Ann deposited $4000into an account with 3.6% interest, compounded annually. Assuming that no withdrawals are made, how much will she have in the account after 9 years?

Let's see

[tex]\\ \rm\rightarrowtail (fog)(x)=f(g(x))[/tex]

[tex]\\ \rm\rightarrowtail f(5x-4)=20x+p[/tex]

[tex]\\ \rm\rightarrowtail 4(5x-4)+3=20x+p[/tex]

[tex]\\ \rm\rightarrowtail 20x-16+3=20x+p[/tex]

[tex]\\ \rm\rightarrowtail -13=p[/tex]

Answer:

[tex]\sf p = -13[/tex]

Step-by-step explanation:

Given functions:

[tex]\sf f(x)=4x+3[/tex]

[tex]\sf g(x)=5x-4[/tex]

[tex]\sf fg(x)=20x+p[/tex]

[tex]\sf fg(x)=f[g(x)][/tex]

This means g(x) is substituted for x in f(x).

[tex]\sf \implies f[g(x)]=4[g(x)]+3[/tex]

                 [tex]\sf =4(5x-4)+3[/tex]

                  [tex]\sf= 20x-16+3[/tex]

                  [tex]\sf =20x-13[/tex]

[tex]\sf f[g(x)]=f[g(x)][/tex]

[tex]\sf \implies 20x+p=20x-13[/tex]

[tex]\sf \implies p=-13[/tex]

The amount of money that Han saves by using a coupon is $1.5%

The percentage off Clare's purchase with the coupon is 25%

A coupon reduces the price at which an item is sold

Amount saved = coupon discouunt x normal price

10% x $15

0.1 x $15 = $1.5

Percent off = (6/24 x 100  = 25%

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Answer:

13+26=39

Step-by-step explanation:

the 7th odd number is 13 and 13th even is 26

Answer:

[tex]\displaystyle 39[/tex]

Step-by-step explanation:

[tex]\displaystyle 39 = 13 + 26 \Rightarrow 39 = 13 + 2[13][/tex]

On a hundred chart, you will see that the seventh odd number is [tex]\displaystyle 13,[/tex]and the thirteenth even number is self-explanatory, so you should know what occurs there.

I am joyous to assist you at any time.

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

As we know ~

Area of the circle is :

[tex]\qquad \sf  \dashrightarrow \:\pi {r}^{2} [/tex]

And radius (r) = diameter (d) ÷ 2

[ radius of the circle = half the measure of diameter ]

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[tex]\qquad \sf  \dashrightarrow \:r = d \div 2[/tex]

[tex]\qquad \sf  \dashrightarrow \:r = 4.4\div 2[/tex]

[tex]\qquad \sf  \dashrightarrow \:r = 2.2 \: mm[/tex]

Now find the Area ~

[tex]\qquad \sf  \dashrightarrow \: \pi {r}^{2} [/tex]

[tex]\qquad \sf  \dashrightarrow \:3.14 \times {(2.2)}^{2} [/tex]

[tex]\qquad \sf  \dashrightarrow \:3.14 \times {4.84}^{} [/tex]

[tex]\qquad \sf  \dashrightarrow \:area \approx 15.2 \: \: mm {}^{2} [/tex]

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[tex]\qquad \sf  \dashrightarrow \:r = d \div 2[/tex]

[tex]\qquad \sf  \dashrightarrow \:r = 3.7 \div 2[/tex]

[tex]\qquad \sf  \dashrightarrow \:r = 1.85 \: \: cm[/tex]

Bow, calculate the Area ~

[tex]\qquad \sf  \dashrightarrow \: \pi {r}^{2} [/tex]

[tex]\qquad \sf  \dashrightarrow \:3.14 \times (1.85) {}^{2} [/tex]

[tex]\qquad \sf  \dashrightarrow \:3.14 \times 3.4225 {}^{} [/tex]

[tex]\qquad \sf  \dashrightarrow \:area \approx 10.75 \: \: cm {}^{2} [/tex]

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[tex]\qquad \sf  \dashrightarrow \:\pi {r}^{2} [/tex]

[tex]\qquad \sf  \dashrightarrow \:3.14 \times (8.3) {}^{2} [/tex]

[tex]\qquad \sf  \dashrightarrow \:3.14 \times 68.89[/tex]

[tex]\qquad \sf  \dashrightarrow \:area \approx216.31 \: \: cm {}^{2} [/tex]

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[tex]\qquad \sf  \dashrightarrow \:r = d \div 2[/tex]

[tex]\qquad \sf  \dashrightarrow \:r = 5.8 \div 2[/tex]

[tex]\qquad \sf  \dashrightarrow \:r = 2.9 \: \: yd[/tex]

now, let's calculate area ~

[tex]\qquad \sf  \dashrightarrow \:3.14 \times {(2.9)}^{2} [/tex]

[tex]\qquad \sf  \dashrightarrow \:3.14 \times 8.41 [/tex]

[tex] \qquad \sf  \dashrightarrow \:area \approx26.41 \: \: yd {}^{2} [/tex]

・ ．━━━━━━━†━━━━━━━━━．・

[tex]\qquad \sf  \dashrightarrow \:r = d \div 2[/tex]

[tex]\qquad \sf  \dashrightarrow \:r = 1 \div 2[/tex]

[tex]\qquad \sf  \dashrightarrow \:r = 0.5 \: \: yd[/tex]

Now, let's calculate area ~

[tex]\qquad \sf  \dashrightarrow \:\pi {r}^{2} [/tex]

[tex]\qquad \sf  \dashrightarrow \:3.14 \times (0.5) {}^{2} [/tex]

[tex]\qquad \sf  \dashrightarrow \:3.14 \times 0.25[/tex]

[tex]\qquad \sf  \dashrightarrow \:area \approx0.785 \: \: yd {}^{2} [/tex]

・ ．━━━━━━━†━━━━━━━━━．・

[tex]\qquad \sf  \dashrightarrow \:\pi {r}^{2} [/tex]

[tex]\qquad \sf  \dashrightarrow \:3.14 \times {(8)}^{2} [/tex]

[tex]\qquad \sf  \dashrightarrow \:3.14 \times 64[/tex]

[tex]\qquad \sf  \dashrightarrow \:area = 200.96 \: \: yd {}^{2} [/tex]

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Answer:

268

Step-by-step explanation:

Divide the figure into 3 pieces, a triangle, a rectangle, and another rectangle.

Triangle area=18*(15+6+1) * 1/2, or 189

Rectangle 1= 4*11= 44

Rectangle 2= 7*5= 35

Add all up, get 268

Answer:

find the product of (4m-n)and(3m-2n)

Answer:

1

Step-by-step explanation:

the number of circles which can pass through three given non-collinear points is exactly one.

the reason is that the center of such a circle must be on the (perpendicular) bisectors of the lines between each pair of these points.

these bisectors intersect at one unique point which is the center of the circle and the distance of any point from the center is the radius. So given three non collinear points fixes the center and the radius thereby giving us one unique circle.

the same way we find the circumscribing circle of a triangle, which is exactly the same situation.

Answer:

x = 9/10

Step-by-step explanation:

This problem features a ratio: 4x/3 = 6/5

By cross multiplying you get that 4x*5 = 3*6 or 20x = 18. By dividing both sides by 20, you get that x = 18/20, and when simplified, 9/10.

Answer:

I have no idea what are you talking about......

Answer:

V = 8 cm³

Step-by-step explanation:

the volume (V) of a cube is calculated as

V = s³ ( s is the side length )

here s = 2 , then

V = 2³ = 8 cm³

Answer:

8 cm

Step-by-step explanation:

Answer:

4 084 080 ways

Step-by-step explanation:

The number of ways is :

[tex]C^{3}_{18}\times C^{6}_{15}\times C^{9}_{9}[/tex]

___________

Since

  18C3=816

  15C6=5 005

  9C9=1

then

[tex]C^{3}_{18}\times C^{6}_{15}\times C^{9}_{9}=816\times5005\times1= 4\ 084\ 080[/tex]

Answer:

craig had played approx 5 games..

Step-by-step explanation:

25 is the amount which craig have. 25-5.2=19.9 and when we divide 19.9 with 4 ie. 19.9÷4 = 4.9

Step-by-step explanation:

C = 2 Pi R = 220

R = 220 / 2 pi = 35 cm

diameter = 35 x 2 = 70 cm

Area = pi r² = pi 35² = 3848.45 cm

The 3,572.1 m³ volume of the hexagon and the 19 m. length of the cars and 3-m connector, gives;

From a similar question, we have;

Side length of the hexagon = 8.1 m

Perpendicular distance from the center to a side of the hexagon = 7 m.

Therefore;

Cross sectional area of the hexagon, A is found as follows;

A = 6 × 0.5 × 7 × 8.1 = 170.1

Length of the tunnel, D = 3572.1 ÷ 170.1 = 21

D = 21 meters

Length of two cars and a connector, L = 8 + 8 + 3 = 19

The tunnel length, D = 21 m. is longer than the length of two cars and the connector, L = 19 m.

Therefore;

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Answer:

height: 125m; distance: 500m

Step-by-step explanation:

x = 250

f(x) = x − 0.002x2

f(x) = 250 − 0.002(250 )2

= 250 − 0.002(62,500)

= 250 − 125

= 125

250 times 2 = 500

[tex]\\ \rm\rightarrowtail y=x-0.002x^2[/tex]

So

[tex]\\ \rm\rightarrowtail y=250-0.002(250)^2[/tex]

[tex]\\ \rm\rightarrowtail y=250-0.002(62500)[/tex]

[tex]\\ \rm\rightarrowtail y=250-125[/tex]

[tex]\\ \rm\rightarrowtail y=125m[/tex]

Distance:-

Answer:

true because the center of the triangle is in the middle

Answer:

long side/length is 63 cm and short side/height is 28 cm

Step-by-step explanation:

18 x 3.5 and 8 x 3.5

To dilate means to grow so you grow the side by 3.5

Answer:

Done. Check comments

Step-by-step explanation:

Inscribed angles are half the measure of the intercepted arcs. This example shows major arc BEA being intercepted by the supplement of 74, which is 106. If you double that, you get the arc, 212.

The next example shows 80 and 154 are intersected by two chords. The angle created by that intersection is the average of the two intercepted arcs. 117.

When two secant lines intersect, they form an angle that is half the measure of the DIFFERENCE of the intercepted arcs. 145 - 53 = 92/2 = 46.

This is similar to the previous one. Except the difference is between 263 and 97. ACTUALLY, I MADE AN ERROR on the comment about this one.

263-97 = 166/2 = 83***

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

Let's solve ~

Assume width of rectangle be " x ", length = 3×width + 8 = 3x + 8 ~

Now, Perimeter of rectangle is :

[tex]\qquad \sf  \dashrightarrow \:2(l + w) = 56[/tex]

[tex]\qquad \sf  \dashrightarrow \:2(3x + 8 + x) = 56[/tex]

[tex]\qquad \sf  \dashrightarrow \:2(4x + 8) = 56[/tex]

[tex]\qquad \sf  \dashrightarrow \:4x + 8 = 56 \div 2[/tex]

[tex]\qquad \sf  \dashrightarrow \:4x + 8 = 28[/tex]

[tex]\qquad \sf  \dashrightarrow \:4x = 28 - 8[/tex]

[tex]\qquad \sf  \dashrightarrow \:x = 20\div 4[/tex]

[tex]\qquad \sf  \dashrightarrow \:x =5 \: cm[/tex]

Hence, width = x = 5 cm

[tex]\qquad \sf  \dashrightarrow \:l = 3w + 8[/tex]

[tex]\qquad \sf  \dashrightarrow \:l = 3(5)+ 8[/tex]

[tex]\qquad \sf  \dashrightarrow \:l = 15+ 8[/tex]

[tex]\qquad \sf  \dashrightarrow \:l =23 \:cm[/tex]

And, length = 26 cm

Answer:

[tex] \sf \: Proved \: \angle \: OAD \: = \angle \: OCD[/tex]

Step-by-step explanation:

Given:

Mn is diameter of circle having centre O

and BD = OD,

To prove that:

[tex]\angle \: OAD \: = \angle \: OCD[/tex]

Solution:

Join the points O and B and draw OB,

On joining the line,

in ∆OCD and ∆OBD,

OC =OB → (Radius of same circle)

BD =CD → (from given)

OD =OD → (Common side in both the triangles)

Hence ∆OCD and ∆OBD are congruent from SSS property.

so we can say that,

[tex]\angle \: OBD \: = \angle \: OCD[/tex]

Consider above prove as statement A

Corresponding angles of congruent traingle.

in ∆ OAB,

OA = OB (radius of same circle)

hence ∆OAB is an isosceles traingle.

We know that opposite angle of isosceles traingle are always equal. hence,

[tex]\angle \: OBD \: = \angle \: OAB \\ \angle \: OAB \: = \angle \: OAD (same \: angles) \\ \angle \: OBD \: = \angle \: OAD[/tex]

Consider above prove as statement B

From Statement A & B we can say that

[tex]\angle \: OAD \: = \angle \: OCD[/tex]

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Answer:

[tex]\sf 3^{-3x}=3^{(-4x+6)}[/tex]

[tex]\sf x=6[/tex]

Step-by-step explanation:

[tex]\sf Given \ equation: \left(\dfrac{1}{27}\right)^x=3^{(-4x+6)}[/tex]

[tex]\sf As \ \dfrac{1}{27}=\dfrac{1}{3^3} \ and \ \dfrac{1}{a^b}=a^{-b} \ then \ \dfrac{1}{27}=3^{-3}[/tex]

Therefore, we can rewrite the given equation with base 3:

[tex]\implies \sf (3^{-3})^x=3^{(-4x+6)}[/tex]

Apply the exponent rule [tex]\sf (a^b)^c=a^{bc}[/tex] :

[tex]\implies \sf 3^{-3x}=3^{(-4x+6)}[/tex]

[tex]\sf If \ a^{f(x)}=a^{g(x)} \ then \ f(x)=g(x)[/tex]

[tex]\implies -3x=-4x+6[/tex]

Add 4x to both sides to solve for x:

[tex]\implies \sf x=6[/tex]

Answer: 108, 132, 156, and 180

Step-by-step explanation:

If you need to add 24 to the first term for the next 4 terms, you would have

84 + 24 = 108

108 + 24 = 132

132 + 24 = 156

156 + 24 = 180

So your sequence would be 108, 132, 156, and 180 for the next 4 terms.

About the minus sign with 16, square root of -1 is an imaginary number i(iota). Also we know that square root of -16 will be multiplication of square root of -1 and square root of 16. So the final answer becomes, + or - 4i. Hope this helps

Answer:

1. any number from 42 to 68

e.g 52

2. any number less than 42 or greater than 68

e.g 76

The circle has a center at the origin (0, 0) and have a radius of 4 units.

A circle is the locus of a point such that its distance from a fixed point (center) is constant.

The standard equation of a circle is:

(x - h)² + (y - k)² = r²

Where (h, k) is the circle center and r is the radius.

The circle given by the equation:

x² + y² = 16

Hence the center is (0, 0) and the radius is 4 units.

The circle has a center at the origin (0, 0) and have a radius of 4 units.

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The center is (0, 0) and the radius is 4 units.

A circle is a round-shaped figure that has no corners or edges. In geometry, a circle can be defined as a closed, two-dimensional curved shape.

The base of a solid is s is the region bounded by the circle is;

[tex]\rm x^2+ y^2=16[/tex]

The standard equation of a circle is:

(x - h)² + (y - k)² = r²

Where (h, k) is the circle center and r is the radius.

On comparing with the standard equation of a circle;

h = 0 , k = 0 and r =4

Hence. the center is (0, 0) and the radius is 4 units.

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Find theFind tha amount of profit per bag by subtracting:

2.99 - 1.90 = 1.09

He made $1.09 profit per bag.

Now multiply the profit per bag by the number of bags:

1.09 x 12 = 13.08

Total profit = $13.08 amount of profit per bag by subtracting:

2.99 - 1.90 = 1.09

He made $1.09 profit per bag.

Now multiply the profit per bag by the number of bags:

1.09 x 12 = 13.08

Total profit = $13.08

Using proportions, it is found that the value of a is of 3.

A proportion is a fraction of a total amount, and the measures are related using a rule of three.

In this problem, a 180º angle has a measure of [tex]\pi[/tex] radians. What is the measure of a 540º angle? The rule of three is given by:

180º - [tex]\pi[/tex] rad.

540º - x [tex]\pi[/tex] rad.

Applying cross multiplication:

180x = 540.

x = 540/180 = 3.

The value of a is of 3.

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Using proportions, the value of a is 3.

Proportion is a mathematical comparison between two numbers.

When going from degrees to radians, 180 degrees is always going to equal π radians.

The conversion of angle from radian to degree;

[tex]\rm 180-\ 1 \\\\540-a[/tex]

The value of a is;

[tex]\rm \dfrac{180}{540}=\dfrac{1}{a}\\\\\dfrac{1}{3}=\dfrac{1}{a}\\\\1 \times a=1\times 3\\\\a=3[/tex]

Hence, the value of a is 3.

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Answer:

[tex]3x - 1 = 5x + 7 \\ \\ 5x - 3x = 7 + 1 \\ \\ 2x = 8 \\ \\ x = \frac{8}{2} \\ \\ x = 4[/tex]

# Jerry Don is here#

The confidence interval for the sample proportion is [tex]CI = 0.16 \pm 0.0545[/tex]

The formula for calculating the confidence interval for proportion is expressed as:

[tex]p \pm z\sqrt{p(1-p)/n[/tex]

p is the proportion

z is the z score at 99% interval

n is the sample size

Given the following parameters

[tex]p=\frac{47}{300} = 0.16\\z=2.576\\n=300[/tex]

Substitute the values

[tex]CI = 0.16 \pm 2.576\sqrt{0.16(0.84)/300}\\CI=0.16\pm2.576(0.02116)\\CI = 0.16 \pm 0.0545[/tex]

Hence the confidence interval for the sample proportion is [tex]CI = 0.16 \pm 0.0545[/tex]

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Linear equations in slope intercept form for the given scenarios:

Juwan: y = 4x + 4

Jamie: y = 3x

Linear equations in standard form for the given scenarios:

Juwan: y - 4x = 4

Jamie: y - 3x = 0

Where, y = current stock value and x = number of times the stock value increased.

Domain: Set of natural number.

Range: Real number greater than or equal to 0.

This scenario is an example of causation.

The standard form of a line is Ax + By = C, where A and B are not both zero.

In the given scenario, we take

y =  the present stock value of Jamie and Juwan.

x = number of times the stock value increased.

where, y depends upon x.

Juwan: y - 4x = 4

Jamie: y - 3x = 0

The slope intercept form of a straight line is one of the most common forms used to represent the equation of a line. The slope intercept formula can be used to find the equation of a line when given the slope of the straight line and the y-intercept( the y-coordinate of the point where the line intersects the y-axis).

Juwan: y = 4x + 4

Slope = 4

y - intercept = 4

Jamie: y = 3x

Slope = 3

y - intercept = 0

The domain is the set of all the input values of a function and range is the possible output given by the function.

In this particular question, x is the number of times stock price increased. Thus, it can be 0, 1, 2, 3....set of natural number. y is the current stock price. The stock price was positive in the beginning and has been increasing since then, therefore y lies in the set of positive real number.

Causation means that one variable causes another to change, which means one variable is dependent on the other. Here, the price of stock depends upon the number of times its value increased.

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Answer:

7.28 in change

Step-by-step explanation:

When she used 20 bucks it gave her 7.28 cents.