Ask question

Ask question

All categories

3 years ago

10

Find the area of the circle

1 answer:

pickupchik [31]3 years ago

4 0

Answer:

254.34

Step-by-step explanation:

You might be interested in

What’s the distance between (5,1)and(3,6)

laiz [17]

<u>How to Find the Distance Between Two Points</u>

Distance formula: d(P, Q) =√(x2 − x1)^2+(y2 − y1)^2

If you use this equation the answer will be 5.

4 0

3 years ago

X squared+ y squared = 2 y = 2x squared – 3 Which of the following describes the system?

cluponka [151]

Answer:

$x=-1,1,-\sqrt{\frac{7}{4} },\sqrt{\frac{7}{4}}$ and $y=-1,\frac{1}{2}$

The ordered pair solutions are $(-\sqrt{\frac{7}{4}},0.5)$ , $(\sqrt{\frac{7}{4}},0.5)$ , $(-1,-1)$ , and $(1,-1)$ .

Step-by-step explanation:

I'm assuming the system is $\left \{ {x^2+y^2=2} \atop {y=2x^2-3}} \right.$ :

$x^2+y^2=2$

$x^2+(2x^2-3)^2=2$

$x^2+(4x^4-12x^2+9)=2$

$x^2+4x^4-12x^2+9=2$

$4x^4-11x^2+9=2$

$4x^4-11x^2+7=0$

$x^4-11x^2+28=0$

$(x^2-7)(x^2-4)=0$

$(4x^2-7)(x^2-1)=0$

$4x^2-7=0$

$4x^2=7$

$x^2=\frac{7}{4}$

$x=\pm\sqrt{\frac{7}{4}}$

$x^2-1=0$

$x^2=1$

$x=\pm1$

$y=2x^2-3$

$y=2(\pm\sqrt{\frac{7}{4}})^2-3$

$y=2({\frac{7}{4}})-3$

$y=\frac{7}{2}-3$

$y=\frac{1}{2}$

$y=2x^2-3$

$y=2(\pm1)^2-3$

$y=2(1)-3$

$y=2-3$

$y=-1$

Therefore, $x=-1,1,-\sqrt{\frac{7}{4} },\sqrt{\frac{7}{4}}$ and $y=-1,\frac{1}{2}$

The ordered pair solutions are $(-\sqrt{\frac{7}{4}},0.5)$ , $(\sqrt{\frac{7}{4}},0.5)$ , $(-1,-1)$ , and $(1,-1)$ .

4 0

2 years ago

5x + 2 (x+7) =14x - 7

qaws [65]

The answer is x=-23/21

5 0

3 years ago

What is spark addition?

Arturiano [62]

Answer:

spark addition is a left to right addition

3 0

2 years ago

I have k quarters, five less quarters than nickels and one more than twice as many dimes as quarters. Find the value of the coin

Alenkinab [10]

Answer:

(35k + 20) cents

Step-by-step explanation:

First of all, let us have the value of each unit:

1 quarter = 25 cents

1 nickel = 5 cents

1 dime = 10 cents

Given that number of quarter = k

Quarters are 5 lesser than Nickels, so number of nickels = k+5

One more than twice as many dimes as quarters:

k = 2 $\times$ Number of Dimes + 1

So, number of dimes = $\frac{1}{2}(k-1)$

Value of quarters = $25 \times k$ cents

Value of nickels = $5 \times (k+5) = (5k+25)\ cents$

Value of dimes = $\frac{1}{2}(k-1) \times 10 = (5k-5)\ cents$

So, total value of coins =

$25k + 5k +25 +5k-5\\\Rightarrow (35k+20)\ cents$

5 0

3 years ago

Read 2 more answers