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gogolik [260]
3 years ago
9

Pls help due in 15 mins

Mathematics
2 answers:
SCORPION-xisa [38]3 years ago
6 0
Y is 8 x is 7.5 for number one
Number two is y is 3 x is 12
solmaris [256]3 years ago
5 0
1) y = 8 x = 7.5
2) y = 3 x = 12
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A: 7,500 miles
kvasek [131]

Answer:

the minimal number is 3

Step-by-step explanation:

6 0
3 years ago
If you can’t see the pic don’t bother answering. if you could please explain !!
il63 [147K]

Answer:

you can split the garden in: 1/2 circles x 4 (because you have 4 half-circles) + a square.

the square's area is 4 x 4 = 16; the circles is pi x R x R (pi = aprox 3.14, R of the cirlce is 4m / 2 = 2m, because the square's side is 4m and is the same size with circle diagonale: Diag of circle = 2 X Radius => Radius = Diag / 2)

than answer will be: 16 + 3.14 x 2 x 2 x 2 (the last 2 is, because you have 4 halves of circles, than you will have 2 full circles) = 45.12, than answer is H, because 41 is the nearest number to 45.12

6 0
3 years ago
Read 2 more answers
(when u solved it, can u show me how you got the answer because my teacher wants us the show the work as well)
Flura [38]
Answer 243mm^2

Explanation:
I first split the shape into two smaller shapes. Then I found those two areas and added them! (18x12)+(9x3)=243

8 0
3 years ago
ASAP ASAP ASAP ASAP!!!!! Pls help me <br><br> I didn’t study r.I.p
kkurt [141]

Answer:

1) Scalene

2) Equilateral

3) Isosceles

4) Right

5)Obtuse

Step-by-step explanation:

4 0
2 years ago
Read 2 more answers
Find the volume of the solid bounded by z = 2 - x2 - y2 and z = 1. Express your answer as a decimal rounded to the hundredths pl
butalik [34]

Answer:

The answer is "(\frac{\pi}{2})".

Step-by-step explanation:

z = 2 - x^2 - y^2.........(1) \\\\z = 1.............(2)  

Let add equation 1 and 2:

Using formula: x^2+y^2=1 \\\\

convert to polar coordinates

r=2

\theta \varepsilon (0=\pi)=z\\\\V=\int^{2\pi}_{\theta=0}\int^{1}_{\pi=0}\int^{z_{2}}_{z_1} r \ dr \d \theta\\\\

    =\int^{2\pi}_{0}\int^{1}_{0}  (Z_2-z_1)r  \ dr \d \theta\\\\=\int^{2\pi}_{0}\int^{1}_{0} 1- (-1-x^2-y^2) r \ dr \d \theta\\\\=\int^{2\pi}_{0}\int^{1}_{0} (+1 \pm r^2) r \ dr \d \theta\\\\=\int^{2\pi}_{0}\int^{1}_{0} (-r^3 + r)  \ dr \d \theta\\\\=\int^{2\pi}_{0} (-\frac{r^4}{4}+\frac{r^2}{1})^{1}_{0}  \d \theta\\\\=\int^{2\pi}_{0} (\frac{1}{4})  \d \theta\\\\=(\frac{2 \pi}{4}) \\\\=(\frac{\pi}{2}) \\\\

8 0
3 years ago
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