URGENT!! ILL GIVE
BRAINLIEST!!!! AND 100 POINTS!!!

Answer:

8 %

Step-by-step explanation:

To find the tax rate, we first have to find the tax

4.32 - 4 = .32

The tax rate is found by the tax / total

.32 /4 =.08

To change this to a percent multiply by 100

.08 * 100% = 8 %

Answer:

2

Step-by-step explanation:

(-11, -16) and (-13, -20)

To find the slope of the line, we use the slope formula: (y₂ - y₁) / (x₂ - x₁)

Plug in these values:

(-20 - (-16)) / (-13 - (-11))

Simplify the parentheses.

= (-20 + 16) / (-13 + 11)

= (-4) / (-2)

Simplify the fraction.

-4/-2

4/2

= 2

This is your slope.

Equation of the line - Slope

The points of the slope are, we also apply the formula:

  [tex]\huge \boxed{\boldsymbol{\sf{\left \{ {{x_1,y_1} \atop {x_2,y_2}} \right. \iff \ m=\frac{y_2-y_1}{x_2-x_1} }}}[/tex]

The points are:

     [tex]\boldsymbol{\sf{\underbrace{x_1=-11} \ \ \ \ \ \ \ \ \ \ \overbrace{y_1=-16} \ \ \ \ \ \ \ \ \ \ \underbrace{x_2=-13} \ \ \ \ \ \ \ \ \ \ \overbrace{y_{2}=-20} }}[/tex]

Substitute data into the formula:

[tex]\boldsymbol{\sf{m=\dfrac{\Delta y}{\Delta x} \iff \ m=\dfrac{y_2-y_1}{x_2-x_1} }}[/tex]

                          [tex]\boldsymbol{\sf{m=\dfrac{-20-(-16)}{-13-(-11)} }} \\ \\ \\ \boldsymbol{\sf{m=\dfrac{-4}{-2} }}\\ \\ \\ \boxed{\boxed{\boldsymbol{\sf{m=2}}}}[/tex]

The slope through the points (-11,-16) and (-13, -20) is 2.

Using the given information, Billy's grade in his Math class will be 94.85%

From the question, we are to determine what the grade of Billy should be.

From the given information,
Billy had 100% in his Math class and he got 27/34 on his midterm

Also,

The midterm is 25%

This means the exam is 75%

Billy's grade will be

100% of 75 + 27/34 of 25

= 75 + 19.85

= 94.85

Hence, Billy's grade will be 94.85%

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The zero of the function in the given table is -1, indicating that the function provided has an x-intercept.

The x-intercept is defined as an intercept that is located at the x-axis of the plane, which is the location or coordinates from where the line crosses.

The table is given in the question,

If the value of x is -1, then the value of y is zero, according to the table. This is the function's x-intercept. The zero of the function depicted in this table is -1, which signifies that the supplied function has an x-intercept.

Therefore, the zero of the function given in this table is -1, indicating that the function provided has an x-intercept.

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The rocket reaches its maximum height at 2.25 second(s) after launch

The maximum height reached by the object is 88 feet.

The function is given as

h(t) = -16t² + 72t + 7

Differentiate the above function

So, we have the following representation

h'(t) = -32t + 72

Set to 0

-32t + 72 = 0

So, we have

32t = 72

Divide by 32

t = 2.25

Hence, the time is 2.25 seconds

In (a), we have

t = 2.25

Substitute t = 2.25 in h(t) = -16t² + 72t + 7

h(2.25) = -16(2.25)² + 72(2.25) + 7

Evaluate

h(2.25) = 88

Hence, the maximum height is 88 feet

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The area of the triangle is 8 unit².

The amount of space a shape takes up is called the area of that space.

Here,

The given vertices are

A= (-4,1)

B= (-7,5)

C= (0,1)

Now, as we know the x and y coordinates of three vertices of triangle.

So, we will use the coordinate geometry formula.

The general formula of the area of triangle is

If the triangle have a vertices (Ax,Ay ),(Bx,By ) and (Cx,Cy ) then the area is

Area=1/2×|Ax (By-Cy )+Bx (Cy-Ay )+Cx (Ay-By )|

Now, we will put the given coordinates values in above equation

Then we will get

Area=(1/2)×|(-4)(5-1)+(-7)(1-1)+0(1-5)|

Area=(1/2)×|(-4)(4)+(-7)(0)+0(-4)|

Area=(1/2)×|-16|

Remove the mode bracket.

Area=1/2×16

Area=8

Thus, the area of the triangle is 8 unit².

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Answer:

[tex]\frac{5}{4} y^1^1[/tex]

Step-by-step explanation:

Simplify...

[tex]\frac{45y^7}{36y^4}[/tex]

[tex]\frac{5}{4} y^1^1[/tex]

Hope this helps! :)

Step-by-step explanation:

Given, [tex]{ \green{ \sf{y = A {e}^{mx} + B {e}^{nx}}}} [/tex]

differentiate with respect to x

this is in the form of [tex]{ \tt{ \frac{dy}{dx} ( { {e}^{a} }^{x}) = {ae}^{ax}}} [/tex]

[tex]{ \red{ \sf{ \frac{dy}{dx} = A {me}^{mx} + B {ne}^{nx}}}} [/tex]

[tex]{ \red{ \sf{ \frac{{d}^{2} y}{d{x}^{2} } = Am.{me}^{mx} + Bn. {ne}^{nx}}}} [/tex]

[tex]{ \red{ \sf{ \frac{{d}^{2} y}{d{x}^{2} } = A {m}^{2} {e}^{mx} + B {n}^{2} {e}^{nx}}}} [/tex]

L.H.S.[tex]{ = \purple{ \sf{ \frac{ {d}^{2} y}{d {x}^{2} } - (m + n) \frac{dy}{dx} + mny}}}[/tex]

[tex]{ \purple{ \sf{A {m}^{2} {e}^{mx} + B {n}^{2} {e}^{nx} - (m + n)( A {me}^{mx} + B {ne}^{nx}) \\ \\ + mn(A {e}^{mx} + B {e}^{nx})}}}[/tex]

[tex]{ \purple{ \sf{ \cancel{A {m}^{2} {e}^{mx} }} + { \cancel{B {n}^{2} {e}^{nx} }- \cancel {A{m}^{2}{e}^{mx}}} { \cancel{-B {mne}^{nx} }} - { \cancel{A {mne}^{mx}}} - { \cancel{B{n}^2 {e}^{nx}}+{ \cancel{Amn{e}^{mx}}+{ \cancel{Bmn{e}^{nx}}}}}}}[/tex]

= 0 → R.H.S

Step-by-step explanation:

I'm assuming "A" and "B" are some constants that are coefficients as well as "m" and "n". Under this assumption we can take the derivative and then take the derivative once more to find the 2nd derivative which is denoted as: [tex]\frac{d^2y}{dx^2}[/tex]

The chain rule can be denoted as: [tex]\frac{dy}{dx}=\frac{dy}{du}*\frac{du}{dx}[/tex]

It can also be denoted in function notation as: [tex]h(x)=f(g(x)) \implies h'(x)=f'(g(x))*g'(x)[/tex]

and in general it deals with composite functions and it it's going to help us take the derivative of the exponentials.

So take for example: [tex]Ae^{mx}[/tex], we can rewrite this as a composite function where: [tex]u=mx\implies Ae^{mx}=Ae^u[/tex]

So from here we can just take the derivative of [tex]Ae^u[/tex] and then multiply it by the derivative of "u". We can also apply the same thing to [tex]Be^{nx}[/tex]

So let's first take the derivative of both sides, with respect to x

[tex]\frac{d}{dx}[y]=\frac{d}{dx}[Ae^{mx} + Be^{nx}][/tex]

From here we can take the derivative of [tex]Ae^{mx} \text{ and }Be^{nx}[/tex] and then add them after taking the derivative individually.

[tex]\frac{dy}{dx}=\frac{d}{dx}[Ae^{mx}]+\frac{d}{dx}[Be^{nx}][/tex]

Assuming that "A" and "B" are constants, we can just take the derivative of: [tex]e^{mx}\text{ and }e^{nx}[/tex] and then multiply by "A" and "B" after.

[tex]\frac{dy}{dx}=A\frac{d}{dx}[e^{mx}]+B\frac{d}{dx}[e^{nx}][/tex]

now from here we apply the chain rule. I'll start with: [tex]e^{mx}[/tex]. First let's rewrite the equation so that we have a composite function: [tex]u=mx\implies e^{mx}=e^u[/tex] now from here, we can apply the chain rule: [tex]\frac{d}{dx}[e^{mx}]=\frac{d}{dx}[e^u]*\frac{d}{dx}[u][/tex]

You may know that the derivative of e to the x is just going to be e to the x, so the derivative of e^u is just e^u. Now substitute mx back in as u, and you get e^(mx). Now take the derivative of "u" which is just mx, and you get "m" since the derivative of any linear equation is just the slope (since this is constant): [tex]\frac{d}{dx}[e^{mx}]=e^{mx}*m[/tex]

Now from here we can apply the same logic to e^(nx) to get: [tex]\frac{d}{dx}[e^{nx}]=e^{nx}*n[/tex]

So now we have the following expression:

[tex]\frac{dy}{dx}=A[e^{mx}*m]+B[e^{nx}*n][/tex]

[tex]\frac{d}{dx}[\frac{dy}{dx}]=\frac{d}{dx}[A(e^{mx}*m)+B(e^{nx}*n)][/tex]

we can apply some of the rules we used before like taking each individual derivative then adding them after, and "factoring out" the A and B (because they're constants). This gives us the following expression:

[tex]\frac{d^2y}{dx^2}=A\frac{d}{dx}[e^{mx}*m]+B\frac{d}{dx}[e^{nx}*n][/tex]

From here we can also think of "factoring out" the m and n because they're constants, so we get the following:

[tex]\frac{d^2y}{dx^2}=Am*\frac{d}{dx}[e^{mx}]+Bn*\frac{d}{dx}[e^{nx}][/tex]

From here we can apply the chain rule... except we already did this work (taking the derivative of e to the mx and e to the nx) so we can just put that here

[tex]\frac{d^2y}{dx^2}=Am*[e^{mx}*m]+Bn*[e^{nx}*n][/tex]

We can further simplify this to:

[tex]\frac{d^2y}{dx^2}=Am^2*[e^{mx}]+Bn^2*[e^{nx}][/tex]

So we know that: [tex]\frac{d^2y}{dx^2}=Am^2*[e^{mx}]+Bn^2*[e^{nx}][/tex]

and that: [tex]\frac{dy}{dx}=A[e^{mx}*m]+B[e^{nx}*n][/tex] or [tex]\frac{dy}{dx}=Am[e^{mx}]+Bn[e^{nx}][/tex]

Plugging these into the expression: [tex]\frac{d^2y}{dx^2} - (m+n)(\frac{dy}{dx}})+mny=0[/tex]

we get the following: [tex]Am^2e^{mx}+Bn^2e^{nx}-(m+n)(Ame^{mx}+Bne^{nx})+mny=0[/tex]

First let's distribute the (m+n) to the first derivative to get:

[tex](m+n)(Ame^{mx}+Bne^{nx}) =(Am^2e^{mx} +Bnme^{nx} + Anme^{mx} + Bn^2e^{nx})[/tex]

so plugging this in we get:

[tex]Am^2e^{mx}+Bn^2e^{nx}-(Am^2e^{mx} +Bnme^{nx} + Anme^{mx} + Bn^2e^{nx})+mny=0[/tex]

from here we can distribute the negative sign and cancel out values to get:

[tex]-Bnme^{nx} -Anme^{mx}+mny=0[/tex]

So we're almost done and it may look a bit weird in this situation since none of the variables look like they align to cancel them out... but notice how there is a "y" variable? We know what that is equal to! We started with that expression: [tex]y=Ae^{mx}+Be^{nx}[/tex]

So let's plug this in to our expression:

[tex]-Bnme^{nx} -Anme^{mx}+mn(Ae^{nx}+Be^{nx})=0[/tex]

Let's distribute the "mn" using the distributive property:

[tex]-Bnme^{nx} -Anme^{mx}+(Anme^{nx}+Bnme^{nx})=0[/tex]

and from here we can finally cancel out all the values! and this simplifies to zero.

[tex]0=0[/tex]

so we have proven that this expression is always equal to zero!

Answer:

Step-by-step explanation:

18 times 9 then -9

Answer:

135

Step-by-step explanation:

First, Times 18 x 8. which gives you 144. Then, subtract 9, giving you 135.

Answer:37

Step-by-step explanation:

The rocket reaches its maximum height at 2.25 second(s) after launch

The maximum height reached by the object is 88 feet.

The function is given as

h(t) = -16t² + 72t + 7

Differentiate the above function

So, we have the following representation

h'(t) = -32t + 72

Set to 0

-32t + 72 = 0

So, we have

32t = 72

Divide by 32

t = 2.25

Hence, the time is 2.25 seconds

In (a), we have

t = 2.25

Substitute t = 2.25 in h(t) = -16t² + 72t + 7

h(2.25) = -16(2.25)² + 72(2.25) + 7

Evaluate

h(2.25) = 88

Hence, the maximum height is 88 feet

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1 km = 1000 m and 1m = 100cm

1 km is the same as 1000 m = 1000 x 100 = 100000 cm

There are 862 kilometres so there is 862 x 100000 = 86200000 cm in 862 km.

Answer:

86200000

Step-by-step explanation:

Answer:

Step-by-step explanation:

No, buying small bag and medium bag of rubber bands will not give him more rubber bands than a large bag.

Combining things and counting them as one big group is done through addition. In math, addition is the process of adding two or more integers together. Addends are the numbers that are added, while the total refers to the outcome of the operation. The combination of at least two numbers constitutes basic addition. Once a student has mastered counting, they can begin to study basic addition. By locating the addend at the top of the chart and tracing down to the row of the second addend, they can utilize an addition chart as a learning tool.

Given,

Small = 25

Medium = 45

Large = 70

Juan wants to buy small bag and medium bag of rubber bands.

So, adding:

Small + Medium

25 + 45

70

No, buying small bag and medium bag of rubber bands will not give him more rubber bands than a large bag.

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According to the population function, the population of mosquitoes in the area at any time is 400000 = 200000e⁷ⁿ

Population function:

The term population function refers the effective population served over the course of a time

Given,

The population of mosquitoes in a certain area increases at a rate proportional to the current population, and in the absence of other factors, the population doubles each week. There are 200000 mosquitoes in the area initially , and predators (birds, bats, and so forth) eat 20000 mosquitoes/day.

Now, we need to determine the population of mosquitoes in the area at any time.

Here we have given that we have two time frames in the information given; weeks and days.

Now, we need the population after predation lets choose time in days.

And we also know that the population doubles every seven days (ignoring predation) so if P(a) is the population function we have IVP

For one week, the population function of the given situation is written as

=> dP/dt = rP

Then the value of P is P(0) = 200000

And through the given question we know that,

=> P(7) = 2P(0)

While we solving the function then we get

P = Ceⁿᵃ

When we apply the values of P(0) = 200,000 then we get the value of C = 200,000

Then, the function is rewritten as

P = 200000eⁿᵃ

Now, we have to apply the value of P(7) = 2P(0), then we get,

400000 = 200000e⁷ⁿ

When we simplify it then we get,

e⁷ⁿ = 1/2

When we apply the number of days on these function, then we can get the number of mosquitoes in the city.

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Answer:

See below

Step-by-step explanation:

Read across the x-axis to '5' then go up to intersect the graph.....then read to the LEFT to find the price on the y-axis

The amount in the fund immediately after the last deposit is $1148830.185

Given,

The amount deposited by Steve annually =  $53,700

Interest rate of the fund = 15%

Time period of the investment = 10 years

We have to find the amount in the fund immediately after the last deposit.

Here,

To calculate the future value after the last deposit, we need to use the following formula:

Future Value = {A[(1+i)^n-1]}/i

A = Annual deposit

Then,

FV = {53,700[(1.15)^9 - 1]}  / 0.15 + 53,700

FV= 1095130.185 + 53,700

FV= $1148830.185

Therefore,

The amount in the fund immediately after the last deposit is $1148830.185

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The expression that can be used to represent this situation is 5p

Expression:

Expression refers the mathematical equation that consists of variables, numbers and mathematical operator.

Given,

Bre'Asia wanted to buy 5 pizzas. She used a coupon for $7 off the entire order. Let p represent the cost of 1 pizza.

Here we need to find the expression that represents the situation.

In the given question, let us consider p represent the cost of 1 pizza and the n refers the number of pizza.

So, the expression to calculate the price of the pizza based on the number of pizza is written as,

=> np

Here Bre'Asia brought 5 pizzas, then the expression is

=> 5p.

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Answer:

A

Step-by-step explanation:

Answer:

880 mango trees and 440 star apple strees

Step-by-step explanation:

when for every 8 mango trees there are 4 star apple trees, then there is a 2:1 ratio of mango : star apple trees.
1320 divided by 3 = 440
and 2*440 = 880 mango trees, 1*440 = 440 star apple trees

Answer:

x-18=x4=-36 is the answer

Answer:

21/2 or 10.5

Step-by-step explanation:

7 = 7/1

When dividing fractions we use KFC :

K - Keep the first fraction :

7/1

F - Flip the second fraction :

2/3 ⇒ 3/2

C - Change the sign :

÷ ⇒ ×

Now we have :

7/1 × 3/2 =

Now we multiply the first numerator with the second numerator and the first denominator with the second denominator :

7×3 = 21

1×2 = 2

So our final answer is 21/2 or 10.5

Hope this helped and have a good day

The green family will double their investment in 34 years after the principal is subject to compound interest continuously.

The principal invested = $2500

Rate of interest annually = 2.75%

Time = t years

Amount  = double of principal = 2500 × 2 = $ 5000

Interest is compounded continuously

Formula to calculate the amount is given by:

[tex]A=Pe^{rt}[/tex]

Now we will use the values to find the exact value of t

or, [tex]5000 = 2500 e^{0.0275t}[/tex]

or, ln 2 = 0.0275t

or, t = 25.205 years

or, t ≈ 25 years

Compound interest allows interest to accumulate over time as opposed to simple interest, which does not compound because prior interest is simply not added to the principal for the current period.

The simple annual interest rate is obtained by multiplying the interest per period by the total number of periods in a year.

Hence the Green family will have doubled the investment in approximately 25 years.

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The rate of change for the given graph is 10.

Rate of change

Rate of change describes how one quantity changes in relation to another quantity.

The formula for the rate of change is,

R = Δx/Δy

where

Δx = x₂ - x₁

Δy = y₂ - y₁

where (x₁, y₁) and (x₂, y₂) refers the points given.

Given,

Here we have the graph that shows the distance, d, meters, travelled by Jaden on her bicycle as a function of the number of wheel revolutions, n, as she rode from Whitehorse to the Grey Mountain Road lookout in the Yukon.

And we need to determine the rate of change of the graph.

In order to find the rate of change , we have to select the points from the given graph.

So, here we have select the following point. (2400, 5000) and (2500, 6000)

Now, as per the rate of change formula, it can be calculated as,

Δx = 2500 - 2400

Δx = 100

And the next one is,

Δy = 6000 - 5000

Δy = 1000

Then the rate of change is,

R = 1000/100

R = 10

Therefore, the rate of change is 10.

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The expression y = 4.998 · [tex]e^{1.098\cdot x}[/tex] is the exponential function that passes through the points (- 1, 5 / 3) and (3, 135).

Herein we find the location of two points set on Cartesian plane that belongs to an exponential function of the form:

[tex]y = A \cdot e^{B \cdot x}[/tex]

Where:

Which is equivalent to the following logarithmic expression:

㏑ y = ㏑ A + B · x

If we know that (x₁, y₁) = (- 1, 5 / 3) and (x₂, y₂) = (3, 135), then the following system of equations is generated:

㏑ (5 / 3) = ㏑ A - B

㏑ 135 = ln A + 3 · B

Then, we solve the system by numerical methods:

㏑ A = 1.609 (A = 4.998), B = 1.098

And the exponential function is equal to y = 4.998 · [tex]e^{1.098\cdot x}[/tex].

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Answer:

Step-by-step explanation:

it’s 5.5

(B) 1/2,1/4,1/10,6/3 are considered benchmark fractions.

Benchmark fractions are regular fractions that we can use to compare or judge other fractions when measuring, comparing, or ranking them. Benchmark fractions are simple to see and recognize, making it easier to estimate the components. We can either make the denominators of two fractions with distinct numerators and denominators the same, or we can compare them to a benchmark fraction like 1/2. The best way to use benchmark fractions is to plot the fractions to be compared on a number line next to the benchmarks.

A ruler is a tool most frequently used for number lines. As reference points, it uses half, fourth, and eighth.

Only option (B) satisfies the meaning of the benchmark fraction.

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