How do you solve this?

The 4-digit positive integers have four different digits where the leading digit is not zero, the integer is a multiple of 5, and 5 is the largest digit is 108.

[tex]5*4*3*1+4*4*3*1=60+28=108[/tex]

Positive integers are the numbers we use to depend: 1,2, 3,4, 1 , 2, three, four, and so forth. Numbers with a fractional component not identical to zero and poor numbers are excluded from a fixed of high-quality integers. The operations of addition, subtraction, multiplication, and division can be finished with tremendous integers.

Positive integers are definitely your counting numbers. wonderful integers are in reality a part of a larger institution of numbers referred to as integers. Integers are all of the whole numbers, each advantageous and poor. by complete numbers, we imply numbers without fractions or decimals.

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We make four cases based on where the multiple of the 5 digits (0 or5) is. The number has to end with either 5 or 0 since it's a multiple of5. In all but the last case, 5 and 0 are used at the end and in another spot that separates the cases.

Case 1: The first digit can't be 0 so it must be 5. There are 4.3 to choose the middle two digits. After that, the last digit has to be 0, so there are a total of 1.4.3.1=12 numbers.

Case 2: The second digit can be 5, leaving 2 choices. The first and third numbers can be chosen in 4.3 ways, like last time. The last digit has to be 0 or 5, but not the one we already used. There are a total of 4.2.3.1=24 numbers.

Case 3: There are the same choices, but the digits 0 and 5 are at the last and second-to-last spots. So there are 4.2.3.1 =24 numbers again.

Case 4: There are 4.3.2 ways to choose the first three numbers. There has to be a 5 in the number because the largest digit is 5. Coincidentally, there are 4.2.3.1 numbers again.

There are a total of 12 + 24.3= (D)84 numbers.

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[tex]x^3-2x^2+x-1[/tex] is one of the prime factors of the polynomial

The question implies that we determine one of the prime factors of the polynomial.

The polynomial is given as:

[tex]x^8 - 3x^6 + x^4 - 2x^3 - 1[/tex]

Expand the polynomial by adding 0's in the form of +a - a

[tex]x^8 - 3x^6 + x^4 - 2x^3 - 1 = x^8 -2x^7 + 2x^7 - 4x^6 +x^6 + 2x^5 -2x^5- 3x^4 + 4x^4 + 2x^3 -6x^3+2x^3- x^2 -3x^2 +4x^2-2x+2x-1[/tex]

Rearrange the terms

[tex]x^8 - 3x^6 + x^4 - 2x^3 - 1 = x^8 -2x^7 + 2x^5 - 3x^4 + 2x^3 - x^2 + 2x^7 - 4x^6 + 4x^4 -6x^3+4x^2-2x+x^6-2x^5+2x^3-3x^2+2x-1[/tex]

Factorize the expression

[tex]x^8 - 3x^6 + x^4 - 2x^3 - 1 = x^2(x^6-2x^5+2x^3-3x^2+2x-1) + 2x(x^6-2x^5+2x^3-3x^2+2x-1) + 1(x^6-2x^5+2x^3-3x^2+2x-1)[/tex]

Factor out x^6-2x^5+2x^3-3x^2+2x-1

[tex]x^8 - 3x^6 + x^4 - 2x^3 - 1 = (x^2+2x + 1)(x^6-2x^5+2x^3-3x^2+2x-1)[/tex]

Express x^2 + 2x + 1 as a perfect square

[tex]x^8 - 3x^6 + x^4 - 2x^3 - 1 = (x+1)^2(x^6-2x^5+2x^3-3x^2+2x-1)[/tex]

Expand the polynomial by adding 0's in the form of +a - a

[tex]x^8 - 3x^6 + x^4 - 2x^3 - 1 = (x+1)^2(x^6- 2x^5+x^4-x^4-x^3 +x^3-2x^3-x^2 -2x^2 +x+x - 1)[/tex]

Rearrange the terms

[tex]x^8 - 3x^6 + x^4 - 2x^3 - 1 = (x+1)^2(x^6- 2x^5+x^4-x^3-x^4-2x^3-x^2+x+x^3-2x^2 +x - 1)[/tex]

Factorize the expression

[tex]x^8 - 3x^6 + x^4 - 2x^3 - 1 = (x+1)^2(x^3(x^3-2x^2+x-1) -x(x^3-2x^2+x-1)+1(x^3-2x^2+x-1))[/tex]

Factor out x^3-2x^2+x-1

[tex]x^8 - 3x^6 + x^4 - 2x^3 - 1 = (x+1)^2(x^3 -x+1)(x^3-2x^2+x-1)[/tex]

One of the factors of the above polynomial is [tex]x^3-2x^2+x-1[/tex].

This is the same as the option (c)

Hence, [tex]x^3-2x^2+x-1[/tex] is one of the prime factors of the polynomial

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Answer:

[tex](8n - 5)(n-1)[/tex]

Step-by-step explanation:

Hello!

We can factor this by grouping. We have to find two numbers that add up to -13 but multiply to 8 * 5 .

The two numbers that work are -8 and -5. Expand -13n to -8n and -5n.

The factored expression is [tex](8n - 5)(n-1)[/tex].

Answer:

Step-by-step explanation:

Answer:

(x² + 2)(2x - 5)

Step-by-step explanation:

assuming the expression has to be factorised

2x³ - 5x² + 4x - 10 ← rearranging terms as indicated

= 2x³ + 4x - 5x² - 10 ( factor the first/second and third/fourth terms )

= 2x(x² + 2) - 5(x² + 2) ← factor out (x² + 2) from each term

= (x² + 2)(2x - 5) ← in factored form

The area of the shaded region shown in the figure is 80π unit²

An equation is an expression that shows the relationship between two or more variables and numbers.

Area of the shaded region = area of semicircle IL - area of semicircle JK + area of semicircle IJ + area of semicircle KL

Area of the shaded region = area of semicircle IL + area of semicircle IJ = (0.5 * π * 12²) + (0.5 * π * (12/3)²) = 80π

The area of the shaded region shown in the figure is 80π unit²

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The value of the stamp collection in terms of b is; 45b + 43 cents

We are given;

If there are x 3-cent stamps, then there are 4x number 10-cent stamps

So the number of 2-cent stamps is 1 less than 3 cent stamps and that will be x - 1, 2 cent stamps

Number of 2 cents stamps = b

Thus;  x - 1 = b

x = b + 1

The value of the stamp collection is as follows;

x number of 3 cents stamps = 3(b + 1) = 3b + 3 cents

4x number of 10 cents stamps = 10(4(b + 1)) = 10(4b + 4) = 40b + 40

The value of the 2 cents stamp = 2b cents

Total value is;

2b + 40b + 40 + 3b + 3 = (45b + 43) cents

Complete Question is;

I have 4 times as many 10-cent stamps as 3-cent stamps. I have one less 2-cent stamp than 3 cent stamp. If I have b2-cent stamps, what is the value of my stamp collection in terms of b ?

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An isosceles triangle is one with two equal-length sides. The correct option is C.

An isosceles triangle is one with two equal-length sides. It is sometimes stated as having exactly two equal-length sides, and sometimes as having at least two equal-length sides, with the latter form containing the equilateral triangle as a particular case.

The diagram as per the given conditions is drawn below.

In ΔUTZ, since the sides UZ and TZ are equal because the point z is equidistance from T and U, therefore, the triangle is an isosceles triangle. Thus, ∠BTZ≅TZB.

Hence, the correct option is C.

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Answer:

1/2

Step-by-step explanation:

The dimensions are cut in half ....scale factor 1/2

Answer:

Step-by-step explanation:

Let the distance between docks be d.

This is the opposite side to 45° angle of triangle with other sides 400 ft and 500 ft.

Use the law of cosines to find the value of d:

Answer:

Answer is 32

Step-by-step explanation:

150-22=128

128÷4=32

Answer: (99/13, 150/13)

Step-by-step explanation:

I'm too lazy to explain so here's a screen shot from desmos.

Answer:

(8,3),(8,3),(-8,-3),(-8,-3)

Step-by-step explanation:

A function is defined as a relation in which each domain element (x-value) is paired with exactly one range element (y-value).

This means that each value we input for x must have only one possible output. If the same x-value yields more than one y-value, it is not a function.

The correct answer, as shown above, meets the criteria of a function, as each x-value is shown to produce exactly one y-value. In this set of ordered pairs, if x=8, then y=3, and if x=-8, then x=-3.

In the other sets of ordered pairs, however, one x-value has more than one possible y-value. Let's use the first set as an example:

(5,1),(5,2),(5,3),(5,4)

Inputting 5 as x produces four different y values; there is no one y value for the x-value. y could equal 1, 2, 3, or 4.

Step-by-step explanation:

so, let me get this straight :

s = sqrt(sa/6)

this is because the surface area of a cube consists of 6 squares (as everybody knows that dice have 6 equal sides).

and the area of a single square is side².

for a cube with sa = 180 m² we have

s = sqrt(180/6) = sqrt(30)

for a cube with sa = 120 m² we have

s = sqrt(120/6) = sqrt(20) = sqrt(4×5) = 2×sqrt(5)

so, the side length of the larger cube is

sqrt(30) - 2×sqrt(5) meters

longer than the side length of the smaller cube.

in other words, the difference between both side lengths.

Answer:

answer above is correct, its B

Step-by-step explanation:

Answer:

Step-by-step explanation:

Comment

You don't have to do anything to the 6 feet. It already is in feet. It's the 2 inches you have to worry about.

2 inches = 2/12 of a foot.

2/12 = 1/6 = 0.167 feet

Answer

6 feet 2 inches = 6.167  feet

Step-by-step explanation:

Assuming the salary is included in the job descriptions, then they all pay the same amount for the same length of time, meaning he will earn the same amount at any of the jobs. This looks like a trick question.

Answer:

last option : vertex is 2, -1 . axis of symmetry is x = 2

Answer:

D

(2 -1)

*(^_^)*

...............

Answer:

Step-by-step explanation:

Answer:10.2km

He took the route eastwards then southwards

Answer:

20 sq in

Step-by-step explanation:

formula is (base * height)/2 so plug it in:

5*8=40

40/2=20

20 sq in

36 miles is the fewest number of miles she can walk in February

There are 28 days in February.

Every third day is 28/3 = 9.333 = 9 days of walking.

since she walks four miles,

9*4 = 36

Hence, 36 miles is the fewest number of miles she can walk in February

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Answer:

Projected height = (85 *21) / 35 = 51 cm

Step-by-step explanation:

Since the images are similar, we can set up the proportional equation like this:

Slide width / Slide height = Projected width / Projected height

35 mm / 21 mm = 85 cm / Projected height (in cm)

Projected height = (85 *21) / 35 = 51 cm

(i) Given that

[tex]\tan^{-1}(x) + \tan^{-1}(y) + \tan^{-1}(xy) = \dfrac{7\pi}{12}[/tex]

when [tex]x=1[/tex] this reduces to

[tex]\tan^{-1}(1) + 2 \tan^{-1}(y) = \dfrac{7\pi}{12}[/tex]

[tex]\dfrac\pi4 + 2 \tan^{-1}(y) = \dfrac{7\pi}{12}[/tex]

[tex]2 \tan^{-1}(y) = \dfrac\pi3[/tex]

[tex]\tan^{-1}(y) = \dfrac\pi6[/tex]

[tex]\tan\left(\tan^{-1}(y)\right) = \tan\left(\dfrac\pi6\right)[/tex]

[tex]\implies \boxed{y = \dfrac1{\sqrt3}}[/tex]

(ii) Differentiate [tex]\tan^{-1}(xy)[/tex] implicitly with respect to [tex]x[/tex]. By the chain and product rules,

[tex]\dfrac d{dx} \tan^{-1}(xy) = \dfrac1{1+(xy)^2} \times \dfrac d{dx}xy = \boxed{\dfrac{y + x\frac{dy}{dx}}{1 + x^2y^2}}[/tex]

(iii) Differentiating both sides of the given equation leads to

[tex]\dfrac1{1+x^2} + \dfrac1{1+y^2} \dfrac{dy}{dx} + \dfrac{y + x\frac{dy}{dx}}{1+x^2y^2} = 0[/tex]

where we use the result from (ii) for the derivative of [tex]\tan^{-1}(xy)[/tex].

Solve for [tex]\frac{dy}{dx}[/tex] :

[tex]\dfrac1{1+x^2} + \left(\dfrac1{1+y^2} + \dfrac x{1+x^2y^2}\right) \dfrac{dy}{dx} + \dfrac y{1+x^2y^2} = 0[/tex]

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

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

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

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

From part (i), we have [tex]x=1[/tex] and [tex]y=\frac1{\sqrt3}[/tex], and substituting these leads to

[tex]\dfrac{dy}{dx} = -\dfrac{\left(1 + \frac13 + \frac1{\sqrt3} + \frac1{\sqrt3}\right) \left(1 + \frac13\right)}{\left(1 + \frac13 + 1 + \frac13\right) \left(1 + 1\right)}[/tex]

[tex]\dfrac{dy}{dx} = -\dfrac{\left(\frac43 + \frac2{\sqrt3}\right) \times \frac43}{\frac83 \times 2}[/tex]

[tex]\dfrac{dy}{dx} = -\dfrac13 - \dfrac1{2\sqrt3}[/tex]

as required.

Answer:

Step-by-step explanation:

You need to find the set of points that will yield a slope that is the negative reciprocal of the slope of Line L because perpendicular lines have negative reciprocal slopes. The negative reciprocal of 13/7 is -7/13. Which set of points will produce this result? The formula for finding the slope is:

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

Consider the second set of coordinates.

(2 - (-5))/(-7 - 6) = (2 + 5)/(-13) = -7/13

The second set of coordinates satisfy the condition.

Answer:

[tex]x=10^{4}[/tex]

Step-by-step explanation:

An exponential base is the inverse of a logarithm with the same base.  Given the equation log x = 4, note that the base of the logarithm isn't written in (it has no subscript).  By default, the base of the logarithm function is base-10.  So, to rewrite the equation using an exponential, we need to undo the logarithm with an exponential of the same base.

[tex]\log(x)=4[/tex]

[tex]10^{\log{(x)}}=10^{4}[/tex]

[tex]x=10^{4}[/tex]

Think about the equation x + 7 = 10.

I could tell you this is the equation in "addition form".  What if I asked you to write an equation in "subtraction form"?  While neither "addition form" nor "subtraction form" have been defined explicitly, one could undo the addition, and get an equation with subtraction in it, which one could argue is a "subtraction form" of the equation.

[tex]x+7=10[/tex]

Subtracting 7 from both sides to undo the addition...

[tex](x+7)-7=(10)-7[/tex]

Simplifying the left side since the "+7" and "-7" cancel...

[tex]x=10-7[/tex]

This is arguably in a "subtraction form" (there's a subtraction in it), whereas the original equation had addition in it.

While we could simplify this particular equation's right-hand side, we might not always be able to (what if the 10 had been a "y"... then the "y" and the "7" aren't like terms, and they would have to remain separate)

Similarly, in the logarithm problem, one could simplify 10^4 (it's 10,000), but one doesn't have to, and one won't always be able to.

For the logarithm problem, x=10^4 is the exponential form, as requested.

Answer:

[tex]y=5[/tex]

Step-by-step explanation:

[tex]\textsf{If }y \textsf{ is \underline{inversely proportional} to the square root of }x, \textsf{ then}:[/tex]

[tex]y \propto\dfrac{1}{\sqrt{x}} \implies y=\dfrac{k}{\sqrt{x}}\quad \textsf{(where k is some constant)}[/tex]

[tex]\textsf{When }x=25, y = 3:[/tex]

[tex]\implies 3=\dfrac{k}{\sqrt{25}}[/tex]

[tex]\implies 3=\dfrac{k}{5}[/tex]

[tex]\implies k=15[/tex]

Inputting the found value of k into the equation:

[tex]\implies y=\dfrac{15}{\sqrt{x}}[/tex]

To find the value of y when x is 9, substitute x = 9 into the found equation:

[tex]\implies y=\dfrac{15}{\sqrt{9}}[/tex]

[tex]\implies y=\dfrac{15}{3}[/tex]

[tex]\implies y=5[/tex]

The description below proves that the perpendicular drawn from the vertex angle to the base bisect the vertex angle and base.

Let ABC be an isosceles triangle such that AB = AC.

Let AD be the bisector of ∠A.

We want to prove that BD=DC

In △ABD & △ACD

AB = AC(Thus, △ABC is an isosceles triangle)

∠BAD =∠CAD(Because AD is the bisector of ∠A)

AD = AD(Common sides)

By SAS Congruency, we have;

△ABD ≅ △ACD

By corresponding parts of congruent triangles, we can say that; BD=DC

Thus, this proves that the perpendicular drawn from the vertex angle to the base bisect the vertex angle and base.

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The expression of 2B - 3C in standard form is -3n^2 + 24n - 2

Given the following expressions

B = 3n - 10

C = n^2-6n-6

Required

2B - 3C

Substitute

2B - 3C = 2(3n -10) - 3(n^2-6n-6)

2B - 3C = 6n - 20 -3n^2+18n+18

2B - 3C = -3n^2 + 24n - 2

Hence expression of 2B - 3C in standard form is -3n^2 + 24n - 2

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The expression that represents the width is w = (P-2l)/2

Given the formula for calculating the perimeter of a triangle expressed as:

P = 2(l + w)

where

l is the length

w is the width

Make w the subject of the formula

Given

P = 2(l + w)

Expand

P = 2l + 2w

Subtract 2l from both sides

P - 2l = 2w

Divide both sides by2

2w/2 = (P-2l)/2

w = (P-2l)/2

Hence the expression that represents the width is w = (P-2l)/2

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Answer:

Yes, it is a linear equation.

Step-by-step explanation:

   An equation of the form ax + by +c = 0, where x and y are two variables and a, b,c are the non-zero real numbers is called a linear equation and the degree of the equation is 1.

       y = 1 - x

x + y - 1 = 0

Where a = 1 ; b=1 and c = -1

So, y = 1 -x is a linear equation.