How do I solve this math problem

Answer:

[tex]r^2=4\sec 2\theta}[/tex]

Step-by-step explanation:

The given equation is:

[tex]x^2-y^2=4[/tex]

We substitute [tex]x=r\cos (\theta)[/tex] and [tex]y=r\sin (\theta)[/tex] to obtain:

[tex]r^2\cos^2\theta-r^2\sin^2\theta=4[/tex]

This implies that:

[tex]r^2(\cos^2\theta-\sin^2\theta)=4[/tex]

Apply double angle identity to obtain:

[tex]r^2\cos 2\theta=4[/tex]

This implies that:

[tex]r^2=\frac{4}{\cos 2\theta}[/tex]

This simplifies to:

[tex]r^2=4\sec 2\theta}[/tex]

Answer:

Step-by-step explanation:

1. Due to the stretchiness of shirt fabric and the waste involved in cutting pattern pieces, it is unrealistic to require that the length of the fabric be specified to a hundredth of a centimeter. Measurement to the nearest half-centimeter is sufficiently accurate for the purpose. The appropriate choice is the measurement with a precision of 1 cm:

  232 cm

__

2. The most appropriate choice is the one that shows cubing 6^(1/3) will result in 6, just as cubing ∛6 will result in 6.

Answer:Vertical Asymptotes: x=0

Horizontal Asymptotes: y=2

Here’s a graph picture...

Answer:

Step-by-step explanation:

(x+3)^2=35+1

(x+3)^2=36

(x+3)^2=6^2

taking squareroot

x+3=+-6

x+3=6,-6

x=6-3=3

or x=-6-3=-9

Answer:

The numbers between 30 and 40 whose only factors are one and themselves, also called prime numbers, are 31 and 37.  :)

Answer:

It is 31 and 37

Numbers that have only one and itself as a factor are called prime numbers.

Answer:

D

Step-by-step explanation:

Since the denominators of both fractions are common, then subtract the numerators leaving the common denominator, that is

[tex]\frac{4a+1-2a-7}{a^2-4}[/tex] = [tex]\frac{2a-6}{a^2-4}[/tex]

Answer:

[tex]\boxed{\bold{\frac{2a-6}{a^2-4}}}[/tex]

Explanation:

Apply Rule [tex]\bold{\frac{a}{c}\pm \frac{b}{c}=\frac{a\pm \:b}{c}}[/tex]

= [tex]\bold{\frac{4a+1-\left(2a+7\right)}{a^2-4}}[/tex]

Expand [tex]\bold{4a+1-\left(2a+7\right): \ 2a-6}[/tex]

= [tex]\bold{\frac{2a-6}{a^2-4}}[/tex]

Mordancy.

Answer:

x = - 2, x = 12

Step-by-step explanation:

Given

x² - 10 x  = 24 ( subtract 24 from both sides )

x² - 10x - 24 = 0 ← in standard form

To factorise the quadratic

Consider the factors of the constant term (- 24) which sum to give the coefficient of the x- term (- 10)

The factors are - 12 and + 2, since

- 12 × 2 = - 24 and - 12 + 2 = - 10, thus

(x - 12)(x + 2) = 0

Equate each factor to zero and solve for x

x - 12 = 0 ⇒ x = 12

x + 2 = 0 ⇒ x = - 2

The solution of this quadratic equation to x² – 10 x = 24 is  x = - 2 & x = 12

A quadratic is a sort of problem that deals with a variable accelerated by using an operation called squaring. This language derives from the area of a rectangular being its facet period expanded with the aid of itself. The phrase "quadratic" comes from quadrant, the Latin word for square.

x² – 10 x = 24

⇒ x² +2X -12X -24 = 0

⇒ x (x - 2) -12(x - 2) =0

⇒ x = 12 OR ( x = -2)

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The expression (4y^6)^3 - (10z^2)^3 is equivalent to 64y^18 - 1000z^6.

Factor out common powers:

64y^18 = (2^6)(y^3)^6

1000z^6 = (10^3)(z^2)^3

Rewrite the expression with factored terms:

64y^18 - 1000z^6 = (2^6)(y^3)^6 - (10^3)(z^2)^3

Apply power of a power rule:

(a^m)^n = a^(mn)

(2^6)(y^3)^6 = 2^(66) * y^(36) = 2^36 * y^18

(10^3)(z^2)^3 = 10^(33) * z^(2*3) = 10^9 * z^6

Substitute back the simplified terms:

2^36 * y^18 - 10^9 * z^6 = (4y^6)^3 - (10z^2)^3

Therefore, (4y^6)^3 - (10z^2)^3 is the equivalent expression to 64y^18 - 1000z^6. Both expressions involve the difference of cubes of binomials, with one focusing on powers of 4y^6 and the other emphasizing powers of 10z^2.

Answer:

£9.6

Step-by-step explanation:

x = the original price of a CD

£x = 100% of the original price

The price of a CD was decreased by 20% to £7.68.

This means:

£7.68 = 100% - 20%

£7.68 = 80% of the original price

From this, we will find 1% of the original price.

£7.68 ÷ 80 = 1%

£0.096 = 1%

Since the original price ( x ) = 100% of the original price, we will find 100% of the original price.

£0.096 × 100 = 100%

£9.6 = 100%

Therefore, the original price of a CD = £9.6

The price before the decrease was approximately £9.60.

To find the price before the decrease, we need to calculate the original price before the 20% decrease. Let's call the original price 'x'. We know that after the 20% decrease, the price is £7.68.

So, if we take 20% of 'x' and subtract it from 'x', we should get £7.68. Mathematically, this can be expressed as:

x - 0.20x = 7.68

Simplifying the equation, we have:

0.80x = 7.68

Dividing both sides of the equation by 0.80, we find that 'x' is approximately 9.60. Therefore, the price before the decrease was approximately £9.60.

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Step-by-step explanation:

*****************here is the answer

area=28.61+12=40.61**************

Answer:  2 years 1 month

Step-by-step explanation:

2 years 1 month at $100 a month = $2,500

$720 x 2 years = $1,440

$720 / 12 months = $60

$1,440 + $60 = $1,500

$4,000 - $1,500 = $2,500

The two investment funds will have the same amount of money after 25 months. This conclusion is reached by setting up and solving an equation where the monthly growth of the first fund, $100x, equals the monthly decrease of the second fund, $4000 - $60x.

Let's denote the time in months that it takes for both investment funds to have the same amount of money as x. The equation that represents the first fund's growth is 100x, as it grows at $100 per month. The equation for the second fund is 4000 - 60x, as it decreases by $720 per year, or $60 per month.

We can set these two equations equal to each other to solve for x: 100x = 4000 - 60x. Solving this equation involves adding 60x to both sides to yield 160x = 4000, and then dividing by 160 to get x = 25. Therefore, after 25 months, both funds will have the same amount of money.

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Answer:

15

Step-by-step explanation:

if you have 2/5 lengths each time, if you divide that by 6 it gives 15.

Or if you go 2/5 to 4/5 to 6/5 until you have 30/5 because that is how long 6 feet is.

Answer:

Surface Area = Sum of area of all sides = 203.2 Units Squared

Step-by-step explanation:

Top and Bottom: A = ((.5*4*6) + (.5*3.6*2)) * 2 => (12+3.6) *2 => 15.6*2 = 31.2

Sides: A = ((5*10)+(3.6*10)) *2 => (50 + 36) *2 => 86*2 = 172

Total Surface Area = Top + Bottom + Sides = 203.2 units squared

Answer might be d I think

Answer: D -2

Step-by-step explanation:

[tex]\bf \cfrac{4^2-20\div 5}{1-5+7}\implies \cfrac{\stackrel{\downarrow }{16}-20\div 5}{1-5+7}\implies \cfrac{16-\stackrel{\downarrow }{4}}{1-5+7}\implies \cfrac{\stackrel{\downarrow }{12}}{1-5+7} \\\\\\ \cfrac{12}{\stackrel{\downarrow }{-4}+7}\implies \cfrac{12}{\stackrel{\downarrow }{3}}\implies 4[/tex]

Answer:

[tex]\left(a+3\right)\left(a-2\right)=a^2+a-6=[/tex]

Step-by-step explanation:

Given expression is [tex]\left(a+3\right)\left(a-2\right)[/tex].

Now we need to multiply this using FOIL.

F = First [tex]=\left(a\right)\left(a\right)= a^2[/tex]

O = Outside [tex]=\left(a\right)\left(-2\right)= -2a[/tex]

I = Inside [tex]=\left(3\right)\left(a\right)= 3a[/tex]

L = Last [tex]=\left(3\right)\left(-2\right)= -6[/tex]

Hence we get :

[tex]\left(a+3\right)\left(a-2\right)=a^2-2a+3a-6=a^2+a-6=[/tex]

Answer:

[tex](a+3)(a-2)=a^2+a-6[/tex]

Step-by-step explanation:

The given expression is:

[tex](a+3)(a-2)[/tex]

Using FOIL, we multiply the;

First terms:[tex]a\times a=a^2[/tex]

Outside terms: [tex]a\times -2=-2a[/tex]

Inner terms:[tex]3\times a=3a[/tex]

Last terms:[tex]3\times -2=-6[/tex]

Putting all together we have:

[tex](a+3)(a-2)=a^2-2a+3a-6[/tex]

This simplifies to [tex](a+3)(a-2)=a^2+a-6[/tex]

Answer: 39$

Step-by-step explanation: calls=x x=30 30*.05= 1.50 40.50-1.50= 39$

Answer:38.55

Step-by-step explanation:

because you would make x=to 30 for the time and that would make the problem

40.05-0.05*30

0.05*30=1.5 so

40.05-1.5=38.55

Answer:It will be 2/12

Step-by-step explanation:

6 blue chips           6 red chips

6+6=12  so 2 red chips to 12 =2/12

The answer is 8x^4y^5.It has the highest degree and that is 4 in x and 5 in y.

Answer:

Step-by-step explanation:

The distribution is very simple. Using FOIL.

It states that [tex](a+b)(c+d)=ac+ad+bc+bd[/tex].

Also note that when multiplying expressions we multiply variables and values differently. If we have variables like [tex]x[/tex] their exponents will add. If we have values like 3 we multiply them normally.

For example your practise 3.

[tex](x+3)(2x^2+4)=2x^2\cdot x+4x+3\cdot2x^2+3\cdot4 \\

\underline{2x^3+4x+6x^2+12}

[/tex]

Now just order the expressions from bigger exponent to smaller and than values. (Usual notation although no need).

And solution is:

[tex]\boxed{2x^3+6x^2+4x+12}[/tex]

Hope this helps.

r3t40

Answer:

See below

Step-by-step explanation:

[tex]a\cdot(b + c) = a\cdot b + a\cdot c[/tex]

3) Practice: Organizing information

[tex]\begin{array}{lll}\qquad \textbf{Steps} & \textbf{Problem: }(x + 3)(2x^{2} + 4) & \\\textbf{1. List variables} & a = x + 3 & \\ & b = 2x^{2} & \\ & c = 4 &\\\\\textbf{2. Distribute} & (x + 3)(2x^{2} + 4)& = (x + 3)(2x^{2}) + (x + 3)(4)\\\\\textbf{3. Redistribute} & (x + 3)(2x^{2})& (x + 3)(4)\\& a = 2x^{2} & a = 4\\& b = x & b = x\\& c = 3 & c = 3\\& 2x^{3} + 6x^{2} & 4x + 12\\\textbf{4. Combine}& & \\\qquad\textbf{terms} & 2x^{3} + 6x^{2}+ 4x + 12 & \\\end{array}[/tex]

4. Practice: Summarizing

[tex]\text{You can use the FOIL method to multiply two }\underline{\text{binomials}}.\\\text{The letters in FOIL stand for }\underline{\text{First, Outer, Inner, Last}}.\\\text{The FOIL method helps you to remember how to multiply each term in one }\\\underline{\text{binomial}} \text{ by each term in the other }\underline{\text{binomial}}.[/tex]

Answer:

295

Step-by-step explanation:

.3 x 295 = 88.5

295 - 88.5 = 206.5

For this case we must find the solutions of the following quadratic equation:

[tex]x ^ 2-5x-1 = 0[/tex]

We solve by means of[tex]x = \frac {-b \pm \sqrt {b ^ 2-4 (a) (c)}} {2a}[/tex]

Where:

[tex]a = 1\\b = -5\\c = -1[/tex]

Substituting:

[tex]x = \frac {- (- 5) \pm \sqrt {(- 5) ^ 2-4 (1) (- 1)}} {2 (1)}\\x = \frac {5 \pm \sqrt {25 + 4}} {2}\\x = \frac {5\pm\sqrt {29}} {2}[/tex]

Finally, the roots are:

[tex]x_ {1} = \frac {5+ \sqrt {29}} {2}\\x_ {2} = \frac {5- \sqrt {29}} {2}[/tex]

Answer:

[tex]x_ {1} = \frac {5+ \sqrt {29}} {2}\\x_ {2} = \frac {5- \sqrt {29}} {2}[/tex]

Answer:

Step-by-step explanation:

x(7x^2+4x) can be expanded by multiplying each of the two terms inside the parentheses by x:

x(7x^2+4x) = 7x^3 + 4x^2

Multiply the numbers by 2

7*2=14

14*2= 28

28*2=56

Answer is 56

ANSWER

56

EXPLANATION

We want to find the next number in the sequence,

7,14,28,

Observe that, there is a common ratio of 2 among the terms.

That is ,

7(2)=14

14(2)=28

Therefore the next term is

28(2)=56

Therefore the number that goes into the box is 56.

Answer:

-13

Step-by-step explanation:

-4-8+3+6-10

Answer:-13

Step-by-step explanation:

Answer:

20.6 (3sf)

Step-by-step explanation:

a^2 + b^2 = c^2

5^2 + 7^2 = 74

c= root 74

5 + 7 + root 74 = 12 + root 74 = 20.6 (3sf)

Answer:

[tex]\large\boxed{m<\dfrac{71}{7}\to\left\{m\ |\ m<\dfrac{71}{7}\right\}\to m\in\left(-\infty,\ \dfrac{71}{7}\right)}[/tex]

Step-by-step explanation:

[tex]170-7m>99\qquad\text{subtract 170 from both sides}\\\\-7m>-71\qquad\text{change the signs}\\\\7m<71\qquad\text{divide both sides by 71}\\\\m<\dfrac{71}{7}[/tex]

10 because if you minus 170 from 10 it would still be less than 99

Answer:

∆AMN ≅ ∆NOA

Step-by-step explanation:

Given:

Quadrilateral AMNO

MN║AO

AM║ON

To prove:∆AMN ≅ ∆NOA

Lets first draw two diagonals represented by lines MO and AN inside the given quadrilateral AMNO

Now we know if lines are parallel then the alternate interior angles are congruent , hence

∠NMO≅∠AOM

∠MNA≅∠NAO

∠AMO≅∠NOM

∠MAN≅∠ANO

Also by Reflexive Property we have

NA≅NA

MO≅MO

From ASA congruence property of triangles that states that if two angles and a side of two triangles are congruent then the two triangle are said to be congruent, hence we have

ΔAMN≅ΔNOA

ΔMAO≅ΔONM  !

Answer:

∆AMN=∆NOA by rule SSS

Answer:

∠EFA

Step-by-step explanation:

Linear pair : A linear pair is a pair of adjacent angles formed when two lines intersect and the sum of these angles is 180°

Vertical angles: The opposite angles formed by the two intersecting lines are called vertical angles.

Option 1) ∠BFC

Line BE and CF intersect at point F

So, the two adjacent angles formed when two lines intersect are ∠BFC and ∠EFC.

These are linear pair.

So,  ∠BFC is a part of linear pair.

Now by the definition of vertical angles , ∠BFC has no vertical pair.

So, ∠BFC is not a part of vertical pair.

Option 2) ∠CFG

According to the definition of linear pair ∠CFG is not a part of linear pair.

According to the definition of vertical pair ∠CFG is not a part of vertical pair.

Option 3) ∠GFD

According to the definition of linear pair ∠GFD has a linear pair ∠AFG

Thus ∠GFD is a part of linear pair

According to the definition of vertical pair ∠GFD is not a part of vertical pair.

Option 4) ∠EFA

According to the definition of linear pair ∠EFA has a linear pair ∠EFD

Thus ∠EFA is a part of linear pair

According to the definition of vertical pair ∠EFA has a ∠BFD vertical pair.

Thus ∠EFA is a part of vertical pair.

Hence ∠EFA is part of a linear pair and part of a vertical pair.

Answer:

D) EFA

Step-by-step explanation:

Let's see the definition of linear pair and vertical angles.

A linear pair of angles are the adjacent angles, when the angles add upto 180°.

Vertical angles are the  opposite angles when the two lines are intersecting. The vertical angles are equal in measure.

In the given figure there are only two lines, they are AD and BE. Othere are just rays.

By look at the figure, ∠EFA is a linear pair to∠EFD and as well as vertical angle to ∠DFB.

Therefore, the answer is D) EFA