A drawer contains 3 white balls and 6 black balls. Two balls are drawn out of the box at random. What is the probability that they both are black

Answer:

1413.7

Step-by-step explanation:

A(sphere) = 4 * π * r².

You can think about it like two times the cap surface area of a hemisphere. Therefore, the hemisphere cap area equals:

Ac = A(sphere) / 2,

Ac = 2 * π * r².

Answer:

See below

Step-by-step explanation:

SAS says the two triangles are congruent

 so   side 6x-4 = 3x+8     then x = 4  

x= 4 true

x = 3 false ( because we just found it = 4)

Pythag theorem says QT

   (6x-4)^2 = QT^2 + 12^2

   ( 6(4)-4)^2 = 12 ^2 + QT^2

       400  -144 = QT^2

       QT = 16   true

The order of the side lengths from least to greatest is; GH< IH< FG< GI< FH.

As with other closed geometric figures, the shortest side length is opposite the smallest angle measure while the longest side length, is opposite the greatest angle measure.

Hence, it follows from the statement above that the order of side lengths is; GH< IH< FG< GI< FH.

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Answer:

-10u^3 - 4

Step-by-step explanation:

-3u^3-7u^3 = -10u^3

then -4

Answer:

0.98 km/h

Step-by-step explanation:

Converting mi to km: 66*1.6 = 105.6 miles

Imani's average: 105.6 miles/ 90 min = 1.73 km/h

Abedi's average: 56 miles/ 75 min = 0.7466 ≈ 0.75 km/h

Difference: 1.73-0.75 = 0.98 km/h

Answer:

x ≈ 14.4

Step-by-step explanation:

using the cosine ratio in the right triangle

cos59° = [tex]\frac{adjacent}{hypotenuse}[/tex] = [tex]\frac{x}{28}[/tex] ( multiply both sides by 28 )

28 × cos59° , then

x ≈ 14.4 ( to the nearest tenth )

Among the congruent quadrilaterals, the segment that must also measure 3 cm is: A. A F.

Congruent quadrilaterals have corresponding congruent angles and sides.

Given that, GHJK ≅ ASDF and GHJK ≅ VBNM, it means that all corresponding sides and angles are equal to each other.

MV corresponds to KG and A F. If MV = 3 cm, then one segment that must measure 3 cm is:

A. A F

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Answer: A. A F

same answer as the other person gave but you don't have to read a bunch

Answer:

35 and 10

Step-by-step explanation:

Let m and n be the numbers

m  is 5 more than 3 times n can be written as

m = 3n + 5    ... (1)

Sum of numbers is 45 so m + n = 45 .or

m = 45 - n ... (2)

Equating RHS of (1) and (2) gives us

3n + 5 = 45 - n

Collecting like terms

3n + n = 45 -5

4n = 40

n = 10

Therefore m = 45-10 = 35

Quick Check

In (1) substitute for m and n,

LHS is m = 35

RHS = 3 * 10 + 5 = 35

Answer:

(- 2, [tex]\frac{5}{3}[/tex] )

Step-by-step explanation:

y = [tex]\frac{2}{3}[/tex] x + 3 → (1)

x = - 2

substitute x = - 2 into (1)

y = [tex]\frac{2}{3}[/tex] × - 2 + 3 = - [tex]\frac{4}{3}[/tex] + [tex]\frac{9}{3}[/tex] = [tex]\frac{5}{3}[/tex]

solution is (- 2, [tex]\frac{5}{3}[/tex] )

fraction 2/9 is twice the value of fraction 1/9.

What are mathematics operations?

• A mathematical operation is a function that converts a set of zero or more input values (also called "b" or "arguments") into a defined output value. The number of operands determines the operation's arity. Most commonly studied operations are binary operations (i.e., operations of arity 2), such as addition and multiplication, and unary operations (i.e., operations of arity 1), such as additive and multiplicative inverses.

• Zero-arity operations, or nullary operations, are constants, and mixed products are arity three operations, or ternary operations.

Here, the given fractions are :

1/9 and 2/9

clearly 2/9 is twice the value of fraction 1/9.

And, there decimal form​ will be :

1/9 = 0.11

2/9 = 0.22

Here, both are multiple of 11.

Therefore, fraction 2/9 is twice the value of fraction 1/9.

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The equations of the perpendicular lines are: y = 1/2x + 6, y = 15, y = -x - 2, y = 6x + 3 and y = 1/3x - 4

When a linear equation is represented as:

Ax + By = C

The slope (m) is:

m = -A/B

When the linear equation is represented as:

y = mx + c

The slope is m

A line perpendicular to a linear equation that has a slope of m would have a slope of -1/m

Using the above highlights, the equations of the lines are:

6. y = -2x + 5; (2, 7)

The slope is:

m = -2

The perpendicular slope is:

n = 1/2

The equation of the perpendicular line is:

y = n(x - x1) + y1

This gives

y = 1/2(x - 2) + 7

Evaluate

y = 1/2x - 1 + 7

This gives

y = 1/2x + 6

7. y = -5; (11, 15)

The slope is:

m = 0

The perpendicular slope is:

n = 1/0 = undefined

The equation of the perpendicular line is:

y = n(x - x1) + y1

This gives

y = 15

8. Graph ; (-12, 10)

The slope is:

m = (y2 - y1)/(x2 - x1)

Using the points on the graph, we have:

m = (2 - 3)/(3 - 4)

m = 1

The perpendicular slope is:

n = -1

The equation of the perpendicular line is:

y = n(x - x1) + y1

This gives

y = -1(x + 12) + 10

y = -x - 12 + 10

Evaluate

y = -x - 2

9. y = -1/6x + 1; (-2, -9)

The slope is:

m = -1/6

The perpendicular slope is:

n = 6

The equation of the perpendicular line is:

y = n(x - x1) + y1

This gives

y = 6(x + 2) - 9

Evaluate

y = 6x + 12 - 9

This gives

y = 6x + 3

10. 6x + 2y = 14; (12, 0)

The slope is:

m = -6/2

m = -3

The perpendicular slope is:

n = 1/3

The equation of the perpendicular line is:

y = n(x - x1) + y1

This gives

y = 1/3(x - 12) + 0

Evaluate

y = 1/3x - 4

Hence, the equations of the perpendicular lines are: y = 1/2x + 6, y = 15, y = -x - 2, y = 6x + 3 and y = 1/3x - 4

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[tex]\Huge\boxed{\textsf{A. Exactly one solution}}[/tex]

We have the following system:

[tex]\begin{cases}4x-2y&=8\\2x+y&=2\end{cases}[/tex]

Here, a simple solution to find the answer is to graph the two lines and see how many times they intersect.

I've attached a graph, with [tex]4x-2y=8[/tex] in red and the other equation in blue.

See that the lines only intersect once, at [tex](1.5,-1)[/tex]. This means the system only has one solution.

Answer: No Solution

Let's bring both equations to slope-intercept form. Then we can think about the slopes and the y-intercepts of the lines represented by each equation.

The slope-intercept form of the first equation 2y=4x+62y=4x+62, y, equals, 4, x, plus, 6 is y=2x+3y=2x+3y, equals, 2, x, plus, 3. The second equation y = 2x+6y=2x+6y, equals, 2, x, plus, 6 is already in slope-intercept form.

Hint #22 / 3

The first equation is y = 2x+3y=2x+3y, equals, 2, x, plus, 3, so the slope of its line is 222 and the yyy-intercept is (0,3)(0,3)left parenthesis, 0, comma, 3, right parenthesis.

The second equation is y = 2x+6y=2x+6y, equals, 2, x, plus, 6, so the slope of its line is 222 and the yyy-intercept is (0,6)(0,6)left parenthesis, 0, comma, 6, right parenthesis.

Since both lines have the same slopes but different yyy-intercepts, they are distinct parallel lines.

Hint #33 / 3

Since distinct parallel lines don't intersect, we conclude that the system has no solutions.

Answer:

The Answer is x>0

Answer:

[tex]\sf \boxed{\bf y =5\sqrt{22}}[/tex]

Step-by-step explanation:

       The ratio of sides of 45 - 45 - 90 triangle is  a : a : a√2.

a is the side opposite to 45°.

From the figure, a = 5√11

The side opposite to 90° is a√2.

y = a√2

  [tex]\sf = 5\sqrt{11} * \sqrt{2}\\\\ = 5*\sqrt{11*2}\\\\ = 5\sqrt{22}[/tex]

The system of equations represented by the attached graph is y = (x + 3)^2 - 2 and y = -x + 1

This question will be answered using the attached graph

The curve

The curve is a quadratic function, and it has the following features:

Vertex, (h, k) = (-3, -2)

Point (x, y) = (-1, 2)

A quadratic function is represented as:

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

So, we have:

y = a(x + 3)^2 - 2

Substitute (x, y) = (-1, 2)

2 = a(-1 + 3)^2 - 2

This gives

2 = 4a-2

Solve for a

a = 1

Substitute a = 1 in y = a(x + 3)^2 - 2

y = (x + 3)^2 - 2

The line

The line is a linear function, and it has the following features:

Point (x1, y1) = (-1, 2)

Point (x2, y2) = (-6, 7)

The linear function is calculated as:

y = (y2 - y1)/(x2 - x1) *(x- x1) + y1

So, we have:

y = (7 -2)/(-6 +1) *(x + 1) + 2

Evaluate the quotient

y =  -1(x + 1) + 2

Expand

y = -x -1 + 2

y = -x + 1

Hence, the system of equations represented by the attached graph is y = (x + 3)^2 - 2 and y = -x + 1

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A true statement about this transformation is: C. It can be a rigid or a nonrigid transformation depending on whether the corresponding side lengths have the same measures.

In Geometry, a transformation can be defined as the movement of a point from its initial position to a new location. Hence, when an object is transformed, all of the points would also be transformed.

In this scenario, we can logically deduce that triangle J'K'L' can either be a rigid or a nonrigid transformation based on the magnitude of the corresponding side lengths in both triangles, considering that their angles are equal in magnitude.

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Complete Question:

Triangle JKL is transformed to create triangle J'K'L'. The angles in both triangles are shown.

J = 90° J' = 90°

K = 65° K' = 65°

L = 25° L' = 25°

Which statement is true about this transformation?

A. It is a rigid transformation because the pre-image and image have the same corresponding angle measures.

B. It is not a rigid transformation because the corresponding side lengths are not equal.

C. It can be a rigid or a nonrigid transformation depending on whether the corresponding side lengths have the same measures.

D. It can be a rigid or a nonrigid transformation because a pair of corresponding angles measures 90°.

It is a rigid transformation because the pre-image and image have the same corresponding angle measures.

The complete question is added as an attachment

From the image and the preimage triangles, we have that:

The corresponding sides of both triangles are equal

This is identified by the marks I, II and III on the side lengths

Equal corresponding sides represent a rigid transformation

Hence, the true statement about the dilation is (a)

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Answer:

Answer:

$183.7

Step-by-step explanation:

To find the deposited amount you need to....

[tex]$532.20-$348.50=$183.70[/tex]

Hello!

Subtract the old balance from new balance to find deposit amount.

⇒ Deposit = $532.20 - $348.50

⇒ Deposit = $183.70

Answer: A

Step-by-step explanation:

A) Correct. [tex]x^2 \geq 0 \implies -x^2 \leq 0[/tex]

B) Wrong. [tex]2^x > 0[/tex], so the range is [tex](4, \infty)[/tex]

C) Wrong. Linear functions have a range of all real numbers.

D) Wrong. Same logic as C.

The valid exclusion of the algebraic fraction is (c) a =0, b =0, a =2b

The expression is given as:

8ab^2x/4a^2b - 8ab^2

Set the denominator to 0

4a^2b - 8ab^2 = 0

Divide through by 4ab

a - 2b = 0

Add 2b to both sides

a = 2b

Hence, the valid exclusion of the algebraic fraction is (c) a =0, b =0, a =2b

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Hello and Good Morning/Afternoon

Let's take this problem step-by-step:

To find the solution to a system of equation

  ⇒ must set the equations equal to each other

     ⇒ and solve

Let's put that into action

  [tex]x^2-2x+3 = -2x + 12\\x^2-2x+2x+3-12=0\\x^2-9=0\\(x+3)(x-3)=0[/tex]

          ⇒ either 'x+3' or 'x-3' equals zero

              ⇒ must find 'x' that satisify either equation

                    [tex]x-3=0\\x=3\\\\x+3=0\\x=-3[/tex]

Let's find the corresponding f(x) to each x-value

  [tex]f(3) = -2(3)+12=-6+12 = 6\\\\f(-3) = -2(-3) + 12=6+12=18[/tex]

Therefore the solutions are (3,6) and (-3,18)

Answer: (3,6), (-3,18)

Hope that helps!

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Answer:

450 people were at the match

Step-by-step explanation:

So let's make equation :

W = women

M = Men

C = Children

First equation :

M = W + 250

Second equation :

C = 2W

Third equation :

M = 2(W+C)

Let's substitute the second equation into the third :

M = 2(W + 2W)

M = 2W + 4W

M = 6W

Let's substitute this equation into the first one :

6W = W + 250

5W = 250

W = 50

Now we know that that there were 50 women at the match and can substitute this value for the first equation and then the second :

M = 50 + 250

M = 300

This means that there must have been 300 men at the match.  

C = 2(50)

C = 100

This means that there must have been 100 children at the match.

Adding all values together gives us :

50 + 300 + 100  =

450 people were at the match

Hope this helped and have a good day

Answer: 27:1

Step-by-step explanation:

The ratio of the volumes is equal to the cube of the ratio of the altitudes.

Answer:

Sarah runs at 14.4km/h

Step-by-step explanation:

36km = 2.5 hours

/ = per or divide

36km/2.5hours = 36 divided by 2.5 = 14.4km/h

[tex]\frac{dy}{dy}=1[/tex], so I assume you mean "find [tex]\frac{dy}{dx}[/tex]".

We can rewrite this as an implicit equation to avoid using too much of the chain rule, namely

[tex]y = \sqrt[3]{\dfrac{e^x (x+1)}{x^2+1}} \implies (x^2+1) y^3 = e^x (x+1)[/tex]

Differentiate both sides with respect to [tex]x[/tex] using the product and chain rules.

[tex]2x y^3 + 3(x^2+1) y^2 \dfrac{dy}{dx} = e^x(x+1) + e^x[/tex]

[tex]\implies 3(x^2+1) y^2 \dfrac{dy}{dx} = e^x (x+2) - 2x y^3[/tex]

[tex]\implies \dfrac{dy}{dx} = \dfrac{e^x (x+2) - 2x y^3}{3(x^2+1) y^2}[/tex]

Now substitute the original expression for [tex]y[/tex].

[tex]\dfrac{dy}{dx} = \dfrac{e^x (x+2) - 2x \left(\sqrt[3]{\frac{e^x(x+1)}{x^2+1}}\right)^3}{3(x^2+1) \left(\sqrt[3]{\frac{e^x(x+1)}{x^2+1}}\right)^2}[/tex]

[tex]\implies \dfrac{dy}{dx} = \dfrac{e^x (x+2) - \frac{2e^x(x^2+x)}{x^2+1}}{3(x^2+1) \left(\frac{e^x(x+1)}{x^2+1}\right)^{2/3}}[/tex]

[tex]\implies \dfrac{dy}{dx} = e^x \dfrac{x^3-x+2}{3(x^2+1)^2 \frac{e^{2x/3}(x+1)^{2/3}}{(x^2+1)^{2/3}}}}[/tex]

[tex]\implies \dfrac{dy}{dx} = e^{x/3} \dfrac{x^3-x+2}{3(x^2+1)^{4/3} (x+1)^{2/3}}[/tex]

Now, since

[tex]y = \sqrt[3]{\dfrac{e^x (x+1)}{x^2+1}} = \dfrac{e^{x/3} (x+1)^{1/3}}{(x^2+1)^{1/3}}[/tex]

we can write

[tex]\dfrac{dy}{dx} = e^{x/3} \dfrac{x^3-x+2}{3(x^2+1)^{4/3} (x+1)^{2/3}} = \dfrac{e^{x/3} (x+1)^{1/3}}{(x^2+1)^{1/3}} \times \dfrac{x^3-x+2}{3(x^2+1)^{3/3} (x+1)^{3/3}}[/tex]

[tex]\implies \dfrac{dy}{dx} = y \dfrac{x^3-x+2}{3(x^2+1)(x+1)}[/tex]

Focusing on the rational expression in [tex]x[/tex], we have the partial fraction expansion

[tex]\dfrac{x^3 - x + 2}{(x^2 + 1) (x+1)} = a + \dfrac{bx+c}{x^2+1} + \dfrac d{x+1}[/tex]

where we have the constant term on the right side because both the numerator and denominator have degree 3.

Writing everything with a common denominator and equating the numerators leads to

[tex]x^3 - x + 2 = a (x^2+1) (x+1) + (bx+c)(x+1) + d(x^2+1) \\\\ = ax^3 + (a+b+d)x^2 + (a+b+c)x + a+c+d[/tex]

[tex]\implies \begin{cases} a = 1 \\ a+b+d=0 \\ a+b+c = -1 \\ a+c+d=2 \end{cases}[/tex]

[tex]\implies a=1, b=-2, c=0, d=1[/tex]

[tex]\implies \dfrac{x^3 - x + 2}{(x^2 + 1) (x+1)} = 1 - \dfrac{2x}{x^2+1} + \dfrac 1{x+1}[/tex]

and it follows that

[tex]\boxed{\dfrac{dy}{dx} = \dfrac y3 \left(1 - \dfrac{2x}{x^2+1} + \dfrac1{x+1}\right)}[/tex]

Here ABC is a right angle triangle because 1 angle is 90°

BY ASP,

90+56+ angle CAB=180

angle CAB=34°

rounded to nearest = 30°

sin A =opp/hyp

sin 30 = 7/x

1/2=7/x

x=14

rounding to nearest 10

14 =10

AB=10

By Pythagoras Theorem,

AB²=BC²+AC²

AC²=AB²-BC²=10² - 7²

100- 49 = 51

AC=root 51

Answer:

390 m^2

Step-by-step explanation:

So let's start by finding the surface area of the two triangles on the side. Generally the area of a triangle can be calculated as: [tex]\frac{1}{2}bh[/tex]. but since there is two of them, we can calculate the area of both of them by simply canceling out the 1/2 to a 1. So the area of both triangles are 12 m * 5m as given in the diagram. This gives you 60m^2 area for both triangles on the side

Now let's calculate the rectangle on the top. It has a width and length of 13 and 11 m. So multiply these together to get: 143 m^2

Now let's calculate the rectangle that's on the bottom, it has the dimensions 11 and 12m and multiplying these together gets you: 132 m^2\

Now calculate the rectangle all the way in the back, it has the dimensions 5 and 11m, multiplying these together gets you 55 m^2.

Now add all these together: 60m^2 + 143m^2 + 132m^2 + 55m^2 = 390m^2

The proof is shown below:

A parallelogram is a quadrilateral with opposite sides parallel (and therefore opposite angles equal).

As,

<A= 104

<B= 76

As AB || CD,

<A + <D =180

104 + <D =180

<D = 76

and <B + <C =180

76 + <C = 180

<C = 104.

As, opposite angles are equals and  AB || CD.

Hence, ABCD is a parallelogram.

For 2 part,

Use m<A = 104° and m<B= 76° to show that <A and <B are same-side interior angles. Then, use AB || CD to show that <A and <D are supplementary angles.

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