19 + 23 + 24 = 66

Answer:

Step-by-step explanation:

A fifth-grade polynomial requires a minimum of 6 different points to create an adequate graph. Let is [tex]X[/tex] the dominion of the polynomial, such that [tex]0[/tex], [tex]1[/tex], [tex]2[/tex], [tex]3[/tex], [tex]4[/tex], [tex]5[/tex] [tex]\in X[/tex]. The values of the function for each value are calculated herein:

x = 0

[tex]f(0) = 6\cdot 0^{5}+8\cdot 0^{4}-7\cdot 0^{3}-5\cdot 0^{2}+10[/tex]

[tex]f(0) = 10[/tex]

x = 1

[tex]f(1) = 6\cdot 1^{5}+8\cdot 1^{4}-7\cdot 1^{3}-5\cdot 1^{2}+10[/tex]

[tex]f(1) = 12[/tex]

x = 2

[tex]f(2) = 6\cdot 2^{5}+8\cdot 2^{4}-7\cdot 2^{3}-5\cdot 2^{2}+10[/tex]

[tex]f(2) = 254[/tex]

x = 3

[tex]f(3) = 6\cdot 3^{5}+8\cdot 3^{4}-7\cdot 3^{3}-5\cdot 3^{2}+10[/tex]

[tex]f(3) = 1882[/tex]

x = 4

[tex]f(4) = 6\cdot 4^{5}+8\cdot 4^{4}-7\cdot 4^{3}-5\cdot 4^{2}+10[/tex]

[tex]f(4) = 7674[/tex]

x = 5

[tex]f(5) = 6\cdot 5^{5}+8\cdot 5^{4}-7\cdot 5^{3}-5\cdot 5^{2}+10[/tex]

[tex]f(5) = 22760[/tex]

The table is now presented:

 x                 y

 0                10

 1                 12

 2               254

 3               1882

 4               7674

 5              22760

Finally, the graphic is now constructed by using an online tool (i.e. Desmos). The image is included below as attachment.

Answer:

The correct statment is B.

Step-by-step explanation:

A. is not correct:  y = 2.4(30) - 1.8 does not equal 70...

B. Is correct because the slope is 2.4 From the equation

C. is not correct because the points have no 2.4 (maybe 2.2)? difference.

D. is not correct. the correlation isn't positive.

Answer:

big area/small area = (big length/small length)^2

24/small area = (6/3)^2

24/sa = 4

sa= 6

Answer:

£0.79 add £1.69 add £1.00 equals £ 3. 4 8

Take away £4.00 and change is 52p

Step-by-step explanation:

An equation is formed of two equal expressions. The change that Chris would receive is $0.52.

An equation is formed when two equal expressions are equated together with the help of an equal sign '='. while an expression is defined as a set of numbers, variables, and mathematical operations formed according to rules dependent on the context.

Given that Chris had $4 in cash. Also, he purchased coffee for $0.79, a bag of chips for $1.69 and a lottery ticket for $1.00. Therefore, the change he would receive is,

Change = Cash Chris has - Cost of coffee - Cost of chips - Cost of lottery

              = $4 - $0.79 - $1.69 - $1.00

              = $0.52

Hence, the change that Chris would receive is $0.52.

Learn more about Equation:

https://brainly.com/question/2263981

#SPJ2

Answer:

[tex](-5)^{11}[/tex]

Step-by-step explanation:

We can use the exponent rules. If we have [tex]\frac{a^b}{a^c}[/tex], then it will simplify to [tex]a^{b-c}[/tex].

b is 5, c is -6, and a is -5 so:

[tex]-5^{5-(-6)}\\-5^{11}[/tex]

Hope this helped!

============================================

Explanation:

The perimeter of the square is 36, so each side length is 36/4 = 9 units.

Point B is located at (5,17). We move down 9 units to get to (5,8), which is the location of point A. Then we move 9 units to the right to arrive at (14,8) which is point D's location.

Or we could go from B = (5,17) to C = (14,17) and then to D = (14,8). Each time we move 9 units.

Hello, because of the geometric sequence we can say that:

[tex]\alpha = \dfrac{b}{a}=\dfrac{c}{b}\\\\\dfrac{c}{a}=\dfrac{c*b}{a*b}=\dfrac{c}{b}\dfrac{b}{a}=\alpha^2\\\\\text{So the equation becomes.}\\\\ax^2+bx+c=0<=>x^2+\dfrac{b}{a}x+\dfrac{c}{a}=0\\\\<=>x^2+\alpha x+ \alpha^2=0\\\\\Delta=b^2-4ac = \alpha^2-4\alpha^2=-3\alpha^2 < 0[/tex]

So there is no real root, so the function does not cut the x axis.

Thank you

For the given quadratic function, the x-axis is not cut by the function because there is no true root.

To determine values for various parameters, quadratic functions are employed in a variety of scientific and engineering disciplines. A parabola is used to graphically illustrate them. The orientation of the curve is defined by the highest degree factor.

As per provided data in question,

α = b/a = c/b

c/a = (c × b)/(a × b) = (c/b) (b/a) = α²

For the equation,

ax² + bx + c = 0

x² + b/a(x) + c/a = 0

⇒ x² + ax + α² =0

Δ = b² - 4 ac = α² - 4α²

Δ = -3α² < 0, which means that no real root is there.

To know more about quadratic functions:

https://brainly.com/question/27958964

#SPJ2

Answer:

if im understanding right the answer should be x = -1

Step-by-step explanation:

5x - 11 = 3x - 9

subtract 3x

2x -11 = -9

add 11

2x = -2

divide by 2

x = -1

Answer:

1.

= Here,

Narasimha invest = 120,000

Madhu 135,000

pavan =

Answer:

3 * 10 ^3

Step-by-step explanation:

Divide the numbers

3.9 / 1.3

3

Subtract the powers of 10

10^5 / 10^2 = 10 ^ (5-2) = 10^3

The result is 3 * 10 ^3

[tex] \frac{3.9 \times {10}^{5} }{1.3 \times {10}^{2} } \\ = \frac{3.9}{1.3} \times {10}^{5 - 2} \\ = 3.0 \times {10}^{3} [/tex]

Hope you could get an idea from here.

Doubt clarification - use comment section.

Answer:

Step-by-step explanation:

Given the expression [tex]sin^{-1} 1/2[/tex], the following steps must be taken to evaluate the expression.

[tex]Let\ y = sin^{-1}1/2[/tex]

[tex]y = sin^{-1} 0.5[/tex]

From the calculator, the value of y wil give;

[tex]y = 30\\\\y = 30.00 (to\ the \ nearest \ hundredth)[/tex]

Hence [tex]sin^{-1} 0.5 = 30.00[/tex].

Answer:

Step-by-step explanation:

The answer is 0.52

Hope that helped

Step-by-step explanation:

Disclaimer: When writing this on the paper use the theta symbol, I'm using x since I'm on mobile.

2.

i).

[tex] \sin(x) \tan(x) \sec(x) = \tan {}^{2} (x) [/tex]

[tex] \sin(x) \sec(x) \tan(x) = \tan {}^{2} (x) [/tex]

[tex] \sin(x) \frac{1}{ \cos(x) } \tan(x) = \tan {}^{2} (x) [/tex]

[tex] \frac{ \sin(x) }{ \cos(x) } \tan(x) = \tan {}^{2} (x) [/tex]

[tex] \tan( x) ) \tan(x) = \tan {}^{2} (x) [/tex]

[tex] \tan {}^{2} (x) = \tan {}^{2} (x) [/tex]

iii).

[tex] \sec {}^{2} (x) (1 - \sin {}^{2} ( x ) ) = 1[/tex]

[tex] \sec {}^{2} (x) ( \cos {}^{2} (x) ) = 1[/tex]

[tex] \frac{1}{ \cos {}^{2} (x) } \cos {}^{2} (x) = 1[/tex]

[tex]1 = 1[/tex]

v).

[tex] \cot {}^{2} (a) - \cos {}^{2} (a) = \cot {}^{2} (a) \cos {}^{2} (a) [/tex]

[tex] \frac{ \cos{}^{2} (x) }{ \sin {}^{2} (x) ) } - \cos {}^{2} (x) [/tex]

Factor out cosine

[tex] \cos {}^{2} (x) ( \frac{1}{ \sin {}^{2} (x) } - 1) [/tex]

Simplify

[tex] \cos {}^{2} (x) ( \frac{1 - \sin {}^{2} (x) }{ \sin(x) } [/tex]

[tex] \cos {}^{2} (x( \frac{ \cos {}^{2} (x) }{ \sin {}^{2} (x) } ) = [/tex]

[tex]( \cos {}^{2} ( x ) ( \cot {}^{2} (x) )[/tex]