What are the coordinates of 1 and 3

Evaluate the expression 4 1/2 ÷ 3/4

Darya [45]

Answer: 6

Step-by-step explanation:

7 0

1 year ago

Situation A

Colt1911 [192]

The equation a = 100 + 75d represents Maria's account balance.

Step-by-step explanation:

Given that,

Maria opened a new savings account with an initial deposit of $100.
She deposits $75 into this savings account every two-weeks.

You need to find the equation to represent Maria's account balance, a after she has completed d deposits.

Let 'd' be the number of deposits Maria made after the initial deposit.
Let 'a' be the total account balance after 'd' deposits.

Therefore, the equation for the total account balance can be framed as the sum of the initial deposit and amount deposited after that.

To find the amount that was deposited after the initial deposit :

⇒ $75 × number of times Maria deposited.

⇒ 75 × d

The equation is written as,

⇒ a = 100 + 75d

Hence the equation a = 100 + 75d represents Maria's account balance.

7 0

3 years ago

Given z1 = -3

trasher [3.6K]

Answer:

See explanation

Step-by-step explanation:

The given complex number are:

$z_1=-3\sqrt{3}+3i$

and

$z_2=6\cos 150\degree+6i\sin 150\degree$

When we rewrite $z_1=-3\sqrt{3}+3i$ in complex form, we obtain;

$z_1=r(\cos \theta+i\sin \theta)$

where

$r=\sqrt{(-3\sqrt{3})^2+3^2 }=\sqrt{36}=6$

and

$\theta=tan^{-1}(\frac{y}{x})$

$\implies \theta=tan^{-1}(\frac{-3\sqrt{3}}{3})=150\degree$

Hence,

$z_1=6(\cos 150\degree+i\sin 150\degree)$

$z_1=6\cos 150\degree+6i\sin 150\degree$

Hence

$z_1=z_2$

8 0

2 years ago

How to work it out logically? Question a !!!

Arturiano [62]

Answer:

Fencing is done along KL which is (1500+520.8=2020.8 m) from the top left corner and divides the property into half.

Step-by-step explanation:

Given the figure with dimensions. we have to find the area of given figure.

Area of figure=ar(1)+ar(2)+ar(3)

Area of region 1 = ar(ANGI)+ar(AIB)

$=L\times B+\frac{1}{2}\times base\times height\\\\=[1500\times (5000-2000-1500)]+\frac{1}{2}\times (3000-1500)\times (5000-2000-1500)\\\\=3375000m^2=337.5ha$

Area of region 2 = ar(DHBC)

$=2000\times1500\\\\=3000000m^2=300ha$

Area of region 3 = ar(GFEH)

$(2000+1500)\times 1000\\\\=3500000m^2=350ha$

Hence, Area of figure=ar(1)+ar(2)+ar(3)=337.5ha+300ha+350ha

=987.5 ha

Now, we have to do straight-line fencing such that area become half and cost of fencing is minimum.

Let the fencing be done through x m downward from B which divides the two into equal area.

⇒ Area of upper part above fencing=Area of lower part below fencing

⇒ $ar(ANGB)+ar(GKLB)=ar(KLCM)+ar(MDCF)\\\\337500+3000x=(3500-x)\times 1000+2000(1500-x)\\\\3375000+3000x=3500000-1000x+3000000-2000x\\\\6000x=315000\\\\x=\frac{315000}{6000}=520.8m$

Hence, fencing is done along KL which is (1500+520.8=2020.8 m) from the top left corner and divides the property into half.

7 0

2 years ago

Read 2 more answers

· In 12 - In 18/ In 2 - In 3 (A) 2 (B) 1 (C) 3 (D) 4.

klemol [59]

Answer:

S

Step-by-step explanation:

SD

8 0

2 years ago