Ask question

Ask question

All categories

2 years ago

10

The previous balance in your check register $234.67 . You make a deposit of 150.00 .You write a check for $63.23 . What is your

new balance ???

2 answers:

olchik [2.2K]2 years ago

7 0

$321.44 is the answer

skelet666 [1.2K]2 years ago

3 0

Answer:

321.44

Step-by-step explanation:

You might be interested in

A tank is filled with 1000 liters of pure water. Brine containing 0.07 kg of salt per liter enters the tank at 9 liters per minu

DiKsa [7]

Answer: A. S(t)= 1000 + 0.63t + 0.48t - 17t

B. S(t)= 984.11

Step-by-step explanation:

7 0

3 years ago

ABCD- parallelogram, If the perimeter of Triangle CPQ is 15cm, Find the perimeter of triangle BAQ. Find the perimeter of triangl

melamori03 [73]

Answer:

The answer is below

Step-by-step explanation:

A parallelogram is a quadrilateral (has 4 sides and 4 angle) with two pair of parallel and opposite sides. Opposite sides of a parallelogram are parallel and equal.

Given parallelogram ABCD:

AB = CD = 18 cm; BC = AD = 8 cm

∠P = ∠P, ∠PDA = ∠PCQ (corresponding angles are equal).

Hence ΔPCQ and ΔPDA are similar by angle-angle similarity theorem. For similar triangles, the ratio of their corresponding sides equal. Therefore:

$\frac{CD}{PC}= \frac{AD}{CQ}\\\\\frac{18}{6}=\frac{8}{x} \\\\x=\frac{6*8}{18}=\frac{8}{3}\ cm$

Perimeter of CPQ = CP + CQ + PQ

15 = 6 + 8/3 + PQ

PQ = 15 - (6 + 8/3)

PQ = 6.33

∠CQP = ∠AQB (vertical angles), ∠QCP = ∠QBA (alternate angles are equal).

Hence ΔCPQ and ΔABQ are similar by angle-angle similarity theorem

$\frac{AQ}{QP}=\frac{AB}{CP} \\\\\frac{AQ}{6.33} =\frac{18}{6} \\\\AQ=\frac{18}{6}*6.33\\\\AQ = 19$

$\frac{BQ}{CQ}=\frac{AB}{CP} \\\\\frac{BQ}{8/3} =\frac{18}{6} \\\\BQ=\frac{18}{6}*\frac{8}{3} \\\\BQ =8$

Perimeter of BAQ = AB + BQ + AQ = 18 + 8 + 19 = 45cm

PA = AQ + PQ = 19 + 6.33 = 25.33

PD = CD + DP = 18 + 6 = 24

Perimeter of PDA = PA + PD + AD = 24 + 25.33 + 8 = 57.33 cm

7 0

3 years ago

Use variation of parameters to find a general solution to the differential equation given that the functions y1 and y2 are linea

Aleks [24]

Recall that variation of parameters is used to solve second-order ODEs of the form

<em>y''(t)</em> + <em>p(t)</em> <em>y'(t)</em> + <em>q(t)</em> <em>y(t)</em> = <em>f(t)</em>

so the first thing you need to do is divide both sides of your equation by <em>t</em> :

<em>y''</em> + (2<em>t</em> - 1)/<em>t</em> <em>y'</em> - 2/<em>t</em> <em>y</em> = 7<em>t</em>

<em />

You're looking for a solution of the form

$y=y_1u_1+y_2u_2$

where

$u_1(t)=\displaystyle-\int\frac{y_2(t)f(t)}{W(y_1,y_2)}\,\mathrm dt$

$u_2(t)=\displaystyle\int\frac{y_1(t)f(t)}{W(y_1,y_2)}\,\mathrm dt$

and <em>W</em> denotes the Wronskian determinant.

Compute the Wronskian:

$W(y_1,y_2) = W\left(2t-1,e^{-2t}\right) = \begin{vmatrix}2t-1&e^{-2t}\\2&-2e^{-2t}\end{vmatrix} = -4te^{-2t}$

Then

$u_1=\displaystyle-\int\frac{7te^{-2t}}{-4te^{-2t}}\,\mathrm dt=\frac74\int\mathrm dt = \frac74t$

$u_2=\displaystyle\int\frac{7t(2t-1)}{-4te^{-2t}}\,\mathrm dt=-\frac74\int(2t-1)e^{2t}\,\mathrm dt=-\frac74(t-1)e^{2t}$

The general solution to the ODE is

$y = C_1(2t-1) + C_2e^{-2t} + \dfrac74t(2t-1) - \dfrac74(t-1)e^{2t}e^{-2t}$

which simplifies somewhat to

$\boxed{y = C_1(2t-1) + C_2e^{-2t} + \dfrac74(2t^2-2t+1)}$

4 0

2 years ago

Nevaeh rolls a standard six-sided die, numbered from 1 to 6. Which word or phrase

liubo4ka [24]

Answer:

Step-by-step explanation:

If she had two dice, a seven is the most common roll you can get, although the odds are still against you.

To get a seven with just 1 die is impossible.

7 0

2 years ago

Katrina wants to estimate the proportion of adults who read at least 10 books last year. to do so, she obtains a simple random

Vaselesa [24]

E=Z*sqrt (p(1-p)/N), where E= error margin, p=proportion, N=sample size

Katrina's margin error at 85% confidence interval: E=1.96*sqrt (p(1-p)/100) = 0.196 sqrt (1(1-p))

Mathew's margin error at 99% confidence interval: E= 2.58*sqrt (p(1-p)/400) = 0.129 sqrt (p(1-p))

Since both obtained same estimate of proportion (that is, value of p), it can be seen that Mathew's estimate will have a small error (That is, 0.129 is smaller than 0.196). This can be attributed to larger sample size although a wider confidence (99%) interval was considered.

3 0

3 years ago