If 35% of your candies was 42 candies, how many candies do you have?

Answer:

124.99 if i am right pls like my answer and rate it

Step-by-step explanation:

Answer:

\(a. \: 30 + a = 40 \\ a = 40 - 30 \\ a = 10 \\ b. \: 500 + b = 200 \\ b = 200 - 500 \\ b = - 300 \\ c. \: - 1 + c = - 2 \\ c = - 2 + 1 \\ c = - 1 \\ d. \: d +3.567=0 \\ d = 0 - 3.567 \\ d = - 3.567\)

hope it helps:)

\(\displaystyle \lim_{x\to 0}~\cfrac{1-cos^2(7x)}{8x^2(~~1-sin^2(7x)~~)}~\hfill \boxed{u=7x\implies \cfrac{u}{7}=x\implies \cfrac{u^2}{49}=x^2} \\\\[-0.35em] ~\dotfill\\\\ \cfrac{1-cos^2(u)}{8\left(\frac{u^2}{49} \right)(~~1-sin^2(u)~~)}\implies \cfrac{1-cos^2(u)}{ ~~ \frac{8u^2(~~1-sin^2(u)~~)}{49} ~~ }\implies \cfrac{[1-cos^2(u)]49}{8u^2(~~1-sin^2(u)~~)} \\\\\\ \cfrac{[1-cos(u)][1-cos(u)]49}{8u^2(~~1-sin^2(u)~~)}\implies \cfrac{1-cos(u)}{u}\cdot \cfrac{49[1-cos(u)]}{8u[1-sin^2(u)]} \\\\[-0.35em] ~\dotfill\)

\(\displaystyle \lim_{x\to 0}~\cfrac{1-cos(u)}{u}\cdot \cfrac{49[1-cos(u)]}{8u[1-sin^2(u)]}\implies \lim_{x\to 0}~\cfrac{1-cos(u)}{u}\cdot \lim_{x\to 0}~\cfrac{49[1-cos(u)]}{8u[1-sin^2(u)]} \\\\\\ \displaystyle \lim_{x\to 0}~\cfrac{1-cos(7x)}{7x}\cdot \lim_{x\to 0}~\cfrac{49[1-cos(7x)]}{8(7x)[1-sin^2(7x)]} \\\\\\ \displaystyle \text{\LARGE 0}\cdot \lim_{x\to 0}~\cfrac{49[1-cos(7x)]}{8(7x)[1-sin^2(7x)]}\implies \text{\LARGE 0}\)

Answer:

The y coordinates will remain the same. Meanwhile, the x coordinates will be the opposite.

Step-by-step explanation:

Here's an example, because no picture was included. If point C was located at (3, 5), point C' would be at (-3, 5).

Hope this helps!! :)

Answer:

7/3>x

Step-by-step explanation:

-5<9-6x

subtract 9 from both sides

-14<-6x

divide by 6

14/6<x

simplified its 7/3<x

but since we divided by a negative we switch the sign

final answer is 7/3>x

Answer:

1/2

Step-by-step explanation:

20 times 1/2 = 10

10-2 = 8

Answer:

You could use a graphing calculator I just used Desmos. Origin is (0, 0). Rotate 90 degrees about the origin.

I used triangles

The rotated point would be at (1, 2)

ANSWER: the question mark = 1

The number of gallons that they use every​ day will be 200 gallons.

The amount of any product is given as though it was a proportion of a hundred. The ratio can be expressed as a quarter of 100. The phrase % translates to one hundred percent. It is symbolized by the character '%'.

The percentage is given as,

Percentage (P) = [Final value - Initial value] / Initial value x 100

People use water to​ cook, clean, and drink every day. An estimated 34.6% of the water used each day is for drinking. If a family uses 69.2 gallons of water a day for drinking.

Then the total amount of water is given as,

⇒ 69.2 / 0.346

⇒ 200 gallons

The number of gallons that they use every​ day will be 200 gallons.

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Answer:

2:3

Step-by-step explanation:

minutes per day on weekdays:minutes per day on weekends

90:135

It can be written in simpler form, because both numbers can be divided by 45

90÷2=45 and 135÷45=3

The ratio is 2:3.

Answer

the best answer is D

Answer:

I would say A is the answer.

By making x the subject of the formula in this equation u = cos(0.5x)​ gives x = 2cos⁻¹(u).

In Mathematics and Geometry, an equation can be defined as a mathematical expression which shows that two (2) or more thing are equal. This ultimately implies that, an equation is composed of two (2) expressions that are connected by an equal sign.

In this exercise, you are required to make "x" the subject of the formula in the given mathematical equation by using the following steps.

By taking the arc cosine of both sides of the equation, we have the following:

u = cos(0.5x)

cos⁻¹(u) = 0.5x

By multiplying both sides of the mathematical equation by 2, we have the following:

2 × cos⁻¹(u) = 0.5x × 2

x = 2cos⁻¹(u)

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The results of operations between vectors are, respectively:

Case A: u + w = <- 3, - 1>

Case B: - 6 · v = <6, 6>

Case C: 3 · v - 6 · w = <- 21, - 15>

Case D: 4 · w + 3 · v - 5 · u = <39, 4>

Case E: |w - v| = √(4² + 3²) = 5

In this problem we must determine the operations between vectors, this can be done by following definitions:

Vector addition

v + u = (x, y) + (x', y') = (x + x', y + y')

Scalar multiplication

α · v = α · (x, y) = (α · x, α · y)

Norm of a vector

|u| = √(x² + y²)

Now we proceed to determine the result of each operation:

Case A:

u + w = <- 6, - 3> + <3, 2>

u + w = <- 3, - 1>

Case B:

- 6 · v = - 6 · <- 1, - 1>

- 6 · v = <6, 6>

Case C:

3 · v - 6 · w = 3 · <- 1, - 1> - 6 · <3, 2>

3 · v - 6 · w = <- 3, - 3> + <- 18, - 12>

3 · v - 6 · w = <- 21, - 15>

Case D:

4 · w + 3 · v - 5 · u = 4 · <3, - 2> + 3 · <- 1, - 1> - 5 · <- 6, - 3>

4 · w + 3 · v - 5 · u = <12, - 8> + <- 3, - 3> + <30, 15>

4 · w + 3 · v - 5 · u = <39, 4>

Case E:

|w - v| = |<3, 2> - <- 1, - 1>|

|w - v| = |<4, 3>|

|w - v| = √(4² + 3²) = 5

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Answer:

0 hundreds

0 tens

7 ones

.

4 tenths

0 hundredths

8 thousandths

Standard form=7.408

Step-by-step explanation:

Lets first solve (7x1)+(4x1/10)+(8x1/1000)

7+0.4+0.008

Simplify:

7.408

The intersection points are (6, 6) and (-6, -6).

What is Intersection points?

The point at which two lines or curves intersect is referred to as the point of intersection. The point at which two curves intersect is crucial because it is the point at which the two curves take on the same value.

The given curves are the polar equations of two circles with radii 6. To find their intersection points, we can set the two equations equal to each other and solve for Ф.

6 sin(Ф) = 6 cos(Ф)

Dividing both sides by 6 and rearranging terms, we get:

tan(Ф) = 1

This equation has infinitely many solutions, but we are only interested in those that lie in the interval [0, π/2].

Ф = π/4 satisfies this condition and corresponds to the point (6, 6) in Cartesian coordinates.

Since the two curves are circles, they are symmetrical about the origin. Therefore, we can deduce that the other intersection point is (-6, -6).

Therefore, the intersection points are (6, 6) and (-6, -6).

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Answer:

2rr

Step-by-step explanation:

Answer:

The domain is the x-values from least to greatest, while the range is the y-values from least to greatest. For this problem, domain would be [3, 18] and range would be [-12, 15].

Step-by-step explanation:

Since we know domain is the x-values listed from smallest to largest, we can order our values that way:

(3, -1), (5, 15), (7, -1), (11, -12), (18, -6).

3<5<7<11<18

That means the range is (3, 18). However, since both 3 and eighteen are included in the set, we use brackets. [3,18], putting the smaller number first, followed by a comma, and then the bigger number.

We will do the same for the range.

(11, -12), (18, -6), (3, -1), (7, -1), (5, 15)

-12<-6<-1=-1<15

This means the range is (-12, 15). Since -12 and 15 are included in the set of values, we use brackets. [-12, 15].

Answer:

\(f(3)=25\)

Step-by-step explanation:

Hi there!

\(f(x)=x^3-2\)

Replace x with 3

\(f(3)=3^3-2\\f(3)=27-2\\f(3)=25\)

I hope this helps!

Rewrite the factorial parts of the summand as

\(\dfrac{(2n+1)!}{n!(n+2)!} = \dfrac{(2n+1)(2n!)}{(n+2)(n+1)(n!)^2} = \dfrac{2n+1}{(n+2)(n+1)} \dbinom{2n}n\)

where \(\binom nk\) is the binomial coefficient, and \(\binom{2n}n\) are the so-called central binomial coefficients.

Expand the rational expression into partial fractions:

\(\dfrac{2n+1}{(n+2)(n+1)} = \dfrac3{n+2} - \dfrac1{n+1}\)

Pull out a constant factor and collect the exponential terms.

\(\dfrac{(-1)^{n+1}}{4^{2n+3}} = -\dfrac1{64} \left(-\dfrac1{16}\right)^n\)

The sum we want is now

\(\displaystyle \frac{13}8 - \frac1{64} \sum_{n=0}^\infty \binom{2n}n \left(\frac3{n+2} - \frac1{n+1}\right) \left(-\frac1{16}\right)^n\)

Let f(x) and g(x) be functions with power series expansions

\(\displaystyle f(x) = \sum_{n=0}^\infty \binom{2n}n \frac{x^n}{n+1}\)

\(\displaystyle g(x) = \sum_{n=0}^\infty \binom{2n}n \frac{x^n}{n+2}\)

and recall the well-known binomial series

\(\displaystyle \dfrac1{\sqrt{1-4x}} = \sum_{n=0}^\infty \binom{2n}n x^n\)

which converges for |x| < 1/4.

Integrating both sides yields

\(\displaystyle \int \frac{dx}{\sqrt{1-4x}} = \int \sum_{n=0}^\infty \binom{2n}n x^n \, dx\)

\(\displaystyle -\frac12 \sqrt{1-4x} = C_1 + \sum_{n=0}^\infty \binom{2n}n \frac{x^{n+1}}{n+1}\)

Taking x = 0 on both sides, it follows that C₁ = -1/2. We then see that

\(\displaystyle f(x) = \frac{1-\sqrt{1-4x}}{2x}\)

Step back and multiply both sides of the binomial series identity by x, then integrate. This yields

\(\displaystyle \int \frac x{\sqrt{1-4x}} \, dx = \int \sum_{n=0}^\infty \binom{2n}n x^{n+1} \, dx\)

\(\displaystyle -\frac1{12} \sqrt{1-4x} (1 + 2x) = C_2 + C_1 x + \sum_{n=0}^\infty \binom{2n}n \frac{x^{n+2}}{n+2}\)

Taking x = 0 again points to C₂ = -1/12. Hence

\(\displaystyle g(x) = \frac{1 - \sqrt{1-4x}(1+2x)}{12x^2}\)

Then the value of the sum we want is

\(\displaystyle \frac{13}8 - \frac1{64} \left(3g\left(-\frac1{16}\right) - f\left(\frac1{16}\right)\right) = \frac{1+\sqrt5}2 = \boxed{\phi}\)

where ɸ ≈ 1.618 is the golden ratio.

Answer: \(7.39466666\)

Step-by-step explanation:

Setting \(x=2.2\) and \(n=3\),

\(e^{2.2} \approx \sum^{3}_{k=0} \frac{2.2^{k}}{k!}=\frac{2.2^0}{0!}+\frac{2.2^1}{1!}+\frac{2.2^2}{2!}+\frac{2.2^3}{3!} \approx 7.39466666\)

Answer:

7.39466667 (8 d.p.)

Step-by-step explanation:

We can approximate the value of eˣ for any x using the following sum formula:

\(\boxed{e^{x} \approx \displaystyle \sum^n_{k=0}\dfrac{x^k}{k!}}\)

To approximate \(e^{2.2}\) with n = 3, substitute x = 2.2 and n = 3 into the given sum formula:

\(e^{2.2} \approx \displaystyle \sum^3_{k=0}\dfrac{2.2^k}{k!}\)

To calculate the sum, substitute k with each value from 0 to 3 and add the results together:

\(\begin{aligned}e^{2.2} &\approx \displaystyle \sum^3_{k=0}\dfrac{2.2^k}{k!}\\\\&= \dfrac{2.2^0}{0!}+\dfrac{2.2^1}{1!}+\dfrac{2.2^2}{2!}+\dfrac{2.2^3}{3!}\\\\&= \dfrac{1}{1}+\dfrac{2.2}{1}+\dfrac{4.84}{2}+\dfrac{10.648}{6}\\\\&= 1+2.2+2.42+1.774666666...\\\\&=7.39466667\; \sf (8\;d.p.)\end{aligned}\)

Therefore, the approximate value of \(e^{2.2}\) is:

\(\large \textsf{$e^{2.2}$}=\boxed{7.39466667}\; \sf (8\;d.p.)}\)

Note: To obtain a more accurate approximate value, increase the value of n.

Answer:

53.12

Step to Step reasoning:

8.3 x 6.4 = 53.12

First of all, let us find the value of x

Notice that the sum of central angles 4x and (2x+24) must be equal to 180° (half of the entire circle)

So, the value of x is 26

The arc UY is given by

So, the arc UY is 76°

The arc VW must be equal to the arc UY since their central angles are vertically opposite angles.

So, the arc VW is 76°

The arc WX must be half of the arc UV

So, the arc WX is 52°

Finally, the arc WUY is given by

So, the arc WUY is 256°

Therefore, the arcs are

By the Mean Value Theorem, there exists a number c in (1, 7) such that ƒ'(c) = 2/c and 2/c = ln7 / 6, and c ≈ 0.909.

The Mean Value Theorem (MVT) states that if a function ƒ(x) is continuous on the closed interval [a, b] and differentiable on the open interval (a, b), then there exists a number c in (a, b) such that:

ƒ'(c) = [ ƒ(b) - ƒ(a) ] / (b - a)

ƒ(x) = 2lnx - 8 is continuous on the closed interval [1, 7], since it is the sum and composition of continuous functions.

ƒ(x) = 2lnx - 8 is differentiable on the open interval (1, 7), since its derivative ƒ'(x) = 2/x is defined and continuous on (1, 7).

Therefore, the Mean Value Theorem can be applied to ƒ(x) on [1, 7]. To find the value of c, we need to solve the equation:

ƒ'(c) = [ ƒ(b) - ƒ(a) ] / (b - a)

Substituting the given values, we get:

2/c = [ 2ln7 - 2ln1 ] / (7 - 1)

2/c = ln7

c = 2 / ln7

c ≈ 0.909

Therefore, by the Mean Value Theorem, there exists a number c in (1, 7) such that ƒ'(c) = 2/c and 2/c = ln7 / 6, and c ≈ 0.909.

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No, the trapezoids are not similar because their corresponding angles do not have the same measures.

The given trapezoids ABCD and EFGH can be analyzed to determine if they are similar. Similarity in geometric figures implies that their corresponding angles are congruent, and their corresponding sides are proportional.

Let's examine the angles of the trapezoids:

In trapezoid ABCD:

Angle A = 137 degrees

Angle B = 90 degrees

Angle C = 90 degrees

Angle D = 43 degrees

In trapezoid EFGH:

Angle E = 136 degrees

Angle F = 90 degrees

Angle G = 90 degrees

Angle H = 44 degrees

By comparing the angles, we can observe that angles B and F are both right angles (90 degrees) in both trapezoids. Additionally, angles C and G are also right angles (90 degrees) in both trapezoids.

However, angles A and E, as well as angles D and H, do not have the same measures. Angle A measures 137 degrees in trapezoid ABCD, while angle E measures 136 degrees in trapezoid EFGH. Similarly, angle D measures 43 degrees in trapezoid ABCD, while angle H measures 44 degrees in trapezoid EFGH.

Since the corresponding angles in the two trapezoids do not have the same measures, we can conclude that trapezoids ABCD and EFGH are not similar.

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Answer:

Step-by-step explanation:

Given is an absolute value function f(x) = |x|

Its vertex is at origin, the function is decreasing at negative values of x and increasing at positive values of x:

Correct answer choice is A

Answer: ∠16 and ∠11

Step-by-step explanation:

      All of these answer options include ∠16, so we know we're looking for an angle that is corresponding to ∠16. A corresponding angle is an angle that is in the same relative position. We will look at ∠9, ∠11, ∠2, and ∠12 since those are the given answer options, and see which is corresponding.

The correct corresponding angles are ∠16 and ∠11.

Liliana is right. A tuple of numbers a, b, c, and d that has the formula a2 + b2 + c2 = d2 is known as a Pythagorean quadruple.

Liliana is right. A tuple of numbers a, b, c, and d that has the formula a2 + b2 + c2 = d2 is known as a Pythagorean quadruple.

They are Diophantine equation solutions, and frequently, only positive integer values are taken into account.

Pythagorean triples are limitless in number. Anytime 2 n + 1 is a square, a Pythagorean triple is formed.

There exists an endless number of Pythagorean triples since 2 n +1 includes all odd integers, every other square number is odd, and there are an unlimited number of odd squares.

(4, 5, 20, 21} should work, or (1, 2, 2, 3}.

\(4^{2} + 5^{2} + 20^{2} = 21^{2}\)

\(1^{2} + 2^{2} + 2^{2} = 3^{2}\)

so Liliana is right.

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