Angeline puts $2000 into savings account which receives 2.6% interest per year. How much interest will she have on her savings at the end of year??

Thank you so much!

The age of Nina and sam are 15years and 10 years respectively now

They consists of equations and unknown variables

Let their current age be x and y

If five years ago, Nina was twice as old as Sam then;

x - 5 = 2(y-5)

If in five years she will be 1.5 times as old as Sam, then;

x = 1.5y

Substitute

1.5y - 5 = 2y -10

-0.5y  = -5

y =5/0.5

y = 10

Since x = 1.5y

x = 15years

Hence the age of Nina and sam are 15years and 10 years respectively now

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Answer:

my best estimate would be 15,116,600

Step-by-step explanation:

,

Answer:

The probability scale for this question is not 100% sure to buy shoes so the scale is given to be even chance(1/2).

Step-by-step explanation:

Answer:

  30°

Step-by-step explanation:

The angle sum theorem tells you the sum of angles in any triangle is 180°. A right triangle has one angle that is 90°, so the sum of the remaining two acute angles is 180° -90° = 90°. That is, the acute angles in a right triangle are complementary.

__

The other acute angle is ...

  90° -60° = 30°

Answer:

2

Step-by-step explanation:

using the rules of logarithms

logx + logy = log(xy)

logx - logy = log ([tex]\frac{x}{y}[/tex] )

log[tex]x^{n}[/tex] ⇔ nlogx

[tex]log_{b}[/tex] x = n ⇒ x = [tex]b^{n}[/tex]

given

3log2 + log200 - log16

= log 2³ + log200 - log16

= log8 + log200 - log16

= log(8 × 200) - log16

= log1600 - log16

= log ([tex]\frac{1600}{16}[/tex] )

= log100

let [tex]log_{10}[/tex] 100 = n , then

100 = [tex]10^{n}[/tex] = 10²

thus value of expression is n = 2 , that is 2

[tex]2 \cdot 1[/tex]Answer:

Equal to 2

Step-by-step explanation:

1. Apply log rule: [tex]alog_c(b) = log_{c}(b^a)[/tex]

[tex]3log_{10}(2) = log_{10}(2^3)[/tex]

[tex]= log_{10}(2^3) + log_{10}(200) - log_{10}(16)[/tex]

2. Apply log rule: [tex]log_c(a) + log_c(b) = log_c(ab)[/tex]

[tex]log_{10}(2^3) + log_{10}(200) = log_{10}(2^3 \cdot 200)[/tex]

[tex]= log_{10}(2^3 \cdot 200) - log_{10}(16)[/tex]

[tex]2^3 \cdot 200 = 1600[/tex]

[tex]= log_{10}(1600) - log_{10}(16)[/tex]

3. Apply log rule: [tex]log_c(a) - log_c(b) = log_c(\frac{a}{b})[/tex]

[tex]log_{10}(1600) - log_{10}(16) = log_{10}(\frac{1600}{16})[/tex]

Divide the numbers: [tex]\frac{1600}{16} = 100[/tex]

[tex]= log_{10}(100)[/tex]

Rewrite 100 in power - base form: [tex]100 = 10^2[/tex]

[tex]= log_{10}(10^2)[/tex]

4. Apply log rule: [tex]log_a(x^b) = b \cdot log_a(x)[/tex]

[tex]log_{10}(10^2) = 2log_{10}(10)[/tex]

[tex]2log_{10}(10)[/tex]

5. Apply log rule: [tex]log_a(a) = 1[/tex]

[tex]log_{10}(10) = 1[/tex]

[tex]= 2 \cdot 1 = 2[/tex]

Answer:

-1/3

Step-by-step explanation:

because it is a negative number and closer to zero than -3 is making it greater, becuase it is closer to a positive number.

Answer:

-1/3 is greater than -3

-3 < -1/3

Answer:

A

Step-by-step explanation:

Close to 0 means there isn't a certain number, and there also isn't a certain answer for what the probability. While all of the others are alreaady stated, this is the only one that is not.

Answer:

angle1=70

angle2=65

angle4=115

angle7=45

angel6=45

angle5=45

Step-by-step explanation:

110+angle1 = 180

angle1 =180-70

angle1 = 70

115+angle2=180

angle2=180-115

angle2=65

angle4=115 because lines are parallel

angle1+angle2+angle7=180 because triangle

70+65+angle7=180

135+angle7=180

angle7=180-135

angle7=45

angel6=45

angle5=45

Linear equations are usually organized in slope-intercept form, which looks like this:

[tex]y=mx+b[/tex]

To find the equation of a line in slope-intercept form:

We're given:

First, plug the given slope into the slope-intercept form equation:

[tex]y=mx+b\\\\y=\dfrac{1}{2}x+b[/tex]

Now, determine the y-intercept.

[tex]y=\dfrac{1}{2}x+b[/tex]

⇒ Plug in the given point (4,1) and solve for b:

[tex]1=\dfrac{1}{2}(4)+b\\1=2+b\\b=-1[/tex]

⇒ Therefore, the y-intercept is -1. Plug this back into our original equation:

[tex]y=\dfrac{1}{2}x-1[/tex]

[tex]y=\dfrac{1}{2}x-1[/tex]

Answer:

See below for answers and explanations

Step-by-step explanation:

Problem 1

Recall that the projection of a vector [tex]u[/tex] onto [tex]v[/tex] is [tex]\displaystyle proj_vu=\biggr(\frac{u\cdot v}{||v||^2}\biggr)v[/tex].

Identify the vectors:

[tex]u=\langle-10,-7\rangle[/tex]

[tex]v=\langle-8,4\rangle[/tex]

Compute the dot product:

[tex]u\cdot v=(-10*-8)+(-7*4)=80+(-28)=52[/tex]

Find the square of the magnitude of vector v:

[tex]||v||^2=\sqrt{(-8)^2+(4)^2}^2=64+16=80[/tex]

Find the projection of vector u onto v:

[tex]\displaystyle proj_vu=\biggr(\frac{u\cdot v}{||v||^2}\biggr)v\\\\proj_vu=\biggr(\frac{52}{80}\biggr)\langle-8,4\rangle\\\\proj_vu=\biggr\langle\frac{-416}{80} ,\frac{208}{80}\biggr\rangle\\\\proj_vu=\biggr\langle\frac{-26}{5} ,\frac{13}{5}\biggr\rangle\\\\proj_vu=\langle-5.2,2.6\rangle[/tex]

Thus, B is the correct answer

Problem 2

Treat the football and wind as vectors:

Football: [tex]u=\langle42\cos172^\circ,42\sin172^\circ\rangle[/tex]

Wind: [tex]v=\langle13\cos345^\circ,13\sin345^\circ\rangle[/tex]

Add the vectors: [tex]u+v=\langle42\cos172^\circ+13\cos345^\circ,42\sin172^\circ+13\sin345^\circ\rangle\approx\langle-29.034,2.481\rangle[/tex]

Find the magnitude of the resultant vector:

[tex]||u+v||=\sqrt{(-29.034)^2+(2.481)^2}\approx29.14[/tex]

Find the direction of the resultant vector:

[tex]\displaystyle \theta=tan^{-1}\biggr(\frac{2.841}{-29.034}\biggr)\approx -5^\circ[/tex]

Because our resultant vector is in Quadrant II, the true direction angle is 6° clockwise from the negative axis. This means that our true direction angle is [tex]180^\circ-5^\circ=175^\circ[/tex]

Thus, C is the correct answer

Problem 3

We identify the initial point to be [tex]R(-2,12)[/tex] and the terminal point to be [tex]S(-7,6)[/tex]. The vector in component form can be found by subtracting the initial point from the terminal point:

[tex]v=\langle-7-(-2),6-12\rangle=\langle-7+2,-6\rangle=\langle-5,-6\rangle[/tex]

Next, we find the magnitude of the vector:

[tex]||v||=\sqrt{(-5)^2+(-6)^2}=\sqrt{25+36}=\sqrt{61}\approx7.81[/tex]

And finally, we find the direction of the vector:

[tex]\displaystyle \theta=tan^{-1}\biggr(\frac{6}{5}\biggr)\approx50.194^\circ[/tex]

Keep in mind that since our vector is in Quadrant III, our direction angle also needs to be in Quadrant III, so the true direction angle is [tex]180^\circ+50.194^\circ=230.194^\circ[/tex].

Thus, A is the correct answer

Problem 4

Add the vectors:

[tex]v_1+v_2=\langle-60,3\rangle+\langle4,14\rangle=\langle-60+4,3+14\rangle=\langle-56,17\rangle[/tex]

Determine the magnitude of the vector:

[tex]||v_1+v_2||=\sqrt{(-56)^2+(17)^2}=\sqrt{3136+289}=\sqrt{3425}\approx58.524[/tex]

Find the direction of the vector:

[tex]\displaystyle\theta=tan^{-1}\biggr(\frac{17}{-56} \biggr)\approx-17^\circ[/tex]

Because our vector is in Quadrant II, then the direction angle we found is a reference angle, telling us the true direction angle is 17° clockwise from the negative x-axis, so the true direction angle is [tex]180^\circ-17^\circ=163^\circ[/tex]

Thus, A is the correct answer

Problem 5

A vector in trigonometric form is represented as [tex]w=||w||(\cos\theta i+\sin\theta i)[/tex] where [tex]||w||[/tex] is the magnitude of vector [tex]w[/tex] and [tex]\theta[/tex] is the direction of vector [tex]w[/tex].

Magnitude: [tex]||w||=\sqrt{(-16)^2+(-63)^2}=\sqrt{256+3969}=\sqrt{4225}=65[/tex]

Direction: [tex]\displaystyle \theta=tan^{-1}\biggr(\frac{-63}{-16}\biggr)\approx75.75^\circ[/tex]

As our vector is in Quadrant III, our true direction angle will be 75.75° counterclockwise from the negative x-axis, so our true direction angle will be [tex]180^\circ+75.75^\circ=255.75^\circ[/tex].

This means that our vector in trigonometric form is [tex]w=65(\cos255.75^\circ i+\sin255.75^\circ j)[/tex]

Thus, C is the correct answer

Problem 6

Write the vectors in trigonometric form:

[tex]u=\langle40\cos30^\circ,40\sin30^\circ\rangle\\v=\langle50\cos140^\circ,50\sin140^\circ\rangle[/tex]

Add the vectors:

[tex]u+v=\langle40\cos30^\circ+50\cos140^\circ,40\sin30^\circ+50\sin140^\circ\rangle\approx\langle-3.661,52.139\rangle[/tex]

Find the magnitude of the resultant vector:

[tex]||u+v||=\sqrt{3.661^2+52.139^2}\approx52.268[/tex]

Find the direction of the resultant vector:

[tex]\displaystyle\theta=tan^{-1}\biggr(\frac{52.139}{-3.661} \biggr)\approx-86^\circ[/tex]

Because our resultant vector is in Quadrant II, then our true direction angle will be 86° clockwise from the negative x-axis. So, our true direction angle is [tex]180^\circ-86^\circ=94^\circ[/tex].

Thus, B is the correct answer

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

Explanation:

Refer to the figure below.

Focus on triangle ABF. Angle AFB = 40 is marked in red. Angle BAF = 90 is marked in blue. The missing angle is 180-90-40 = 50 which is angle ABF.

Inscribed angle ABF = 50 doubles to arc AC since this inscribed angle subtends the arc in question. I'm using the inscribed angle theorem.

arc AC = 2*(angle ABF) = 2*50 = 100 degrees

Note: We don't need the fact that angle AGD = 86

Answer:

carbon (C), hydrogen (H), nitrogen (N), and oxygen (O)—make up 96.5% of an organism's weight.

Step-by-step explanation:

hope this helps.

Answer:

Θ = 84°

Step-by-step explanation:

using the identity

cosx = sin(90 - x) , then

cos6° = sin(90 - 6)° = sin84°

that is Θ = 84°

Pi, or π, is a number that is used to represent the ratio between a circle's circumference and its diameter.

Step-by-step explanation:

Answer:

a) m problems, d days to complete.

m/d

b) m/15d

c)-if 15 was connected; 15d

if 15 WAS NOT connected, m problems

Answer:

hopefully this helps :] 16th line down I think

[tex]x^2+y^2=r^2~\hfill x=rcos(\theta )~\hfill y=rsin(\theta ) \\\\[-0.35em] ~\dotfill\\\\ r=3sin(A)\implies \stackrel{\textit{multiplying both sides by "r"}}{r^2~~ = ~~3~rsin(A)}\implies x^2+y^2=3y[/tex]

r = 3sinA is written in rectangular form as[tex]x^2+y^2 = 3y.[/tex]

Any complex number z = a + ib can be written in polar form as:

[tex]z = r(\cos(\theta) + i\sin(\theta))[/tex]

Complex numbers are those numbers that contain the imaginary and the real part.

Raising this to nth power (n being an integer), we get:

[tex]z ^n = r^n (\cos(n\theta) + i\sin(n\theta))[/tex]

We need to find r = 3sinA is written in rectangular form as[tex]x^2+y^2[/tex].

We know that

x = r cosA

y = r sin A

We have been given ;

r = 3sinA

Multiply by r on both sides;

[tex]r^2 = 3r sin A\\\\r^2 = 3 y[/tex]

therefore,

[tex]x^2+y^2 = 3y[/tex]

Hence, r = 3sinA is written in rectangular form as [tex]x^2+y^2 = 3y[/tex].

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Answer:

42 in

Step-by-step explanation:

The lengths of the tangent segments drawn from an external point to a circle are equal.

AT and RT meet at point T

⇒ AT = RT = 9 in

RE and GE meet at point E

⇒ RE = GE = 5 in

AN and NG meet at point N

⇒ AN = GN = 7 in

Total perimeter = AT + RT + RE + GE + GN + AN

                          = 9 + 9 + 5 + 5 + 7 + 7

                          = 42 in

Answer:

1 / 216, or 0.00462...

Step-by-step explanation:

Answer:

the area of the rhombus is 56 square inches

Answers:

Area = 254.34 square feet

Circumference = 56.52 feet

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

Work Shown:

Area = pi*(radius)^2

A = pi*r^2

A = 3.14*9^2

A = 254.34 square feet

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

Circumference = 2*pi*(radius)

C = 2*pi*r

C = 2*3.14*9

C = 56.52 feet

The two results are approximate since pi = 3.14 is approximate.

1st area of circle:

[tex]\bold{\ area \ of \ circle \ : \pi r^2}[/tex]

[tex]\hookrightarrow \sf (3.14)(9)^2[/tex]

[tex]\hookrightarrow \sf 254.34 \ ft^2[/tex]

2nd circumference of circle:

[tex]\bold{circumference \ of \ circle \ : \ 2\pi r}[/tex]

[tex]\hookrightarrow \sf 2(3.14)(9)[/tex]

[tex]\hookrightarrow \sf 56.52 \ ft[/tex]

Answer:

13 pints

Step-by-step explanation:

6 1/2 (6.50) quarts is equal to 13 pints

Answer:

13pt

to go from quarts to pints multiply the value by 2

Answer:

A. evaluate progress

Step-by-step explanation:

common sense

Answer:

F= -4t+61

T= -1/4f + 61/4

Answer:

No solution

Step-by-step explanation:

(I am going to use substitution)

2x−10y=−2

y=x/5−1

(Substitute the y into the other problem)

2x - 10 (x/5 -1) = -2

2x - 2x + 10 = -2

10 = -2

No solution

Hope this helps :)

Answer: dude idek how to do this

Step-by-step explanation:

not lying still dont know

The total balance in Raul's account after 40 years when he retires is $65,714.90.

The formula that can be used to determine the balance of the accout is: monthly amount saved x annuity factor.

Annuity factor = {[(1+r)^n] - 1} / r

Where:

$100 x [(1.00125^480) - 1 ] / 0.00125 = $65,714.90

Here is the complete question:

Raul is a saver. He sets aside $100 per month during his career of 40 years to prepare for retirement. He does not like the idea of investing because he prefers to minimize his risk as much as possible, so he puts his money in a savings account which earns 1.5% interest per year. What is the balance in the account after 40 years?

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Answer:

[tex]m=\frac{6}{7}[/tex]

Step-by-step explanation:

[tex]m=\frac{10-4}{9-2}=\frac{6}{7}[/tex]

Answer:  [tex]\frac{6}{7}[/tex]

Explanation:

In the numerator we have [tex]y_2 - y_1 = 10-4 = 6[/tex] to represent the change in y, aka the rise. This is the vertical change. It's how much we go up.

The denominator has [tex]x_2 - x_1 = 9-2 = 7[/tex] to represent the run. It is the horizontal change, or how much we go to the right.

We go up 6 and to the right 7 to have a slope of [tex]m = \frac{\text{rise}}{\tex{run}} = \frac{6}{7}[/tex]

Here's how you can show the steps

[tex]m = \frac{y_2-y_1}{x_2-x_1}\\\\m = \frac{10-4}{9-2}\\\\m = \frac{6}{7}\\\\[/tex]

Note: To write [tex]\frac{6}{7}[/tex] on the keyboard, you can type 6/7