The following table shows the first segment of a five-year amortization schedule. a 5-year amortization schedule. the amount of interest paid for months 1 through 12 are: 99.03, 97.73, 96.42, 95.10, 93.78, 92.44, 91.09, 89.73, 88.36, 86.98, 85.58, 84.18. after one year of payments, how much has been paid to interest? a. $1,100.42 b. $1,010.16 c. $1,188.34 d. $1,241.18

Answers

Answer 1

After one year of payments, the amount need to be paid to interest is $1100.42 if the first segment of a five-year amortization schedule is a 5-year amortization schedule option (a) is correct.

What is a loan amortization schedule?

It is defined as the systematic way of representation of loan payments according to the time in which the principal amount and interest mentioned in a list manner.

We have a table given that showing a  first segment of a five-year amortization schedule that has a 5-year amortization schedule and the amount of interest paid for months 1 through 12.

As the loan amount is not mentioned in the question, we are assuming the loan amount is $1184.6

Therefore, loan amount = $1184.6

The outstanding amount at the end of one year, which is 12 months, is shown in the table as $84.18

After one year of payments, the amount need to be paid to interest:

= Loan amount - Outstanding amount after 12 months

= 1184.6 - 84.18

= $1100.425

Thus, after one year of payments, the amount need to be paid to interest is $1100.42 if the first segment of a five-year amortization schedule is a 5-year amortization schedule option (a) is correct.

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Related Questions

A right triangle includes one algae that measures 14º. what is the measure of the third angle

A 14º C 90º

B 76º D 104º

Answers

Answer: B. 76

Step-by-step explanation: A right triangle is 180 degrees. It has an angle of 90 since it is a right triangle. 90 + 14 = 104. 180 - 104 = 76

Answer:

B. 76 degrees

Step-by-step explanation:

EVERY triangle's angles add up to 180 degrees. We already know that since it's a right triangle, one of the angles equals 90 degrees (that's a right angle) and they give us the second angle measurement, 14 degrees. If we add those two angle measures together and subtract them from 180, we should get the measure of the third angle as our answer.

14 + 90 = 104

180 - 104 = 76

Therefore, the third and final angle in this right triangle equals 76, so your answer is B. I hope this helps! Have a lovely day!! :)

Which value is not a solution of the inequality x-4 symbol 2

Answers

The inequality x -4 > 2 uses a greater than symbol

All numbers lesser or equal to 6 are not a solution of the inequality x -4 > 2

How to determine the value not in the solution?

The inequality is given as:

x -4 > 2

Add 4 to both sides of the inequality

x - 4 + 4 > 2 + 4

Evaluate the sum

x > 6

The above means that only numbers greater than 6 are in the solution of the inequality.

Since the options are not given, I will give a general solution that all numbers lesser or equal to 6 are not a solution of the inequality x -4 > 2

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Find the circumference of the circle of 13 inches use 3.14 for pi and round to the nearest whole number

Answers

Answer:

ohh just use this formula

this is for Area- A= π r^2

this is for Circumference- C= 2 π r

Step-by-step explanation:

The circumference of the circle of the radius of 13 inches is, 41 inches

What is circumference ?

Circumference is the distance around the perimeter of a circular object. It is defined as the length of the circle that is found by multiplying the diameter of the circle by π (pi), which is approximately equal to 3.14.

The formula for the circumference of a circle is given by: C = 2πr,

Given that,

The diameter of the circle is 13 inches,

The radius of the circle can be calculated as follows:

r = d/2

  = 13 inches / 2

  = 6.5 inches

Using the formula for the circumference, we can calculate the circumference as follows:

C = 2πr

   = 2 × 3.14 × 6.5 inches

   = 40.76 inches

Rounding to the nearest whole number, the circumference of the circle is 41 inches.

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8. Daisies and tulips are planted in a
garden. There are 11 fewer tulips
planted than daisies.
a. Write an expression that represents
the number of tulips in terms of the
number of daisies. Define any
variables used.
b. If 18 daisies are planted, how many
tulips are planted?

Answers

Answer:

a) [tex]d-11 = t[/tex]

b) 7

Step-by-step explanation:

Let [tex]d[/tex] = daisies

Let [tex]t[/tex] = tulips

a) 11 fewer tulips than daisies: [tex]d-11 = t[/tex]

b) Substitute 18 into [tex]d[/tex] and solve for [tex]t[/tex].

[tex]18-11= t[/tex]

[tex]7 = t[/tex]

The temperature is dropping at a rate of five degrees per hour.

Let d represent the number of degrees the temperature drops.
Let t represent the number of hours that pass.

Which is the dependent variable?

Answers

Answer:

The number of degrees the temperature drops°

Step-by-step explanation:

hope this helps

and hope this is the answer you was looking for

pls mark brainliest

7. What value of c will make x2 – 20x + c
a perfect square trinomial?

Answers

29x for this answer

Which pair of expressions has equivalent values?
1^13 and 1^15
6^1and 9^1
7^8and 8^7
9- and 4^3

Answers

Answer:

1^13 and 1^15

Step-by-step explanation:

1 raised to anything is still just 1

so, 1^13 = 1 and 1^15 =1

I need this for school, please help!!

Answers

C. Firstly, add up the amount of students and you should get 352. I did the easy route and divided 352 by 8 and got 44. From there, I added the amount of teachers on each graph and C is the only one with the amount of 44 teachers.

Mr. Fuller wants to put fencing around his rectangular-shaped yard. the width of the yard is 55 feet and the length is 75 feet. how many feet of fencing does Mr. Fuller

Answers

If mr fuller wants to put up fencing you need to find the area

The total length of the fencing is the perimeter of the rectangular yard which is P = 260 feet

What is the Perimeter of a Rectangle?

The perimeter P of a rectangle is given by the formula, P=2 ( L + W) , where L is the length and W is the width of the rectangle.

Perimeter P = 2 ( Length + Width )

Given data ,

Let the perimeter of the rectangular yard be = P

Now , the equation will be

The width of the rectangular yard W = 55 feet

The length of the rectangular yard L = 75 feet

And , the required fencing = Perimeter of rectangular yard

Perimeter of rectangular yard P = 2 ( L + W )

Substituting the values in the equation , we get

Perimeter of rectangular yard P = 2 ( 55 + 75 )

Perimeter of rectangular yard P = 2 ( 130 )

Perimeter of rectangular yard P = 260 feet

Therefore , the value of P is 260 feet

Hence , the perimeter of yard is 260 feet

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Arrange the expressions below in order from least to greatest. place the least at the top and greatest at the bottom. ( 72 ÷ 8 ) − 2 × 3 1 72 ÷ ( 8 − 2 ) × 3 1 72 ÷ ( 8 − 2 ) × ( 3 1 ) 72 ÷ 8 − 2 × ( 3 1 )

Answers

The expressions in order from least to greatest 72 / 8 - 2 x (3 + 1), (72 / 8) - 2 x 3 + 1, 72 / (8 - 2) x 3 + 1 and 72 / (8 - 2) x (3 + 1).

What is BODMAS?

BODMAS stands for B - Bracket, O - order of Power, D - Division, M - Multiplication, A - Addition, and S - Subtraction.

To Arrange the expressions below in order from least to greatest. place the least at the top and the greatest at the bottom

72 / 8 - 2 x (3 + 1) equals 1

(72 / 8) - 2 x 3 + 1 equals 4

72 / (8 - 2) x 3 + 1 equals 37

72 / (8 - 2) x (3 + 1) equals 48

Thus, The expressions in order from least to greatest 72 / 8 - 2 x (3 + 1), (72 / 8) - 2 x 3 + 1, 72 / (8 - 2) x 3 + 1 and 72 / (8 - 2) x (3 + 1).

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What are range, index of qualitative variation (IQV), interquartile range (IQR), standard deviation, and variance

Answers

Answer:

To find the interquartile range (IQR), ​first find the median (middle value) of the lower and upper half of the data. These values are quartile 1 (Q1) and quartile 3 (Q3). The IQR is the difference between Q3 and Q1.

Step-by-step explanation:

The scale factor for a model is 8 cm m Model : 11.23cm actual: 40.2 m​

Answers

the answer is = 19 m

Identify the surface area of the cylinder to the nearest tenth. Use 3.14 for π.

Answers

Answer:

967.6

Step-by-step explanation:

967.6

967.12 in

Step-by-step explanation:

the formula for the area (surface area) of a cylinder is: A=2πrh+2πr2

to solve we need to determine the values

R= radius = half the diameter = 14/2 =7

D= diameter =14inches

H= height= 15 inches

plug in

A=2πrh+2πr2 = A=2π([tex]\frac{d}{2}[/tex])h+2π([tex]\frac{d}{2}[/tex])^2

= 2[tex]\pi[/tex]([tex]\frac{14}{2}[/tex])(15) + 2[tex]\pi[/tex]([tex]\frac{14}{2}[/tex])^2

= 2[tex]\pi[/tex](7)(15) + 2[tex]\pi[/tex](7)^2

=[tex]\pi[/tex]((2x7x15)+(2x7^2))

=[tex]\pi[/tex](210+98)

=[tex]308\pi[/tex]

=967.12 in

Samir recorded the grade-level and instrument of everyone in the middle school School of Rock below. Seventh Grade Students Instrument # of Students Guitar 6 Bass 4 Drums 6 Keyboard 7 Eighth Grade Students Instrument # of Students Guitar 9 Bass 9 Drums 9 Keyboard 10 Based on these results, express the probability that a seventh grader chosen at random will play an instrument other than drums as a fraction in simplest form.

Answers

Using it's concept, it is found that there is a [tex]\frac{17}{23}[/tex] probability that a seventh grader chosen at random will play an instrument other than drums.

What is a probability?

A probability is given by the number of desired outcomes divided by the number of total outcomes.

In this problem:

There is a total of 6 + 4 + 6 + 7 = 23 seventh graders.Of those, 23 - 6 = 17 play instruments that are not the drum.

Hence:

[tex]p = \frac{17}{23}[/tex]

There is a [tex]\frac{17}{23}[/tex] probability that a seventh grader chosen at random will play an instrument other than drums.

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Prove that: a + b + c / a^-1+ b^-1+ c^-1 = abc


Answers

Answer:

It's right

Step-by-step explanation:

(dk how to show prove but thank?

NEED HELP ASAP!!!! Will give brainiest

Answers

2%(random sentence bc I need 20 characters)

A kitchen can be broken into 2 rectangles. One rectangle has a base of 7 feet and height of 5 feet. The second rectangle has a base of 2 feet and height of 2 feet. One package of tile will cover 3 square feet. How many packages of tile will she need? 8 13 15 39

Answers

Answer:

its 13 or B

Step-by-step explanation:

Wind is
• air moving from areas of high pressure to areas of low pressure.
• air moving from areas of low pressure to areas of high pressure.
air moving from areas of high temperature to areas of low temperature.

Answers

Wind is caused by differences in the atmospheric pressure. When a difference in atmospheric pressure exists, air moves from the higher to the lower pressure area, resulting in winds of various speeds. On a rotating planet, air will also be deflected by the Coriolis effect, except exactly on the equator.

Hope this helped!

Need help with this problem

Answers

Answer:

A)

Step-by-step explanation:

Sum of all angles of triangle = 180

53 + 68  + x = 180

        121 + x  = 180

                 x = 180 - 121

                 x = 59°

Answer:

180 - 53 - 68 = 59

The third angle is 59°.

Step-by-step explanation:

∠A and \angle B∠B are vertical angles. If m\angle A=(7x-6)^{\circ}∠A=(7x−6)



and m\angle B=(8x-27)^{\circ}∠B=(8x−27)



, then find the measure of \angle B∠B

Answers

keeping in mind that vertical angles are always congruent.

[tex]\stackrel{\measuredangle A}{7x-6}~~ = ~~\stackrel{\measuredangle B}{8x-27}\implies -6=x-27\implies 21=x~\hfill \underset{\measuredangle B}{\stackrel{8(21)~~ - ~~27}{141}}[/tex]

15 × [-5] +15 × [-3] = solution

Answers

[tex]Heyo![/tex]

SaddySokka is here to help!!

Let's do this step-by-step explanation!

[tex](15)(-5)+(15)(-3)[/tex]

[tex]=-75+(15)(-3)[/tex]

[tex]=-75+-45[/tex]

[tex]=-120[/tex]

Answer:

[tex]-120[/tex]

Hopefully, this helps you!!

Have a great day!!

SaddySokka~

Answer:

-120

Step-by-step explanation:

15×[-5]+15×[-3]

Use BODMAS

-75+-45

-120

Which expression is equal to 0.75×0.09

Answers

The answer is C 75/100 * 9/10

Number 21 please help me solve it thank youu

Answers

Answer:

64 pi

Step-by-step explanation:

32 is diameter

diameter is circumference

2 circles

so 2*32=64

Hello Calculus!

Find the value

[tex]\\ \rm\Rrightarrow {\displaystyle{\int\limits_3^5}}(e^{3x}+7cosx-3tan^3x)dx[/tex]

Note:-

Answer with proper explanation required and all steps to be mentioned .

Answers

Answer:

[tex]\displaystyle \int\limits^5_3 {\bigg( e^{3x} + 7 \cos x - 3 \tan^3 x \bigg)} \, dx = \frac{e^{15} - e^9}{3} + 7 \bigg( \sin 5 - \sin 3 \bigg) - 3 \bigg( \frac{\sec^2 5 - \sec^2 3}{2} - \ln \bigg| \frac{\cos 3}{\cos 5} \bigg| \bigg)[/tex]

General Formulas and Concepts:
Calculus

Differentiation

DerivativesDerivative Notation

Derivative Property [Multiplied Constant]:
[tex]\displaystyle (cu)' = cu'[/tex]

Derivative Property [Addition/Subtraction]:
[tex]\displaystyle (u + v)' = u' + v'[/tex]

Derivative Rule [Basic Power Rule]:

f(x) = cxⁿf’(x) = c·nxⁿ⁻¹

Integration

Integrals

Integration Rule [Reverse Power Rule]:
[tex]\displaystyle \int {x^n} \, dx = \frac{x^{n + 1}}{n + 1} + C[/tex]

Integration Rule [Fundamental Theorem of Calculus 1]:
[tex]\displaystyle \int\limits^b_a {f(x)} \, dx = F(b) - F(a)[/tex]

Integration Property [Multiplied Constant]:
[tex]\displaystyle \int {cf(x)} \, dx = c \int {f(x)} \, dx[/tex]

Integration Property [Addition/Subtraction]:
[tex]\displaystyle \int {[f(x) \pm g(x)]} \, dx = \int {f(x)} \, dx \pm \int {g(x)} \, dx[/tex]

Integration Method: U-Substitution + U-Solve

Step-by-step explanation:

Step 1: Define

Identify given.

[tex]\displaystyle \int\limits^5_3 {\bigg( e^{3x} + 7 \cos x - 3 \tan^3 x \bigg)} \, dx[/tex]

Step 2: Integrate Pt. 1

[Integral] Rewrite [Integration Rule - Addition/Subtraction]:
[tex]\displaystyle \int\limits^5_3 {\bigg( e^{3x} + 7 \cos x - 3 \tan^3 x \bigg)} \, dx = \int\limits^5_3 {e^{3x}} \, dx + \int\limits^5_3 {7 \cos x} \, dx - \int\limits^5_3 {3 \tan^3 x} \, dx[/tex][Integrals] Rewrite [Integration Property - Multiplied Constant]:
[tex]\displaystyle \int\limits^5_3 {\bigg( e^{3x} + 7 \cos x - 3 \tan^3 x \bigg)} \, dx = \int\limits^5_3 {e^{3x}} \, dx + 7 \int\limits^5_3 {\cos x} \, dx - 3 \int\limits^5_3 {\tan^3 x} \, dx[/tex][3rd Integral] Rewrite:
[tex]\displaystyle \int\limits^5_3 {\bigg( e^{3x} + 7 \cos x - 3 \tan^3 x \bigg)} \, dx = \int\limits^5_3 {e^{3x}} \, dx + 7 \int\limits^5_3 {\cos x} \, dx - 3 \int\limits^5_3 {\tan^2 x \bigg( \sec^2 x - 1 \bigg)} \, dx[/tex]

Step 3: Integrate Pt. 2

Identify variables for u-substitution and u-solve.

1st Integral

Set u:
[tex]\displaystyle u = 3x[/tex][u] Differentiate [Derivative Properties and Rules]:
[tex]\displaystyle du = 3 \, dx[/tex][Bounds] Swap:
[tex]\displaystyle \left \{ {{x = 5 \rightarrow u = 3(5) = 15} \atop {x = 3 \rightarrow u = 3(3) = 9}} \right.[/tex]

3rd Integral

Set v:
[tex]\displaystyle v = \sec x[/tex][v] Differentiate [Trigonometric Differentiation]:
[tex]\displaystyle dv = \sec x \tan x \, dx[/tex][dv] Rewrite:
[tex]\displaystyle dx = \frac{1}{\sec x \tan x} \, dv[/tex]

Step 4: Integrate Pt. 3

Let's focus on the 3rd integral first.

Apply Integration Method [U-Solve]:
[tex]\displaystyle 3 \int\limits^5_3 {\tan^2 x \bigg( \sec^2 x - 1 \bigg)} \, dx = 3 \int\limits^{x = 5}_{x = 3} {\frac{v^2 - 1}{v}} \, dv[/tex][Integral] Rewrite [Integration Property - Addition/Subtraction]:
[tex]\displaystyle 3 \int\limits^5_3 {\tan^2 x \bigg( \sec^2 x - 1 \bigg)} \, dx = 3 \Bigg( \int\limits^{x = 5}_{x = 3} {v} \, dv - \int\limits^{x = 5}_{x = 3} {\frac{1}{v}} \, dv \Bigg)[/tex][Integrals] Apply Integration Rules [Reverse Power Rule and Logarithmic Integration]:
[tex]\displaystyle 3 \int\limits^5_3 {\tan^2 x \bigg( \sec^2 x - 1 \bigg)} \, dx = 3 \Bigg( \frac{v^2}{2} \bigg| \limits^{x = 5}_{x = 3} - \ln | v | \bigg| \limits^{x = 5}_{x = 3} \Bigg)[/tex][v] Back-Substitute:
[tex]\displaystyle 3 \int\limits^5_3 {\tan^2 x \bigg( \sec^2 x - 1 \bigg)} \, dx = 3 \Bigg( \frac{\sec^2 x}{2} \bigg| \limits^{5}_{3} - \ln | \sec x | \bigg| \limits^{5}_{3} \Bigg)[/tex]Evaluate [Integration Rule - Fundamental Theorem of Calculus 1]:
[tex]\displaystyle 3 \int\limits^5_3 {\tan^2 x \bigg( \sec^2 x - 1 \bigg)} \, dx = 3 \Bigg( \frac{\sec^2 5 - \sec^2 3}{2}- \ln \bigg| \frac{\cos 3}{\cos 5} \bigg| \Bigg)[/tex]

Step 5: Integrate Pt. 4

Focus on the other 2 integrals and solve using integration techniques listed above.

1st Integral:

[tex]\displaystyle\begin{aligned}\int\limits^5_3 {e^{3x}} \, dx & = \frac{1}{3} \int\limits^5_3 {3e^{3x}} \, dx \\& = \frac{1}{3} \int\limits^{15}_9 {e^{u}} \, du \\& = \frac{1}{3} e^u \bigg| \limits^{15}_9 \\& = \frac{1}{3} \bigg( e^{15} - e^9 \bigg)\end{aligned}[/tex]

2nd Integral:
[tex]\displaystyle\begin{aligned}7 \int\limits^5_3 {\cos x} \, dx & = 7 \sin x \bigg| \limits^5_3 \\& = 7 \bigg( \sin 5 - \sin 3 \bigg)\end{aligned}[/tex]

Step 6: Integrate Pt. 5

[Integrals] Substitute in integrals:
[tex]\displaystyle \int\limits^5_3 {\bigg( e^{3x} + 7 \cos x - 3 \tan^3 x \bigg)} \, dx = \frac{e^{15} - e^9}{3} + 7 \bigg( \sin 5 - \sin 3 \bigg) - 3 \bigg( \frac{\sec^2 5 - \sec^2 3}{2} - \ln \bigg| \frac{\cos 3}{\cos 5} \bigg| \bigg)[/tex]

∴ we have evaluated the integral.

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Topic: AP Calculus AB/BC (Calculus I/I + II)

Unit: Integration

Answer:

1086950.36760

Formula's used:

[tex]\rightarrow \sf \int sin(ax+b)=-\dfrac{1}{a} cos(ax+b)+c[/tex]

[tex]\rightarrow \sf \int cos(ax+b)=\dfrac{1}{a} sin(ax+b)+c[/tex]

[tex]\rightarrow \sf \int \dfrac{1}{ax+b} =\dfrac{1}{a} ln|ax+b|+c[/tex]

[tex]\rightarrow \sf \int e^{ax+b}=\dfrac{1}{a} e^{ax+b} + c[/tex]

[tex]\rightarrow \bold{ ln|a| - ln|b| = ln|\frac{a}{b} | }[/tex]

Explanation:

[tex]\dashrightarrow \sf \int \left(e^{3x}+7cos\left(x\right)-3tan^3\left(x\right)\right)[/tex]

                        apply sum rule: [tex]\bold{\int f\left(x\right)\pm g\left(x\right)dx=\int f\left(x\right)dx\pm \int g\left(x\right)dx}[/tex]

[tex]\dashrightarrow \sf \int \:e^{3x}dx+\int \:7\cos \left(x\right)dx-\int \:3\tan ^3\left(x\right)dx[/tex]

                Integrate simple followings first, using formula's given above

[tex]\dashrightarrow \sf \dfrac{1}{3}e^{3x}+7\sin \left(x\right)-\int 3tan^3x[/tex]

                        Breakdown the component

[tex]\dashrightarrow \sf \dfrac{1}{3}e^{3x}+7\sin \left(x\right)-3\int tan^2x(tanx)[/tex]

                                                         [ tan²x = sec²x - 1 ]

[tex]\dashrightarrow \sf \dfrac{1}{3}e^{3x}+7\sin \left(x\right)-3\int (sec^2x-1)(tanx)[/tex]

===========================================================

for integration of [tex]\bold{\int (sec^2x-1)(tanx)}[/tex]

                                                  apply substitution ... u

[tex]\dashrightarrow \int \dfrac{-1+u^2}{u}[/tex]

[tex]\dashrightarrow \sf \int \:-\dfrac{1}{u}+udu[/tex]

[tex]\dashrightarrow \sf - \int \dfrac{1}{u}du+\int \:udu[/tex]

[tex]\dashrightarrow -\ln \left|u\right|+\dfrac{u^2}{2}[/tex]

substitute back u = sec(x)

[tex]\dashrightarrow \sf-\ln \left|\sec \left(x\right)\right|+\dfrac{\sec ^2\left(x\right)}{2}[/tex]

================================================= insert back

[tex]\dashrightarrow \sf \dfrac{1}{3}e^{3x}+7\sin \left(x\right)-3\left(-\ln \left|\sec \left(x\right)\right|+\dfrac{\sec ^2\left(x\right)}{2}\right)[/tex]   outcome after integrating

Now apply the given limits

[tex]\sf \hookrightarrow \sf \dfrac{1}{3}e^{3(5)}+7\sin \left(5\right)-3\left(-\ln \left|\sec \left(5\right)\right|+\dfrac{\sec ^2\left(5\right)}{2}\right) - (\sf \dfrac{1}{3}e^{3(3)}+7\sin \left(3\right)-3\left(-\ln \left|\sec \left(3\right)\right|+\dfrac{\sec ^2\left(3\right)}{2}\right))[/tex]

                                                                   simplify

[tex]\sf \hookrightarrow \sf \dfrac{1}{3}e^{15}+7\sin \left(5\right)-3\left(-\ln \left|\sec \left(5\right)\right|+\dfrac{\sec ^2\left(5\right)}{2}\right) - (\sf \dfrac{1}{3}e^{9}+7\sin \left(3\right)-3\left(-\ln \left|\sec \left(3\right)\right|+\dfrac{\sec ^2\left(3\right)}{2}\right))[/tex]

                          and group the variables

[tex]\sf \hookrightarrow \dfrac{e^{15}-e^9}{3}-\dfrac{3}{2\cos ^2\left(5\right)}+\dfrac{3}{2\cos ^2\left(3\right)}+7\sin \left(5\right)-7\sin \left(3\right)+3\ln \left(\dfrac{1}{\cos \left(5\right)}\right)-3\ln \left(-\dfrac{1}{\cos \left(3\right)}\right)[/tex]

value:

[tex]\sf \hookrightarrow 1086950.36760[/tex]

Tom wants the scale model to be 9 inches tall.
How wide should the scale model be?
A. 1.7 inches
B. 5.4 inches
O c. 15 inches
OD. 20 inches

Answers

Answer:

C 15

Step-by-step explanation:

9 =  12 3/4

20 3/4 = 15

ASAP PLS
Write an equation to represent the following scenario: Ms. Cloutier’s wedding photographer requires a $1000 deposit, and then $250 for every hour she is working.

Answers

1000+(250•x)= y

x being the number of hours she’s working

What is 40 x 40 x 40 please help fast ASAP

Answers

40^3 or 4 times 4 times 4 times 10 times 10 times 10
Which is
64 times 1000
Which is
64000

What are the solutions of the equation (x + 2)2 + 12(x + 2) – 14 = 0? Use u substitution and the quadratic formula to
solve.

-8+5√2
O x=-6252
O x=-4+5√2
x=-2 +5√2

Answers

Answer:

-8+5√2

Step-by-step explanation:

(x+2)^2+12(x+2)–14=0

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

12(x+2)=12x+24

x^2+4+4x+12x+24-14=0

x^2+4x+12x+4+24-14=0

x^2+16x+14=0

quadratic formula

x = {-b +- square root of (b^2 – 4ac)} ÷ {2a}

a= 1

b = 16

c = 14

x = {-16 +- square root of (16^2 – 4*1*14)} ÷ {2*1}

x = {-16 +- square root of (256 – 56)} ÷ {2*1}

x = ((-16 +- square root of (200)) ÷ (2)

x = ((-16 +- 10√2)) ÷ (2)

x= -8+-5√2

The length of a rectangle is 7 cm less than four times its width. The area of the rectangle is 36 square cm

Answers

Answer:

W = 4 cm and  L = 9 cm

Step-by-step explanation:

I don't see a question, but will assume the problem wants the length(L) and width(W) of the described rectangle.

Let L and W stand for Length and Width.

Area of a rectangle is given by L*W

We are told that L*W = 36 cm^2

We are also told that L = 4W-7 ["length of a rectangle is 7 cm less than four times its width"]

Substituting the second into the first equation:

L*W = 36 cm^2

(4W-7)*W = 36 cm^2          [L = 4W-7]

4W^2-7W - 36 cm^2 = 0

(W-4)(4W+9) = 0

The roots are:  4 and -(9/4)

We'll use the positive value:  W = 4

Since L = 4W-7:

   L = 4(4)-7

   L = 16-7

  L = 9 cm

help me please need help​

Answers

Answer:

Step-by-step explanation:

1.  x -> opposite side of 48°

   o → hypotenuse

   b → adjacent side of  48°

[tex]\sf Sin \ 48^\circ = \dfrac{opposite \ side }{hypotenuse}\\\\\\0.7431 = \dfrac{15}{o}\\\\\\0.74 * o = 15\\\\\\ o = \dfrac{15}{0.74}\\\\\\[/tex]

o = 20.27

[tex]\sf cos \ 48^\circ = \dfrac{adjacent \ side }{hypotenuse}\\\\\\0.67 =\dfrac{b}{o}\\\\\\0.67=\dfrac{b}{20.27}[/tex]

b = 0.67*20.27

b = 13.58

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

2) i → opposite side of 25°

n → adjacent side of 25°

[tex]\sf Sin \ 25 =\dfrac{i}{t}\\\\\\0.42=\dfrac{i}{30}\\\\\\0.42*30=i[/tex]

i = 12.6

[tex]\sf Cos \ 30^\circ =\dfrac{n}{t}\\\\0.91=\dfrac{n}{30}\\\\\\0.91*30 = n[/tex]

n = 27.3

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

3) a → opposite side of 70°

e → adjacent side of 70°

[tex]Sin \ 70^\circ =\dfrac{a}{l}\\\\0.94 =\dfrac{a}{25}\\\\0.94*25=a[/tex]

a = 23.5

[tex]\sf Cos \ 70^\circ =\dfrac{e}{l}\\\\0.34=\dfrac{e}{25}\\\\0.34*25=e[/tex]

e = 8.5

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

4)

[tex]\sf Sin \ 52^\circ = \dfrac{x}{75}\\\\0.79*75=x\\[/tex]

x = 59.25

[tex]\sf Cos \ 52^\circ = \dfrac{z}{75}\\\\0.62*75 =z[/tex]

z = 46.5

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