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Sever21 [200]
3 years ago
8

Please help me asap if u can.

Mathematics
2 answers:
Alexandra [31]3 years ago
5 0
Yes photo math works have a great day
kkurt [141]3 years ago
3 0

try that app photo math or math tutor

Step-by-step explanation:

just download the app

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the mass of one California condors is 9 kilograms. what is the mass of twelve 9-kilogram California condors?
serious [3.7K]
9×12=108 kilograms of codors
4 0
4 years ago
N+(-n) = 3+ 15<br><br> Solve
Len [333]

Answer:

no solutions


Step-by-step explanation:

n+(-n) = 3+ 15

n+-n = 0

0= 18

False

There are no solutions

7 0
3 years ago
If an input is -9 what is the output for y = x(to the 2nd power) + 3?
Whitepunk [10]

Answer:

Step-by-step explanation:

6 0
4 years ago
Which equation is represented by the graph below?
LuckyWell [14K]

Answer:

Step-by-step explanation:

Hello

A (5;2)

B (-1;-7)

y = ax + b

2 = 5a + b

-7 = -a + b

2 - (-7) = 5a - (-a) + b - b

2 + 7 = 5a + a

9 = 6a

a = 9/6

a = 3/2

2 = 3/2 * 5 + b

b = 2 - 15/2

b = 4/2 - 15/2

b = 11/2

y = 3/2 x + 11/2

y+1= 3/2(x-4)

y + 1 = 3/2 x - 6

y = 3/2 x - 6 - 1

y = 3/2 x - 7 => no

y-4= 3/2(x+1)

y - 4 = 3/2 x + 3/2

y = 3/2 x + 3/2 + 4

y = 3/2 x + 3/2 + 8/2

y = 3/2 x + 11/2 => yes

y+4= 3/2(x-1)

y + 4 = 3/2 x - 3/2

y = 3/2 x - 3/2 - 4

y = 3/2 x - 3/2 - 8/2

y = 3/2 x - 11/2 => no

y-1= 3/2 (x+4)

y - 1 = 3/2 x + 6

y = 3/2 x + 6 + 1

y = 3/2 x + 7 => no

5 0
3 years ago
Read 2 more answers
In Exercise 4, find the surface area of the solid<br> formed by the net.
Fittoniya [83]

Answer:

3. 150.72 in²

4. 535.2cm²

Step-by-step Explanation:

3. The solid formed by the net given in problem 3 is the net of a cylinder.

The cylinder bases are the 2 circles, while the curved surface of the cylinder is the rectangle.

The surface area = Area of the 2 circles + area of the rectangle

Take π as 3.14

radius of circle = ½ of 4 = 2 in

Area of the 2 circles = 2(πr²) = 2*3.14*2²

Area of the 2 circles = 25.12 in²

Area of the rectangle = L*W

width is given as 10 in.

Length (L) = the circumference or perimeter of the circle = πd = 3.14*4 = 12.56 in

Area of rectangle = L*W = 12.56*10 = 125.6 in²

Surface area of net = Area of the 2 circles + area of the rectangle

= 25.12 + 125.6 = 150.72 in²

4. Surface area of the net (S.A) = 2(area of triangle) + 3(area of rectangle)

= 2(0.5*b*h) + 3(l*w)

Where,

b = 8 cm

h = \sqrt{8^2 - 4^2} = \sqrt{48} = 6.9 cm} (Pythagorean theorem)w = 8 cm[tex]S.A =  2(0.5*8*6.9) + 3(20*8)

S.A =  2(27.6) + 3(160)

S.A =  55.2 + 480

S.A = 535.2 cm^2

6 0
4 years ago
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