Suppose the reserve requirement is changed to 4 percent. Reconstruct the balance sheet of the total banking system after all banks have fully utilized their lending capacity.





c. By how much has the money supply changed as a result of the lower reserve requirement (part b)?



$
billion

Answer:  d: 1:52pm

Step-by-step explanation:  Since Beatrice finished 22 minutes earlier, we subtract 22 minutes from 2:14. 2:14 - 14 is 2:00. 22-14 is 8. 2:00 - 8 is 1:52.

Answer: (5,3)

Step-by-step explanation:

Substituting into point-slope form, the equation of the line is

[tex]y-5=-2(x-4)[/tex]

Which rearranges as follows:

[tex]y-5=-2x+8\\\\y=-2x+13[/tex]

To determine if a point lies on a line, you can see if its coordinates satisfy the equation.

Of all the options, only (5,3) works.

Answer:

Step-by-step explanation:

Answer:

35

Step-by-step explanation:

first, you look at 7x, from the previous equation, you know that x=3, so you take 7x3=21 then you evaluate -3y. as you did with x on the last one you will look at the equation for y and see that it's -4. A negative times a negative is a positive, so -4x(-3)= 12. Then you add them all together, since 12 is a positive, the equation would now look like 21+12+2. After adding all three numbers together, you get 12.

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

Let's solve ~

Equation of directrix is : y = 1, so we can say that it's a parabola of form : -

[tex]\qquad \sf  \dashrightarrow \: (x - h) {}^{2} = 4a(y - k)[/tex]

and since the focus is above the directrix, it's a parabola with upward opening.

[tex]\qquad \sf  \dashrightarrow \: (x - ( - 4)) {}^{2} = 4(2)(y - 5)[/tex]

[tex]\qquad \sf  \dashrightarrow \: (x + 4) {}^{2} = 8(y - 5)[/tex]

[tex]\qquad \sf  \dashrightarrow \: {x}^{2} + 8x + 16 = 8y - 40[/tex]

[tex]\qquad \sf  \dashrightarrow \: 8y = {x}^{2} + 8x + 56[/tex]

[tex]\qquad \sf  \dashrightarrow \: y = \cfrac{1}{8} {x}^{2} + x + 7[/tex]

Directrix

Focus

Focus lies in Q3 and above y=1

Then

Perpendicular distance

Find a for the equation

Now the equation is

[tex]\\ \rm\dashrightarrow 4a(y-k)=(x-h)^2[/tex]

[tex]\\ \rm\dashrightarrow 4(2)(y-5)=(x+4)^2[/tex]

[tex]\\ \rm\dashrightarrow 8(y-5)=x^2+8x+16[/tex]

[tex]\\ \rm\dashrightarrow 8y-40=x^2+8x+16[/tex]

[tex]\\ \rm\dashrightarrow 8y=x^2+8x+16+40[/tex]

[tex]\\ \rm\dashrightarrow 8y=x^2+8x+56[/tex]

[tex]\\ \rm\dashrightarrow y=\dfrac{x^2}{8}+x+7[/tex]

The arc length is given by the definite integral

[tex]\displaystyle \int_1^3 \sqrt{1 + \left(y'\right)^2} \, dx = \int_1^3 \sqrt{1+9x} \, dx[/tex]

since by the power rule for differentiation,

[tex]y = 2x^{3/2} \implies y' = \dfrac32 \cdot 2x^{3/2-1} = 3x^{1/2} \implies \left(y'\right)^2 = 9x[/tex]

To compute the integral, substitute

[tex]u = 1+9x \implies du = 9\,dx[/tex]

so that by the power rule for integration and the fundamental theorem of calculus,

[tex]\displaystyle \int_{x=1}^{x=3} \sqrt{1+9x} \, dx = \frac19 \int_{u=10}^{u=28} u^{1/2} \, du = \frac19\times\frac23 u^{1/2+1} \bigg|_{10}^{28} = \boxed{\frac2{27}\left(28^{3/2} - 10^{3/2}\right)}[/tex]

Answer:

the function is an exponential funtion.

Step-by-step explanation:

learned it

Triangle 1, triangle 3 and triangle 4 are congruent triangles bases on side-side-side and side-angle-side congruency.

Triangle is a polygon that has three sides and three angles. Types of triangles are isosceles, equilateral and scalene triangle.

Two triangles are said to be congruent if they have the same shape and their corresponding sides are congruent to each other. Also, their corresponding angles are congruent.

Triangle 1, triangle 3 and triangle 4 are congruent triangles bases on side-side-side and side-angle-side congruency.

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a) The monthly payment for each loan is as follows:

30-year $1,954.93

15-year $3,053.25

b) The savings in interest by using the 15-year loan is $154,189,20 ($286,774.20 - $132,585).

c) No. the monthly payment for the 15-year loan is not twice that of the 30-year loan as the loan term.

d) The interest savings for the 15-year loan are more than one-half of the interest paid on the 30-year loan.

The calculations for the monthly payments, including interests can be carried out using an online finance calculator, as follows:

N (# of periods) = 360 months (12 x 30 years)

I/Y (Interest per year) = 3.85%

PV (Present Value) = $417000

FV (Future Value) = $0

Results:

PMT = $1,954.93

Sum of all periodic payments = $703,774.80 ($1,954.93 x 360)

Total Interest = $286,774.20 ($703,774.80 - $417,000)

N (# of periods) = 180 months (12 x 15 years)

I/Y (Interest per year) = 3.85%

PV (Present Value) = $417000

FV (Future Value) = $0

Results:

PMT = $3,053.25

Sum of all periodic payments = $549,585 ($3,053.25 x 180)

Total Interest = $132,585 ($549,585 - $417,000)

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The psi that the TPMS would trigger a warning for this car is = 23.36 psi

The target tire pressure of the car is = 32 psi (pounds per square inch.)

The Tire pressure monitoring systems (TPMS) warns the car below 27% of 32psi

That is , 27/100 × 32

= 864/100

= 8.64psi

Therefore, 32 - 8.64 = 23.36. When the car is below 23.36psi, TPMS would trigger a warning for this car.

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Complete question:

Tire pressure monitoring systems (TPMS) warn the driver when the tire pressure of the vehicle is 27% below the target pressure. Suppose the target tire pressure of a certain car is 32 psi (pounds per square inch.)

At what psi will the TPMS trigger a warning for this car? (Round your answer to 2 decimal place.) When the tire pressure is above or below?

Answer:

y= -1

Step-by-step explanation:

(2) (0) − y = 1

0 + − y = 1

(−y) + (0) = 1

−y = 1

Step 2: Divide both sides by -1

Y = −1

Answer:

yes

Step-by-step explanation:

25x² - 40xy + 16y² can be factored as

(5x - 4y)² ← a perfect square

Answer:

Step-by-step explanation:

The correct value of w is 20 as the width of a rectangle must be positive. A quadratic function always has 2 zeroes and in a case like this the negative one is ignored.

Answer:

The Co-ordinate of C is (4/3, -1/2)

Step-by-step explanation:

We have two equation one is of straight line equation which is:

                                              y=2x-3           (i)

Other equation is of quadratic function which is:

                                         y=-3x^2+5         (ii)

Put the value of y from equation (i) in equation (ii)

So, we have:

                                     2x-3=-3x^2+5

                                      3x^2+2x-8=0

By factorization:

                                    3x^2+6x-4x-8=0

                                   3x(x+2)-4x(x+2)=0

                                        (x+2)(3x-4)=0

              x+2=0                             ;                      3x-4=0

              x=-2                                ;                       x=4/3

Put first x=-2 in equation (i)

                                                y=2(-2)-3

                                                  y=-4-3

                                                    y=-7

Now Put x=4/3 in equation (i)

                                             y=2(4/3)-3

                                                 y=8/3-3

                                                   y=-1/2

So, we have two Order pair One is (-2 , -7) and Second one is (4/3 , -1/2)

Hence the Co-ordinate of C is:

                                               C=(4/3 , -1/2)

Answer:

Point C:  (3, 3)

Point D:  (3, -22)

Step-by-step explanation:

If the distance between points C and D is 25 units, the y-value of point D will be 25 less than the y-value of point C.  The x-values of the two points are the same.

Therefore:

[tex]\textsf{Equation 1}: \quad y=2x-3[/tex]

[tex]\textsf{Equation 2}: \quad y-25=-3x^2+5[/tex]

As the x-values are the same, substitute the first equation into the second equation and solve for x to find the x-value of points C and D:

[tex]\implies 2x-3-25=-3x^2+5[/tex]

[tex]\implies 3x^2+2x-33=0[/tex]

[tex]\implies 3x^2-9x+11x-33=0[/tex]

[tex]\implies 3x(x-3)+11(x-3)=0[/tex]

[tex]\implies (x-3)(3x+11)=0[/tex]

[tex]\implies x=3, -\dfrac{11}{3}[/tex]

From inspection of the given graph, the x-value of points C and D is positive, therefore x = 3.

To find the y-value of points C and D, substitute the found value of x into the two original equations of the lines:

[tex]\begin{aligned} \textsf{Point C}: \quad 2x-3 & =y\\2(3)-3 & =3\\ \implies & (3, 3)\end{aligned}[/tex]

[tex]\begin{aligned} \textsf{Point D}: \quad -3x^2+5 & = y \\ -3(3)^2+5 & =-22\\ \implies & (3, -22)\end{aligned}[/tex]

Therefore, point C is (3, 3) and point D is (3, -22).

The proportion of the population exists less than 30 then

(x< 30) = 0.986.

To estimate the z-score using the formula, z = (x - µ)/σ

Where, x be the randomly chosen values = 30

µ be the mean = 31

σ be the standard deviation = 10

Proportion of the population that exists less than 18 = P(x < 30)

Plug in the values into z = (x - µ)/σ, to get z-score.

Substitute the values in the above equation, we get

z = (30 - 31)/10

z = -1/10 = -0.1

Therefore, the value of z = - 0.1.

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Answer:

yes yes yes yes yes Yes Yes you are cute

Answer:

4

Step-by-step explanation:

this is the same question as what number can divide

13-1 = 12, 17-1 = 16 and 21-1 = 20 and has 0 remainder ?

the greatest number that can do that is 4.

we can easily see that, but formally, let's do prime factorization :

12 ÷ 2 = 6

6 ÷ 2 = 3

3 ÷ 2 no

3 ÷ 3 = 1 finished

12 = 2×2×3

16 ÷ 2 = 8

8 ÷ 2 = 4

4 ÷ 2 = 2

2 ÷ 2 = 1 finished

16 = 2×2×2×2

20 ÷ 2 = 10

10 ÷ 2 = 5

5 ÷ 2 no

5 ÷ 3 no

5 ÷ 5 = 1 finished

20 = 2×2×5

so, the largest common factor is the combination of the longest streaks per factor they have in common.

they only have 2s in common.

and the longest common streak is 2×2 = 4.

hence the answer

Answer:

9

Step-by-step explanation:

To find the rate of change we use the formula

f(x2) - f(x1)

------------------

x2 -x1

f(x) = 9x

x2 = 9  and x1 = 0

f(x2) = 9( 8) = 72

f(x1) = 9(0) =0

The rate of change is

72 - 0

------------

8-0

72

----

8

9

The rate of change is 9

Answer:

9

Step-by-step explanation:

The average rate of change of function f(x) over the interval a ≤ x ≤ b is given by:

[tex]\dfrac{f(b)-f(a)}{b-a}[/tex]

Given:

Therefore:

Substitute the given values into the average rate of change formula:

[tex]\begin{aligned}\implies \dfrac{f(8)-f(0)}{8-0} & = \dfrac{9(8)-9(0)}{8-0}\\\\& = \dfrac{72-0}{8-0}\\\\& = \dfrac{72}{8}\\\\& = 9\end{aligned}[/tex]

Therefore, the average rate of change of the function f(x) over the interval 0 ≤ x ≤ 8 is 9.

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Answer: 10,204 feet

Step-by-step explanation: i would assume you would add the two together, seeing as if the mountain is 10,093 above sea level and the valley is 111 below, the difference in elevation is also the distance between each other.

The solution to the given expression is x = -20/11

Logarithmic function are inverse of exponential functions. Given the equation below;

log (6 x + 10) = log 1/2 x

In order to determine the solution to the given logarithmic equation, we will first have to cancel the logarithm on both sides to have

6x + 10 = 1/2x

Collect the like terms

6x - 1/2x = 0 - 10

Find the LCD

12x-x/2 = -10
11x/2 = -10

Cross multiply

11x = -2 * 10

11x = -20

Divide both sides by 11

11x/11 = -20/11

x = -20/11

Hence the solution to the given expression is x = -20/11

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Answer:

Consider the table for y= |x+1| :

x   |   y

---------

0      1

1       2

2      3

-1      2

-2     3

This would give us the parent function of y=|x| but translated up one unit. It should look like a v starting at (0, 1)

The second derivative of the implicit function x · y - 1 = 2 · x + y² is equal to y'' = [2 / (2 · y - x)] · [(2 - y) / (x - 2 · y)] · [1 - [(2 - y) / (x - 2 · y)]].

In this problem we have a function in implicit form, that is, an expression of the form: f(x, y, c) = 0, where c is a constant. Then, we should apply implicit differentiation twice to determine the second derivative of the function:

Original expression

x · y - 1 = 2 · x + y²

First derivative

y + x · y' = 2 + 2 · y · y'

(x - 2 · y) · y' = 2 - y

y' = (2 - y) / (x - 2 · y)

Second derivative

y' + y' + x · y'' = 2 · (y')² + 2 · y · y''

2 · y' - 2 · (y')² = (2 · y - x) · y''

y'' = 2 · [y' - (y')²] / (2 · y - x)

y'' = [2 / (2 · y - x)] · [(2 - y) / (x - 2 · y)] · [1 - [(2 - y) / (x - 2 · y)]]

The second derivative of the implicit function x · y - 1 = 2 · x + y² is equal to y'' = [2 / (2 · y - x)] · [(2 - y) / (x - 2 · y)] · [1 - [(2 - y) / (x - 2 · y)]].

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Answer:

x = 48°

Step-by-step explanation:

Complementary angles

Angles that sum to 90°.

Vertical Angle Theorem

When two straight lines intersect, the vertical angles are congruent (equal).

Therefore, angle x is equal to the angle that is complementary to 42°.

To find x, subtract 42° from 90°:

⇒ x = 90° - 42°

⇒ x = 48°

Answer:

See below for proof.

Step-by-step explanation:

Given:

[tex]y=\left(x+\sqrt{1+x^2}\right)^m[/tex]

First derivative

[tex]\boxed{\begin{minipage}{5.4 cm}\underline{Chain Rule for Differentiation}\\\\If $f(g(x))$ then:\\\\$\dfrac{\text{d}y}{\text{d}x}=f'(g(x))\:g'(x)$\\\end{minipage}}[/tex]

[tex]\boxed{\begin{minipage}{5 cm}\underline{Differentiating $x^n$}\\\\If $y=x^n$, then $\dfrac{\text{d}y}{\text{d}x}=xn^{n-1}$\\\end{minipage}}[/tex]

[tex]\begin{aligned} y_1=\dfrac{\text{d}y}{\text{d}x} & =m\left(x+\sqrt{1+x^2}\right)^{m-1} \cdot \left(1+\dfrac{2x}{2\sqrt{1+x^2}} \right)\\\\ & =m\left(x+\sqrt{1+x^2}\right)^{m-1} \cdot \left(1+\dfrac{x}{\sqrt{1+x^2}} \right) \\\\ & =m\left(x+\sqrt{1+x^2}\right)^{m-1} \cdot \left(\dfrac{x+\sqrt{1+x^2}}{\sqrt{1+x^2}} \right)\\\\ & = \dfrac{m}{\sqrt{1+x^2}} \cdot \left(x+\sqrt{1+x^2}\right)^{m-1} \cdot \left(x+\sqrt{1+x^2}\right)\\\\ & = \dfrac{m}{\sqrt{1+x^2}}\left(x+\sqrt{1+x^2}\right)^m\end{aligned}[/tex]

Second derivative

[tex]\boxed{\begin{minipage}{5.5 cm}\underline{Product Rule for Differentiation}\\\\If $y=uv$ then:\\\\$\dfrac{\text{d}y}{\text{d}x}=u\dfrac{\text{d}v}{\text{d}x}+v\dfrac{\text{d}u}{\text{d}x}$\\\end{minipage}}[/tex]

[tex]\textsf{Let }u=\dfrac{m}{\sqrt{1+x^2}}[/tex]

[tex]\implies \dfrac{\text{d}u}{\text{d}x}=-\dfrac{mx}{\left(1+x^2\right)^\frac{3}{2}}[/tex]

[tex]\textsf{Let }v=\left(x+\sqrt{1+x^2}\right)^m[/tex]

[tex]\implies \dfrac{\text{d}v}{\text{d}x}=\dfrac{m}{\sqrt{1+x^2}} \cdot \left(x+\sqrt{1+x^2}\right)^m[/tex]

[tex]\begin{aligned}y_2=\dfrac{\text{d}^2y}{\text{d}x^2}&=\dfrac{m}{\sqrt{1+x^2}}\cdot\dfrac{m}{\sqrt{1+x^2}}\cdot\left(x+\sqrt{1+x^2}\right)^m+\left(x+\sqrt{1+x^2}\right)^m\cdot-\dfrac{mx}{\left(1+x^2\right)^\frac{3}{2}}\\\\&=\dfrac{m^2}{1+x^2}\cdot\left(x+\sqrt{1+x^2}\right)^m+\left(x+\sqrt{1+x^2}\right)^m\cdot-\dfrac{mx}{\left(1+x^2\right)\sqrt{1+x^2}}\\\\ &=\left(x+\sqrt{1+x^2}\right)^m\left(\dfrac{m^2}{1+x^2}-\dfrac{mx}{\left(1+x^2\right)\sqrt{1+x^2}}\right)\\\\\end{aligned}[/tex]

              [tex]= \dfrac{\left(x+\sqrt{1+x^2}\right)^m}{1+x^2}\right)\left(m^2-\dfrac{mx}{\sqrt{1+x^2}}\right)[/tex]

Proof

  [tex](x^2+1)y_2+xy_1-m^2y[/tex]

[tex]= (x^2+1) \dfrac{\left(x+\sqrt{1+x^2}\right)^m}{1+x^2}\left(m^2-\dfrac{mx}{\sqrt{1+x^2}}\right)+\dfrac{mx}{\sqrt{1+x^2}}\left(x+\sqrt{1+x^2}\right)^m-m^2\left(x+\sqrt{1+x^2\right)^m[/tex]

[tex]= \left(x+\sqrt{1+x^2}\right)^m\left(m^2-\dfrac{mx}{\sqrt{1+x^2}}\right)+\dfrac{mx}{\sqrt{1+x^2}}\left(x+\sqrt{1+x^2}\right)^m-m^2\left(x+\sqrt{1+x^2\right)^m[/tex]

[tex]= \left(x+\sqrt{1+x^2}\right)^m\left[m^2-\dfrac{mx}{\sqrt{1+x^2}}+\dfrac{mx}{\sqrt{1+x^2}}-m^2\right][/tex]

[tex]= \left(x+\sqrt{1+x^2}\right)^m\left[0][/tex]

[tex]= 0[/tex]

Answer:

Hello,

Step-by-step explanation:

All is in the picture.

B=(-2,8), O=(-2,-5)

b=BO=8+5=13

F_1=(-2,4)   O=(-2,5)  Focus distance=4+5=9
Horizontal half axis=√(b²-f²)=√88

Answer:

t2=8/5

Step-by-step explanation:

using this formula

t1/s1 =t2/s2

4/5=t2/2

cross multiply

5t2=8

t2=8/5

The correct answer for the value of t₂ is [tex]1.6[/tex].

Given:

Time t₁ = 4,

Distance s₂ =2

Distance s₁ = 5.

To find value of t₂ , use the concept of proportion:

[tex]\dfrac{t_1}{s_1} = \dfrac{t_2}{s_2}[/tex]

Put value of [tex]t_1 ,s_1 ,s_2[/tex]:

[tex]\dfrac{t_2}{2} =\dfrac{4}{5}\\\\t_2 =\dfrac{8}{5}\\\\ t_2 = 1.6[/tex]

The correct value of [tex]t_2[/tex] is [tex]1.6[/tex].

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the sample space is:

{ (1, 1), (1, 2), (1, 3), (1, 4), (2, 1), (3, 1), (4, 1), (2, 2), (2, 3), (3, 2)}

Event A is rolling a sum less than 6.

Let's define the possible elements in this experiment as:

(outcome of dice 1, outcome of dice 2)

The outcomes where the sum is less than 6 are:

dice 1    dice 2     sum

  1               1           2

  1               2           3

  1               3          4

  1               4          5

  2               1          3

  3              1           4

  4               1          5

  2              2          4

  3              2          5

  2              3           5

So there are 10 outcomes, then the sample space is:

{ (1, 1), (1, 2), (1, 3), (1, 4), (2, 1), (3, 1), (4, 1), (2, 2), (2, 3), (3, 2)}

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Answer:

B) {(1, 1), (1, 2), (1, 3), (1, 4), (2, 1), (2, 2), (2, 3), (3, 1), (3, 2), (4, 1)}

Step-by-step explanation:

If the question said less than 6 meaning you have to find all possible solution that are 5 or lower.

However, if the problem said equal or less than 6 then you have to find all possible solution that are 6 or lower.

B option is only option that don't have sum of 6. Therefore, option B is correct.

The solutions for the given equations are:

Writing a number or an equation as a product of its factors is said to be the factorization.

A linear equation has only one factor, a quadratic equation has 2 factors and a cubic equation has 3 factors.

1. Solving x² - 49 = 0; (quadratic equation)

⇒ x² - 7² = 0

This is in the form of a² - b². So, a² - b² = (a + b)(a - b)

⇒ (x + 7)(x - 7) =0

By the zero-product rule,

x = -7 and 7.

2. Solving 3x³ - 12x = 0

⇒ 3x(x² - 4) = 0

⇒ 3x(x² - 2²) = 0

⇒ 3x(x + 2)(x - 2) = 0

So, by the zero product rule, x = -2, 0, 2

3. Solving 12x² + 14x + 12 = 18; (quadratic equation)

⇒ 12x² + 14x + 12 - 18 = 0

⇒ 12x² + 14x - 6 = 0

⇒ 2(6x² + 7x - 3) = 0

⇒ 6x² + 9x - 2x - 3 = 0

⇒ 3x(2x + 3) - (2x + 3) = 0

⇒ (3x - 1)(2x + 3) = 0

∴ x = 1/3, -3/2

4. Solving -x³ + 22x² - 121x = 0

⇒ -x³ + 22x² - 121x = 0

⇒ -x(x² - 22x + 121) = 0

⇒ -x(x² - 11x - 11x + 121) = 0

⇒ -x(x(x - 11) - 11(x - 11)) = 0

⇒ -x(x - 11)² = 0

∴ x = 0, 11, 11

5. Solving x² - 4x = 5; (quadratic equation)

⇒ x² - 4x - 5 = 0

⇒ x² -5x + x - 5 = 0

⇒ x(x - 5) + (x - 5) = 0

⇒ (x + 1)(x - 5) =0

∴ x = -1, 5

Hence all the given equations are solved.

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The lowest common multiple of the expressions 3xyz^2 and 9x^2y + 9x^2 is 9x^2z^2(y + 1)

The expressions are given as:

3xyz^2 and 9x^2y + 9x^2

Factorize the expressions

3xyz^2 = 3 * x * y * z * z

9x^2y + 9x^2 = 3 * 3 * x * x * (y + 1)

Multiply the common factors, without repetition

LCM = 3 * 3 * x * x * (y + 1) * z* z

Evaluate the product

LCM = 9x^2z^2(y + 1)

Hence, the lowest common multiple of the expressions 3xyz^2 and 9x^2y + 9x^2 is 9x^2z^2(y + 1)

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