Find the area of each circle. Round to the nearest tenth. Use 3.14 or 22/7 for Pi.

Mathematics

1 answer:

maria [59]3 years ago

6 0

Answer: the 1 with 6cm the area is 113.1 and the one with 11ft is 380.2

Step-by-step explanation: use this formula A=πr2

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I will do anything if you answer!!! There's a photo!!! PLEASE!!!

Makovka662 [10]

A, it's always true
remember commutative property
a+b=b+a
a and b can be negativ numbers

so equatinos
-5.2+7.71=2.51
7.71-5.2=2.51

-8.65+3.94=-4.62
3.94-8.65=-4.62

B.
-5.2-7.71=-12.91
7.71-(-5.2)=12.91

-8.65-3.94=-12.59
3.94-(-8.65)=12.59

always false

C. when you change the order, you are subtracting a different number from a different number than before

8 0

3 years ago

◆ Quadratic Equations ◆ Please help !

Mila [183]

I'm sure there's an easier way of solving it than the way I did, but I'm not sure what it could be. Never dealt with a problem like this before.

Anyway, I just plugged in and tested. Chose random values for a, b, c, and d, which follow the rule 0 < a < b < c < d:

a = 1
b = 2
c = 3
d = 4

$\sf ax^2+(1-a(b+c))x+abc-d)$

$\sf 1x^2+(1-1(2+3))x+(1)(2)(3)-(4))$

Simplify into standard form:

$\sf x^2+(1-1(5))x+6-4$

$\sf x^2+(1-5)x+2$

$\sf x^2-4x+2$

Use the quadratic formula to solve:

$\sf x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}$

For functions in the form of $\sf ax^2+bx+c$ . So in this case:

a = 1
b = -4
c = 2

Plug them in:

$\sf x=\dfrac{4\pm\sqrt{(-4)^2-4(1)(2)}}{2(1)}$

Solve for 'x':

$\sf x=\dfrac{4\pm\sqrt{16-8}}{2}$

$\sf x=\dfrac{4\pm\sqrt{8}}{2}$

$\sf x\approx\dfrac{4\pm 2.83}{2}$

$\sf x\approx 0.59,3.41$

So the answer would be A.

3 0

4 years ago

The critical value t* gets larger as the confidence level increases. True or false?

posledela

Answer:

We can find the critical value $t_{\alpha/2}$