PLEASE HELP ME WITH THIS !!!

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

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

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:

abc is a triangle so ,

a is ( 9,6 )

b is ( 9,3 )

and c is ( 3,3 )

[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]

Answer:

Ekta and Preyal

Step-by-step explanation:

Answer:

Slope is (1/4)

Step-by-step explanation:

The slope is calculated by (6-5)/(5-1)=1/4

please mark this answer as brainlist

Answer:

first odd integer=x

second odd integer=x+2

third odd integer=x+4

x+x+2+x+4=2097

x+x+x+2+4=2097

3x+6=2097

3x=2097-6

3x=2091

3x/3=2091/3

x=697

therefore, x=697

x+2=697+2=699

x+4=697+4=701

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

Answer:

x=12

Step-by-step explanation:

LM = RS

9x+27 = 135

Subtract 27 from each side

9x+27-27 =135-27

9x=108

Divide each side by 9

9x/9 = 108/9

x = 12

the boxes are equivalent so the one with a single dash is equal to the other with a single dash.

the one with 2 dashes is equal to the other with 2 dashes so on and so forth

SR=LM

LM=9x+27

RS=135

9x+27=135

so I solve it in my own weird way but you can solve it differently. 135-27=108

108/9=12

so your answer is 12

Answer:

y = 10

Step-by-step explanation:

To find the y coordinate of the midpoint, take the y coordinates of the endpoints and average

(0+y)/2 = 5

Multiply each  die by 2

0+y = 10

y = 10

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

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

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?

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

3x+x^2=3x+x^2

3x-3x=x^2-x^2

which means x=0

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.

Answer:

1320

Step-by-step explanation:

Use the formula for sum of series, s(a) = n/2(2a + (n-1)d)

The terms increase by 6, so d is 6

a is the first term, -56

n is the terms you want to find, 33

Plug in the numbers, 33/2 (2(-56)+(32)6)

Simplify into 33(80)/2 and you get 1320

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

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.

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

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:

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]

Step-by-step explanation:

Given that,

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]

~

I hope it's right have a great day :)

Answer:

divide, 2x+9

Step-by-step explanation:

got it right

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

Step-by-step explanation:

A natural number is a positive whole number.

A whole number is a positive number with no fractions or decimals.

A interger is a whole number negative or positive.

A rational number is a number that terminates or continue with repeating digits.

A irrational number is a number that doesn't terminate or continue with repeating digits.

1. Rational Number

2. Natural,Whole,Interger,Rational

3. Whole,Rational,Interger

4. Rational

5.Irrational

6.Rational

7.Natural,Whole,Interger,Rational

8.Interger,Rational

9.Irrational