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3 years ago

Pre-algebra

Mathematics

2 answers:

ElenaW [278]3 years ago

7 0

37 metershgyvyvyvyvvygyvyvygygybygyb

Vladimir79 [104]3 years ago

4 0

Write down the numbers in the chart, write them on a number line, put a point on the number farthest from 0, Closer

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I will give brainiest to whoever answers correctly !!

Karo-lina-s [1.5K]

Answer:

Step-by-step explanation:

a) The interest generated over the 9 months is how much the woman paid for use of the money, then:

$I=C.t.i\\C=3000\\t=9\\i=3.5%=0.035\\I=3000.9.0.035=945\\I=945$

Therefore , the woman paid $945 for the use of the money.

b) The total amount that she repaid the bank is:

$C_{t} =C+I=3000+945=3945\\C_{t} =3945$

Finally we have that the woman paid $3945 to the bank

PLEASE MARK ME AS BRAINLIEST

7 0

3 years ago

Solve 2x - 1 < 7 and 5x + 3 < 3.

Nimfa-mama [501]

1) 2x - 1 < 7
move the 1 over by addition
2x < 8
divide by 2
x < 4

2) 5x + 3 < 3
move the 3 over by subtraction
5x < 0
divide by 5
x < 0/5
x < 0

3 0

4 years ago

Read 2 more answers

Convert the integral below to polar coordinates and evaluate the integral.

Lelechka [254]

Answer:

$\int\limits_{0}^{4/\sqrt{2}}\int\limits_{y}^{\sqrt{16-y^2}} xy \, dxdy = \int\limits_{0}^{\pi/4} \, \int\limits_{0}^{4} r^3 cos(\theta)\sin(\theta) \,\, drd\theta = 16$

Step-by-step explanation:

We are trying to evaluate this integral.

$\int\limits_{0}^{4/\sqrt{2}}\,\,\int\limits_{y}^{\sqrt{16-y^2}} xy \,\,dxdy$

The first thing that we have to do is understand this region in the plane.

$\{ (x,y) \in \mathbb{R} : 0\leq y \leq \frac{4}{\sqrt{2}} \,\, , y \leq x \leq \, \sqrt{16-y^2} \}$

If you graph it looks something like the photo I join.

Now we need to describe that same region in polar coordinates.

That same region in polar coordinates would be

$\{ (r,\theta) : \,\, 0 \leq \theta \leq \frac{\pi}{4} \,\,\, 0\leq r \leq 4 \}$

Now remember that when we do the polar transformation we use the following formula

$\int\limits_{a}^{b} \, \int\limits_{c}^{d} f(x,y) \,dxdy = \int\limits_{\theta_1}^{\theta_2} \, \int\limits_{r_1}^{r_2} r* f(rcos(\theta),rsin(\theta)) \,drd\theta$

Then our integral would be

$\int\limits_{0}^{4/\sqrt{2}}\int\limits_{y}^{\sqrt{16-y^2}} xy \, dxdy = \int\limits_{0}^{\pi/4} \, \int\limits_{0}^{4} r^3 cos(\theta)\sin(\theta) \,\, drd\theta = 16$

6 0

3 years ago

What is the median score achieved by a student who recorded the following scores on 10 math quizzes 68 55 70 62 71 58 81 82 63 7

Vinil7 [7]

Answer:

b. 68

Step-by-step explanation:

7 0

3 years ago

Pls someone help me these's problem's!

Iteru [2.4K]

Answer:

15. a because b is being squared by a variable meaning it is exponential and goes forever, but the answer choice B is set only to 3 squared, so a is the answer.

16. D because square root of 96 MINUS square root of 54 on calculator is 2.44948974278, BUT the square root of 6 is ALSO 2.44948974278, so the answer is D or square root of 6

Step-by-step explanation:

my explanation is all on the top

3 0

3 years ago