Prove: An odd number squared is an odd number. = (2n + 1)2 ​

Answer:

yes , because -3 is the whole number

Answer:

y = 10

Step-by-step explanation:

if the figure is to be a parallelogram, then

∠ ACB and ∠ DAC are alternate angles and are congruent , that is

5y - 13 = 7 + 3y ( subtract 3y from both sides )

2y - 13 = 7 ( add 13 to both sides )

2y = 20 ( divide both sides by 2 )

y = 10

Answer:

Step-by-step explanation:

A prism has a uniform cross-section equal in area to its base area. The volume of a prism is the product of this cross-section area and the length between the parallel bases.

__

A pyramid or cone has a cross section whose linear dimension is directly proportional to the distance from the peak. The volume is 1/3 of the product of the base area and the height.

  V = (1/3)Bh . . . . . formula for volume of a pyramid

__

Since the pyramid volume is 1/3 that of a prism with the same overall dimensions, the prism has 3 times the volume of the pyramid.

Answer:

Step-by-step explanation:

If m and n are both 1 then

Minimum value = 10^(1+1) * 3^(1-1) * 225

= 325.

If m and n have to be different, then m = 2 and n = 1

Minimum value = 10^3 * 3^1 * 15^2

= 675,000.

An example of how to sketch the Ferris Wheel diagram is given in the image attached.

To solve for the angular speed, you can use the formula of"

Given that the diameter is 80 feet, so we know the radius will be 40 feet.

The Ferris wheel bottom is 20 feet high, so we add the radius we get that the middle of the Ferris wheel is 60 feet high.

Finally, the Ferris wheel completes 1 ½ rotation (2π radians) in 1 minutes. So we say that the Ferris wheel is said to spins at π rad/min.

Angular velocity ( 1 ½ rev/1min)( 1min/60 sec)

C = 2π(60)

C ≈ 377.04 feet

540 revolution (1 ½ rotations) is 60 seconds, the speed is then

s = d/t

s = 377.06 ft / 60 sec

s = 6.28 ft/sec

To plot the graph, the  final equation will be h(t) = 40sin(πt + π/2) + 60.

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Answer:

9y(c=7) witch means the answer is 8,6

Step-by-step explanation:

I took it and got 5 out of 5 right ✅

jt- 22x+9..

It clearly says JT=22X+9

try 22+9..?

Answer: Reflection, translation

Step-by-step explanation:

Answer:

-7

1) 1/7. 3). 1. 5). 49. yes this is right answer

The recursive function is h(n) = -1/3h(n - 1) where h(0) = 63

The function is given as:

[tex]h(n) = 63* (-\frac{1}{3})^n[/tex]

Set n to 0

[tex]h(0) = 63* (-\frac{1}{3})^0[/tex]

Evaluate

h(0) = 63

This means that the function h(n) has the following parameters:

First term = 63

Rate = -1/3

Hence, the recursive function is h(n) = -1/3h(n - 1) where h(0) = 63

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Answer: The answer is three

Step-by-step explanation:

Answer:

67600

Step-by-step explanation:

1300x52=67600

Answer:

$67,600

Step-by-step explanation:

1300 x 52 = 67600

Answer:

the answer is 4×b=4b

Step-by-step explanation:

product=×

4×b=4b

[tex]~~~~~~ \textit{Compound Interest Earned Amount} \\\\ A=P\left(1+\frac{r}{n}\right)^{nt} \quad \begin{cases} A=\textit{accumulated amount}\dotfill &\$68000\\ P=\textit{original amount deposited}\\ r=rate\to 3.575\%\to \frac{3.575}{100}\dotfill &0.03575\\ n= \begin{array}{llll} \textit{times it compounds per year}\\ \textit{quarterly, thus four} \end{array}\dotfill &4\\ t=years\dotfill &30 \end{cases}[/tex]

[tex]68000=P\left(1+\frac{0.03575}{4}\right)^{4\cdot 30}\implies 68000=P(1.0089375)^{120} \\\\\\ \cfrac{68000}{1.0089375^{120}}=P\implies 23377\approx P[/tex]

Answer: Is (B) 23,377

Step-by-step explanation:

Approximately how much principal would need to be placed into an account earning 3.575% interest compounded quarterly so that it has an accumulated value of $68,000 at the end of 30 years?

a.

$23,706

b.

$23,377

c.

$52,069

d.

$58,944

Please select the best answer from the choices provided

A

( B )

C

D

Answer:

Step-by-step explanation:

I calculated it.

Answer:

the answer is 21

21

A rhombus is a quadrilateral whose all sides are of equal length. The measure of the ∠QPR is 42°.

A rhombus is a quadrilateral whose all sides are of equal length, and the opposite are parallel to each other.

The diagonals of a rhombus intersect each other at 90°. Therefore, the measure of ∠QTR can be written as,

∠QTR = 90°

(6y+6) = 90°

6y = 90 - 6

6y = 84

y = 14

Also, the diagonals of the rhombus divide the angle into two equal angles, therefore,

∠QPR = ∠SPR

∠QPR = 3y

∠QPR = 3(14)

∠QPR = 42°

Hence, the measure of the ∠QPR is 42°.

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Since the graph was not given, answer will be subjective. The graph that shows the solution to the system of linear inequalities of  2x – 3y <12 y < –3 is the one that has It is option B.

A linear inequality is  said to be a kind of an inequality that entails a linear function.

Note that a linear inequality is said to be made up of one of the symbols that connote inequality. It is the one that depicts the data that is not equal on a graph.

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Answer:

The correct answer is Graph B

Step-by-step explanation:

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Answer:

500 ÷ 3650 = 0.13698630136986 = 13.698630136986%

Answer:

1. [tex]\normalsize \boxed{\textsf{$4\sqrt{2}\cdot\sqrt{2}$}} \implies \boxed{8}[/tex]

2. [tex]\normalsize \boxed{\textsf{$3\sqrt{7}-2\sqrt{7}$}} \implies \normalsize \boxed{\textsf{$\sqrt{7}$}}[/tex]

3. [tex]\normalsize \boxed{\textsf{$\dfrac{\sqrt{7}}{2\sqrt{7}}$}} \implies \normalsize \boxed{\textsf{$\dfrac{1}{2}$}}[/tex]

4. [tex]\normalsize \boxed{\textsf{$2\sqrt{5}\cdot 2\sqrt{5}$}} \implies \boxed{20}[/tex]

Step-by-step explanation:

The Radical Rules by Lial et al. (2017) state that:

Product rule: [tex]\large \textsf{$\sqrt[n]{a}\cdot\sqrt[n]{b}=\sqrt[n]{ab}$}[/tex]

Quotient rule: [tex]\large \textsf{$\sqrt[n]{\dfrac{a}{b}}=\dfrac{\sqrt[n]{a}}{\sqrt[n]{b}}$}\ \ \textsf{$(b \neq 0)$}[/tex]

1.

[tex]\implies \normalsize \textsf{$4\sqrt{2}\cdot\sqrt{2}$}\\\\\normalsize \implies \textsf{$4\sqrt{2\cdot2}$}\\\\ \implies \normalsize \textsf{$4\sqrt{4}$}\\\\ \implies\normalsize \textsf{$4\cdot 2 = 8$}[/tex]

[tex]\normalsize \boxed{\textsf{$4\sqrt{2}\cdot\sqrt{2}$}} \implies \boxed{8}[/tex]

2.

[tex]\implies \normalsize \textsf{$3\sqrt{7}-2\sqrt{7}$}\\\\ \implies\normalsize \textsf{$(3-2)\sqrt{7}$}\\\\ \implies\normalsize \textsf{$1\sqrt{7}$}\\\\ \implies\normalsize \textsf{$\sqrt{7}$}[/tex]

[tex]\normalsize \boxed{\textsf{$3\sqrt{7}-2\sqrt{7}$}} \implies \normalsize \boxed{\textsf{$\sqrt{7}$}}[/tex]

3.

[tex]\implies \normalsize \textsf{$\dfrac{\sqrt{7}}{2\sqrt{7}}$}\\\\\\ \implies\normalsize \textsf{$\dfrac{1\sqrt{7}}{2\sqrt{7}}$}\\\\\\ \implies\normalsize \textsf{$\dfrac{1}{2}$}\\\\[/tex]

[tex]\normalsize \boxed{\textsf{$\dfrac{\sqrt{7}}{2\sqrt{7}}$}} \implies \normalsize \boxed{\textsf{$\dfrac{1}{2}$}}[/tex]

4.

[tex]\implies \normalsize \textsf{$2\sqrt{5}\cdot 2\sqrt{5}$}\\\\ \implies\normalsize \textsf{$2\cdot2\sqrt{5\cdot5}$}\\\\ \implies \normalsize \textsf{$4\sqrt{25}$}\\\\ \implies \normalsize \textsf{$4\cdot5=20$}\\\\[/tex]

[tex]\normalsize \boxed{\textsf{$2\sqrt{5}\cdot 2\sqrt{5}$}} \implies \boxed{20}[/tex]

Reference:

Lial, M., Hornsby, J., Schneider, D., & Daniels, C. (2017). College Algebra and Trigonometry, Global Edition (6th ed., p. 94).

Reason:

Distance = rate*time

time = distance/rate

time = (38 miles)/(2 mph)

time = 38/2 hours

time = 19 hours

Answer:

Step-by-step explanation:

The measure of angle ∠CAB will be 68 degrees and the measure of angle ∠CAD will be 22 degrees.

It is the center of an equidistant point drawn from the center. The radius of a circle is the distance between the center and the circumference.

In the circle below, stack AD with a bar on top is diameter and stack AB with a left-right arrow on top is tangent at A. Suppose m stack ADC with over parenthesis on top equals 224 degrees.

Then the measures of m angle ∠CAB and m angle ∠CAD will be  

Then we have

We know that

mADC = mAD + mDC

  224° = 180° + mDC

 mDC = 44°

Then we know the theorem, A chord's degree in the centre is double that of the chord's degree at the perimeter. Then we have

2 ∠CAD = 44°

  ∠CAD = 22°

Then the angle ∠CAB and angle ∠CAD are complementary angles. Then we have

∠CAB + ∠CAD = 90°

   ∠CAB + 22° = 90°

            ∠CAB = 68°

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Answer:

If we square both sides then we get

x2-8x+16 = (x-4)2

Now if we expand the right side of the equation we get

x2-8x+16 = (x-4)(x-4)

x2-8x+16 = x2-8x+16

So no matter what x is, the right side will be equal to the left side.

However, if we look at the original equation on the left side, √(x2-8x+16) , the value under the square root symbol has to be greater than or equal to zero (otherwise we get an imaginary number and you might not be working with those yet)

Step-by-step explanation:

So in order for √(x2-8x+16) to be greater than or equal to zero, x has to be greater than or equal to 4. If x is less than 4 you get a negative number. If x is greater than 4, you'll get the same answer on both sides no matter what x is

[tex]\qquad\qquad\huge\underline{{\sf Answer}}[/tex]

Let's solve ~

[tex]\qquad \sf  \dashrightarrow \: \sqrt{ {x}^{2} - 8x + 16} [/tex]

[tex]\qquad \sf  \dashrightarrow \: \sqrt{ {x}^{2} - 4x - 4x + 16} [/tex]

[tex]\qquad \sf  \dashrightarrow \: \sqrt{ {x}^{} (x - 4) - 4(x - 4)} [/tex]

[tex]\qquad \sf  \dashrightarrow \: \sqrt{ (x - 4)(x - 4)} [/tex]

[tex]\qquad \sf  \dashrightarrow \: \sqrt{ (x - 4) {}^{2} } [/tex]

[tex]\qquad \sf  \dashrightarrow \: x - 4[/tex]

Now, it's given that ~

[tex]\qquad \sf  \dashrightarrow \:x \geqslant 4[/tex]

plug the value of x as 4 :

[tex]\qquad \sf  \dashrightarrow \:4 - 4[/tex]

[tex]\qquad \sf  \dashrightarrow \:0[/tex]

So, we can infer that :

Value of the required expression lies in :

[tex]\qquad \sf  \dashrightarrow \: \sqrt{ {x}^{2} - 8x + 16} \geqslant 0[/tex]

it's value is always greater than pr equal to 0, satisfying the given condition ~

ΔAGF ≅ ΔCGE because they are both created from perpendicular diagonals that are intersected at the midpoint by line FE.

A perpendicular line is a straight line that is at right angle with another line. That is both lines intersect each other at right angles.

Lines EG and FG  are ≅ because the are dissected by a line that runs through the diagonal of rectangle ABCD

Lines JK ≅ HL because they are both created by parallel lines JL and HI.

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Answer:

5600x10=56000

56000=B ten thousands

Answer:

Ten Thousands

Step-by-step explanation:

5,600x10=56000

So the 5 will be in the ten thousand place if u get a chart/

ignore the numbers on the chart

From the calculations, we can see that the number of beetles after three years is 11618 beetles.

The term "exponential problem" has to do with any problem in which the increase or decrease occurs exponentially.

In this case we have P = Poe^rt

Po = 10,000

r = 0.05

t = 3

P =  10,000 e^(0.05 * 3)

P= 11618 beetles

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Answer:

the value of x=9 and AB(h)=9.396