Answer:
5,310.
Step-by-step explanation:
Answer:
3690
Step-by-step explanation:
Simplify 18% to 0.18
4500-(4500×0.18)
4500-810
=3690
Evaluate: ab for a = 2 and b = 5
please mark this answer as brainlist
The ratio of Mitchell's age to Connor's age is 8:5. In thirty years, the ratio of their ages will be 6:5. How much older is Mitchell than Connor now?
Answer:
9 years older
Step-by-step explanation:
The ratio of their ages is 8 : 5 = 8x : 5x ( x is a multiplier )
In 30 years their ages will be 8x + 30 and 5x + 30 and the ratio 6 : 5 , so
[tex]\frac{8x+30}{5x+30}[/tex] = [tex]\frac{6}{5}[/tex] ( cross- multiply )
5(8x + 30) = 6(5x + 30) ← distribute parenthesis on both sides
40x + 150 = 30x + 180 ( subtract 30x from both sides )
10x + 150 = 180 ( subtract 150 from both sides )
10x = 30 ( divide both sides by 10 )
x = 3
Then
Michell is 8x = 8 × 3 = 24 years old
Connor is 5x = 5 × 3 = 15 years old
Mitchell is 24 - 15 = 9 years older than Connor
solue for &
X(3 + X) = 3x + x²
3x+x^2=3x+x^2
3x-3x=x^2-x^2
which means x=0
help me pls??????? :)
Answer:4 in each bad 2 left over
Step-by-step explanation:
Answer:
4 in each bag and 2 left over
Step-by-step explanation:
divide 14 by 3
3 goes into 14, 4 times
14 - 12 = 2
4 in each bag and then 2 left over
Select the correct answer from each drop-down menu.
A company makes cylindrical vases. The capacity, in cubic centimeters, of a cylindrical vase the company produces is given by the
function C() = 6.2873 + 28.26x2, where x is the radius, in centimeters. The area of the circular base of a vase, in square
centimeters, is given by the function A () = 3.14.2
To find the height of the vase, divide
represents the height of the vase.
the expressions modeling functions C(x) and A(z). The expression
Answer:
divide, 2x+9
Step-by-step explanation:
got it right
factor and solve problem in picture pleaseeee
I hope it's right have a great day :)
Please help I’ll mark as brainlist
Answer:
Ekta and Preyal
Step-by-step explanation:
Which function represents g(x), a refection of f(x)=1/2(3)^x across the y-axis
Answer: g(x) = (1/2)3^-x reflection over y axis yields (-x,y)
Use the information in the figure. If F=116, find E
58
32
116
64
Step-by-step explanation:
Given that,
m∠F = 116°We have to find the value of m∠E.
Here, two sides are equal, thus it is an isosceles triangle. As the two sides are equal, so their angles must be equal. So, ∠E and ∠D will be equal. Let us assume the measures of both ∠E and ∠D as x.
→ Sum of all the interior angles of ∆ = 180°
→ ∠E + ∠D + ∠F = 180°
→ 116° + x + x = 180°
→ 2x = 180° – 116°
→ 2x = 64°
→ x = 64° ÷ 2
→ x = 32°
Henceforth,
→ m∠E = x
→ m∠E = 32°
[tex] \\ [/tex]
~
in 16 years, Marissa will be five times older than she is today. How old is she?
Answer:
4
Step-by-step explanation:
1 5
2 10
3 15
4 20
i wrote it out
x+16=5x
16=4x
4=x
Marissa is currently four years old.
What is algebra?Algebra is a discipline of mathematics that deals with symbols and the rules that govern their use.
Equations in mathematics express relationships between variables in the same way that sentences indicate relationships between specific words.
Let's represent Marissa's current age "x".
According to the given question, in 16 years she will be five times older than she is today, which we can express as:
x + 16 = 5x
Solving for x, we can subtract x from both sides:
16 = 4x
Dividing both sides by 4, we get:
x = 4
Therefore, Marissa is currently 4 years old.
To learn more about the Algebra link is given below.
brainly.com/question/953809
#SPJ6
Brody works part-time at a veterinarian's office in addition to going to college, and he is paid twice a month. Which type of budget would likely work best for Brody?
The type of budget that would likely work best for Brody is biweeky budget.
Budget is an economic term that refers to the planning and advance formulation of expenses and income. The budget is a tool to organize expenses depending on the amount of money available.
The type of budget that would be best for Brody is a biweekly budget because he receives his payment every fifteen days (twice a month). So, he can schedule his expenses each time he receives his payment, in this way he does not spend all his money before he receives the next payment.
Additionally, weekly, monthly, and dairy are not correct options because they do not fit the time periods in which Brody receives payment for his services.
Learn more in: https://brainly.com/question/141889
Note:
This question is incomplete because options are missing, here are the options.
Daily budget
Biweekly budget
Monthly budget
Weekly budget
Explain why they substituted cos(60) with 1/2 ?
(Look at image)
9514 1404 393
Answer:
equals can be substituted anytime anywhere
Step-by-step explanation:
cos(60°) = 1/2, so wherever one appears, the other can be substituted. This is allowed by the substitution property of equality.
__
If you don't substitute at some point, you find the answer to be ...
x = 10/cos(60°)
Most of us are interested in a numerical value for x, so we prefer that cos(60°) be replaced by a numerical value.
PLZ HELP!! ASAP PLZ!! NO FILES.
Answer:
Slope is (1/4)
Step-by-step explanation:
The slope is calculated by (6-5)/(5-1)=1/4
Find the length of the side and area ot the square whose perimeter is given below a)44cm b)80cm
9514 1404 393
Answer:
a) 11 cm
b) 20 cm
Step-by-step explanation:
A square has four equal-length sides, so the perimeter is 4 times the side length. Then the side length is 1/4 of the perimeter.
a) s = (1/4)(44 cm) = 11 cm
b) s = (1/4)(80 cm) = 20 cm
Consider these four statements about a line that passes through two points on a plane.
Statement A: The lines on a different plane from the points.
Stament B: The line lies in only one plane.
Statement C: The line is on the same plane as the points.
Statement D: The line passes through only one plane.
Which statement is true?
Answer:
4
Step-by-step explanation:
Help pleaseee, I’ll give brainly!
Answer:
1) 6r+7=13+7r —> 7r–6r=7–13 —> r = – 6
2) 13–4x=1–x —> 4x–x=13–1 —> 3x=12 —> x=12/3 —> x=4
3)–7x–3x+2=–8x–8 —> –8x+7x+3x=2+8 —> 2x=10 –> x= 10/2 –> x= 5
4)–8–x=x–4x —> –x–x+4x=8 —> 2X=8 —> x= 8/2 —> x= 4
5) –14+6b+7-2b=1+5b —> 5b +2b –6b = –14+7–1 —> b=–8
6) n+2=–14–n —> n+n=–14–2 —> 2n = –16 —> n = – 16/ 2 —> n = – 8
7) n – 3n = 14 –4n —> n –3n + 4n = 14 —> 2n = 14 —> n = 14/ 2 —> n = 7
8) 7a – 3 = 3 + 6a —> 7a – 6a = 3 + 3 —> a = 6
9) 3(1–3x ) =2(–4x+7) —> 3 –9x = –8x+14 —> 9x–8x = 3–14 —> x = –11
10) –10 +x+4–5 =7x –5 —> 7x–x = –10+4–5 +5 —> 6x = –6 —> x= –6/6 —> x = –1
11) –8n +4(1+5n)=–6n–14 —> –8n +4 + 20n = – 6n– 14
20n –8n +6n= –14 –4 —> 18n = – 18 —> n = –18/18 —> n = –1
12) –6n–20=–2n +4(1–3n) —> –6n –20 = – 2n +4 –12n —> 12n +2n –6n = 4 +20 —> 8n =24 —> n = 24/8 —> n =3
I hope I helped you^_^
Answer:
1.
6r + 7 = 13 +7r
6r - 7r = 13-7
-r = 6
r = 6
2.
13 - 4x = 1-x
-4x +x = 1 -13
-3x = -12
x = -12 / -3
x = 4
3.
-7x - 3x + 2 = -8x -8
-10x +2 = -8x -8
-10x +8x = -8 -2
-2x = -6
x = -6 / -2
x = 3
4.
-8 - x = x- 4x
-8 - x = -3x
-x + 3x = -8
2x = -8
x = -8 / 2
x = -4
5.
-14 + 6b + 7 -2b = 1 + 5b
-7 + 4b = 1 + 5b
4b - 5b = 1 + 7
-b = 8
b = -8
6.
n + 2 = -14 -n
n + n = -14 -2
2n = - 16
n = -16 / 2
n = -8
7.
n - 3n = 14 -4n
-2n = 14 - 4n
-2n +4n = 14
2n = 14
n = 14 /2
n = 7
8.
7a - 3 = 3 + 6a
7a - 6a = 3 +3
a = 6
9.
3 ( 1 - 3x ) = 2 (-4x + 7)
3 - 9x = -8x +14
-9x +8x = 14 - 3
-x = 11
x = -11
10.
-10 + x + 4 - 5 = 7x - 5
-10 +x -1 = 7x - 5
-11 + x = 7x - 5
-11 + 5 = 7x -x
-6 = 6x
x = -6/6
x = -1
11.
-8n + 4 ( 1 + 5n ) = -6n -14
-8n + 4 + 20n = -6n -14
12n +4 = -6n -14
12n + 6n = -14 -4
18n = -18
n = -18/18
n = -1
12.
-6n - 20 = -2n + 4 ( 1 - 3n)
-6n - 20 = -2n + 4 - 12n
-6n - 20 = -14n +4
-20 -4 = -14n +6n
-24 = 8n
n = -24/8
n = -3
7t + 6 + 3v + 6v
Hey can someone help ne
Answer:
7t + 6 + 9v
Step-by-step explanation:
7t + 6 + 3v + 6v (since 3v and 6v are like terms you will add them both.)
7t + 6 + 9v
Hope this helps, thank you :) !!
Answer:
7t+6+9v
Step-by-step explanation:
7t+6+3v+6v
7t has no opponent it is =7t
6 is on it own =6
3v+6v=9v,reason is 3v has an opponent which is 6v so addition of 3v and 6v is =9v
so ur ans. is =7t+6+9v
Solve the equation |8y+4|=2|y−1|.
Explain.
[tex]\\ \sf\longmapsto |8y+4|=2|y-1|[/tex]
[tex]\\ \sf\longmapsto 8y+4=2y-1[/tex]
[tex]\\ \sf\longmapsto 8y-2y=-1-4[/tex]
[tex]\\ \sf\longmapsto 6y=-5[/tex]
[tex]\\ \sf\longmapsto y=\dfrac{-5}{6}[/tex]
Solve for p.
–
19p–2p+16p+12=
–
18
p=
Answer:
6
BRAINLIEST, PLEASE!
Step-by-step explanation:
-19p - 2p + 16p + 12 = -18
-5p + 12 = -18
-5p = -30
p = 6
Answer:
p = 6
Step-by-step explanation:
Given
- 19p - 2p + 16p + 12 = - 18 ( simplify left side )
- 5p + 12 = - 18 ( subtract 12 from both sides )
- 5p = - 30 ( divide both sides by - 5 )
p = 6
pls help me ASAP !!!!!!!!!
What is | −34 | ?
-------
l l
-------
l l
--------
Step-by-step explanation:
Here,
| −34 |
= 34
As,
| | are absolute symbols so it's denote only the neutral number such as, | - 1 | = 1
Answer:
34
Step-by-step explanation:
|-34|
The absolute value bars means taking the non-negative value
|-34| = 34
Answer ASAP please!
………
Answer:
47.746, answer choice C
Step-by-step explanation:
47.746
help help help help
Answer:
abc is a triangle so ,
a is ( 9,6 )
b is ( 9,3 )
and c is ( 3,3 )
Write the following expression as a simplified polynomial in standard form.
(x-4)^2+3(x-4)+6
Answer:
x6−24x5+240x4−1280x3+3840x2−6144x+4102
Step-by-step explanation:
I don't know if this is right or not but there ig?
if the ordered pairs (x-2,3y+1) and (y+1,x+3) are equal,find x and y
plz help me
Write and solve a word problem that can be modeled by addition of two negative integers.
Answer:
Step-by-step explanation:
Question:
Max needs to purchase a car and withdraws $100 from his bank. In a few days he withdraws another $50 to make same repairs. In total what is the change in his bank balance from theese two costs?
Solution:
(-100) + (-50) =
-150
Answered by G a u t h m a t h
Given the following coordinates complete the glide reflection transformation.
hopefully this answer can help you to answer the next question.
Daniel buys a new car.
In the first year, the value of the car decreases by 12% of its original value.
The value of the car at the end of the first year is £9680.
(a) Work out the original value of the car.
The value of the car at the end of the first year is £9680.
In each of the second year, the third year, fourth year and the fifth year, the value of the car decreases by
x% of its value at the beginning of each year.
The value of the car at the end of the fifth year is £5000.
(b) Work out the value of x.
Give your answer correct to 3 significant figures.
Answer:
Step-by-step explanation:
Value of the car at the end of the first year = £9680
Depreciation = 12 %
Original price = £ x
If we reduce 12% of original price from x, we will get £9680
x - 12% of x = £ 9680
[tex]x-\frac{12}{100}x=9680\\\\\\\frac{88}{100}x=9680\\\\x=9680*\frac{100}{88}\\\\x = 11000[/tex]
Original price = £ 11000
3. Find the product, using suitable properties :
a) 26 x (-48) + (-48) x (-36)
b) 625 x (-35) + (-625) x 65
please answer fast 10 marks
a) 26 x (-48) + (-48) x (-36) = ( –1248) + ( + 1728) = – 1248+ 1728 = 480
b) 625 x (-35) + (-625) x 65 = ( –21875) + ( –40625) = – 21875 –40625 = –62500
I hope I helped you^_^
100 POINTS AND BRAINLIEST FOR THIS WHOLE SEGMENT
a) Find zw, Write your answer in both polar form with ∈ [0, 2pi] and in complex form.
b) Find z^10. Write your answer in both polar form with ∈ [0, 2pi] and in complex form.
c) Find z/w. Write your answer in both polar form with ∈ [0, 2pi] and in complex form.
d) Find the three cube roots of z in complex form. Give answers correct to 4 decimal
places.
Answer:
See Below (Boxed Solutions).
Step-by-step explanation:
We are given the two complex numbers:
[tex]\displaystyle z = \sqrt{3} - i\text{ and } w = 6\left(\cos \frac{5\pi}{12} + i\sin \frac{5\pi}{12}\right)[/tex]
First, convert z to polar form. Recall that polar form of a complex number is:
[tex]z=r\left(\cos \theta + i\sin\theta\right)[/tex]
We will first find its modulus r, which is given by:
[tex]\displaystyle r = |z| = \sqrt{a^2+b^2}[/tex]
In this case, a = √3 and b = -1. Thus, the modulus is:
[tex]r = \sqrt{(\sqrt{3})^2 + (-1)^2} = 2[/tex]
Next, find the argument θ in [0, 2π). Recall that:
[tex]\displaystyle \tan \theta = \frac{b}{a}[/tex]
Therefore:
[tex]\displaystyle \theta = \arctan\frac{(-1)}{\sqrt{3}}[/tex]
Evaluate:
[tex]\displaystyle \theta = -\frac{\pi}{6}[/tex]
Since z must be in QIV, using reference angles, the argument will be:
[tex]\displaystyle \theta = \frac{11\pi}{6}[/tex]
Therefore, z in polar form is:
[tex]\displaystyle z=2\left(\cos \frac{11\pi}{6} + i \sin \frac{11\pi}{6}\right)[/tex]
Part A)
Recall that when multiplying two complex numbers z and w:
[tex]zw=r_1\cdot r_2 \left(\cos (\theta _1 + \theta _2) + i\sin(\theta_1 + \theta_2)\right)[/tex]
Therefore:
[tex]\displaystyle zw = (2)(6)\left(\cos\left(\frac{11\pi}{6} + \frac{5\pi}{12}\right) + i\sin\left(\frac{11\pi}{6} + \frac{5\pi}{12}\right)\right)[/tex]
Simplify. Hence, our polar form is:
[tex]\displaystyle\boxed{zw = 12\left(\cos\frac{9\pi}{4} + i\sin \frac{9\pi}{4}\right)}[/tex]
To find the complex form, evaluate:
[tex]\displaystyle zw = 12\cos \frac{9\pi}{4} + i\left(12\sin \frac{9\pi}{4}\right) =\boxed{ 6\sqrt{2} + 6i\sqrt{2}}[/tex]
Part B)
Recall that when raising a complex number to an exponent n:
[tex]\displaystyle z^n = r^n\left(\cos (n\cdot \theta) + i\sin (n\cdot \theta)\right)[/tex]
Therefore:
[tex]\displaystyle z^{10} = r^{10} \left(\cos (10\theta) + i\sin (10\theta)\right)[/tex]
Substitute:
[tex]\displaystyle z^{10} = (2)^{10} \left(\cos \left(10\left(\frac{11\pi}{6}\right)\right) + i\sin \left(10\left(\frac{11\pi}{6}\right)\right)\right)[/tex]
Simplify:
[tex]\displaystyle z^{10} = 1024\left(\cos\frac{55\pi}{3}+i\sin \frac{55\pi}{3}\right)[/tex]Simplify using coterminal angles. Thus, the polar form is:
[tex]\displaystyle \boxed{z^{10} = 1024\left(\cos \frac{\pi}{3} + i\sin \frac{\pi}{3}\right)}[/tex]
And the complex form is:
[tex]\displaystyle z^{10} = 1024\cos \frac{\pi}{3} + i\left(1024\sin \frac{\pi}{3}\right) = \boxed{512+512i\sqrt{3}}[/tex]
Part C)
Recall that:
[tex]\displaystyle \frac{z}{w} = \frac{r_1}{r_2} \left(\cos (\theta_1-\theta_2)+i\sin(\theta_1-\theta_2)\right)[/tex]
Therefore:
[tex]\displaystyle \frac{z}{w} = \frac{(2)}{(6)}\left(\cos \left(\frac{11\pi}{6} - \frac{5\pi}{12}\right) + i \sin \left(\frac{11\pi}{6} - \frac{5\pi}{12}\right)\right)[/tex]
Simplify. Hence, our polar form is:
[tex]\displaystyle\boxed{ \frac{z}{w} = \frac{1}{3} \left(\cos \frac{17\pi}{12} + i \sin \frac{17\pi}{12}\right)}[/tex]
And the complex form is:
[tex]\displaystyle \begin{aligned} \frac{z}{w} &= \frac{1}{3} \cos\frac{5\pi}{12} + i \left(\frac{1}{3} \sin \frac{5\pi}{12}\right)\right)\\ \\ &=\frac{1}{3}\left(\frac{\sqrt{2}-\sqrt{6}}{4}\right) + i\left(\frac{1}{3}\left(- \frac{\sqrt{6} + \sqrt{2}}{4}\right)\right) \\ \\ &= \boxed{\frac{\sqrt{2} - \sqrt{6}}{12} -\frac{\sqrt{6}+\sqrt{2}}{12}i}\end{aligned}[/tex]
Part D)
Let a be a cube root of z. Then by definition:
[tex]\displaystyle a^3 = z = 2\left(\cos \frac{11\pi}{6} + i\sin \frac{11\pi}{6}\right)[/tex]
From the property in Part B, we know that:
[tex]\displaystyle a^3 = r^3\left(\cos (3\theta) + i\sin(3\theta)\right)[/tex]
Therefore:
[tex]\displaystyle r^3\left(\cos (3\theta) + i\sin (3\theta)\right) = 2\left(\cos \frac{11\pi}{6} + i\sin \frac{11\pi}{6}\right)[/tex]
If two complex numbers are equal, their modulus and arguments must be equivalent. Thus:
[tex]\displaystyle r^3 = 2\text{ and } 3\theta = \frac{11\pi}{6}[/tex]
The first equation can be easily solved:
[tex]r=\sqrt[3]{2}[/tex]
For the second equation, 3θ must equal 11π/6 and any other rotation. In other words:
[tex]\displaystyle 3\theta = \frac{11\pi}{6} + 2\pi n\text{ where } n\in \mathbb{Z}[/tex]
Solve for the argument:
[tex]\displaystyle \theta = \frac{11\pi}{18} + \frac{2n\pi}{3} \text{ where } n \in \mathbb{Z}[/tex]
There are three distinct solutions within [0, 2π):
[tex]\displaystyle \theta = \frac{11\pi}{18} , \frac{23\pi}{18}\text{ and } \frac{35\pi}{18}[/tex]
Hence, the three roots are:
[tex]\displaystyle a_1 = \sqrt[3]{2} \left(\cos\frac{11\pi}{18}+ \sin \frac{11\pi}{18}\right) \\ \\ \\ a_2 = \sqrt[3]{2} \left(\cos \frac{23\pi}{18} + i\sin\frac{23\pi}{18}\right) \\ \\ \\ a_3 = \sqrt[3]{2} \left(\cos \frac{35\pi}{18} + i\sin \frac{35\pi}{18}\right)[/tex]
Or, approximately:
[tex]\displaystyle\boxed{ a _ 1\approx -0.4309 + 1.1839i,} \\ \\ \boxed{a_2 \approx -0.8099-0.9652i,} \\ \\ \boxed{a_3\approx 1.2408-0.2188i}[/tex]