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alexandr1967 [171]
3 years ago
7

What’s the area and the circumference of a circle with diameter 8m?

Mathematics
1 answer:
MatroZZZ [7]3 years ago
8 0

Answer:

50.24

Step-by-step explanation:

8÷2 is the radius. 3.14×4^2=50.24 m^2

^= where the exponent is supposed to go

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How many 0.3's would it take to make 3? Explain your reasoning. ​
grin007 [14]
10. Since it you want to find how many 0.3s it would take to make 3, you would divide 3 by 0.3 or move the decimal over one spot to the right making it 10.
3 0
3 years ago
Write a quadratic equation given the roots -1/3 and 5, show your work
Travka [436]

\boxed{(x - a)(x - b) = 0}

The equation above is the intercept form. Both a-term and b-term are the roots of equation.

x =  -  \frac{1}{3}  \\ x = 5

These are the roots of equation. Therefore we substitute a = - 1/3 and b = 5 in the equation.

(x +  \frac{1}{3} )(x -  5) = 0

Here we can convert the expression x+1/3 to this.

x +  \frac{1}{3}  = 0 \\  3x + 1 = 0

Rewrite the equation.

(3x + 1)(x - 5) = 0

Simplify by multiplying both expressions.

3 {x}^{2}  - 15x + x - 5 = 0 \\ 3 {x}^{2}  - 14x  - 5 = 0

<u>Answer</u><u> </u><u>Check</u>

Substitute the given roots in the equation.

3 {(5)}^{2}  - 14(5)  - 5 = 0 \\ 75 - 70 - 5 = 0 \\ 75 - 75 = 0 \\ 0 = 0

3( -  \frac{1}{3} )^{2}  - 14( -  \frac{1}{3}) - 5 = 0 \\ 3( \frac{1}{9} ) +  \frac{14}{3}  - 5 = 0 \\  \frac{1}{3}  +  \frac{14}{3}  -  \frac{15}{3}  = 0 \\  \frac{15}{3}  -  \frac{15}{3}  = 0 \\ 0 = 0

The equation is true for both roots.

<u>Answer</u>

\large \boxed {3 {x}^{2}  - 14x - 5 = 0}

8 0
3 years ago
Plzzzzz help , really easy im just du mb
ra1l [238]

Answer:

Step-by-step explanation:

They are about 3 meters away from each other

and the correct area is 15.5-12 and the correct answer is 3.5 meters apart

3 0
3 years ago
Read 2 more answers
Eden and Courtney each have a bag of candy. There is a mixture of chocolate and sour candy. Eden has a total of 5 pieces of cand
Svetradugi [14.3K]

2

Step-by-step explanation:

8 0
3 years ago
Solve the system of equations using the substitution method.
Karolina [17]
Substitution is where we first Isolate one of the unknowns, express it in terms of the other unknown, and replace the isolated unknown with the other unknown in another equation. So that each time we only need to deal with one unknown. I think you'll get a better idea here:

First name these 2 equations with 1 and 2.
4x + 5y = 7 (1)
y = 3x + 9 (2)

Since y is already isolated in (2), so we can skip the isolation step and continue to substitute.

Substitute (2) into (1).
4x + 5(3x+9) = 7

Expand.
4x + 15x + 45 = 7

Group.
19x + 45 = 7

Shift +45 to the other side and turn it into -45.

19x = 7 - 45
19x = -38

Shift x19 to the other side, turn it into /19.
X = - 38/19
X = - 2

Now we solved x already, we can just substitute x= - 2 back to equation (2).

y = 3(-2) + 9
y = - 6 + 9
y = 3

So, the answers are
x = - 2
y = 3
3 0
3 years ago
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