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3 years ago

The circle shown has a radius of 4 cm.

Mathematics

1 answer:

puteri [66]3 years ago

5 0

Answer:

8 cm

Step-by-step explanation: diamater is half the radius

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What are two integers whose sum is -2 and product is -80?

astra-53 [7]

Answer:

We can write:

x + y = -2

xy = -80

We can rewrite the first equation as x = -y - 2 and then plug that into the second equation to get (-y-2) * y = -80 → -y² - 2y = -80 → y² + 2y - 80 = 0 → (y - 8)(y + 10) = 0 → y = 8, -10. Substituting these values into the first equation we get x = -10, 8 so the answer is (x₁, y₁) = (-10, 8) or (x₂, y₂) = (8, -10).

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3 years ago

Which ordered pairs are solutions to the inequality 2y-x ≤ -6 ?

Inga [223]

Answer:

c (2, -2)

Step-by-step explanation:

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3 years ago

Which ordered pair is a solution to this inequality? -2/3x-4y > 0

Lynna [10]

c i think (answer needs to be 20 characters long)

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3 years ago

Solving simultaneous equations with one quadratic, x^2+y^2=16 and y=x+3

Svetlanka [38]

If $y=x+3$ , then in the first equation you have

$x^2+(x+3)^2=x^2+(x^2+6x+9)=2x^2+6x+9=16$

Rewriting a bit, you get

$x^2+3x+\dfrac92=8$

$x^2+3x+\dfrac94+\dfrac94=8$

$\left(x+\dfrac32\right)^2=\dfrac{23}4$

$x=-\dfrac32\pm\dfrac{\sqrt{23}}2$

Now, since $y=x+3$ , you also get

$y=-\dfrac32\pm\dfrac{\sqrt{23}}2+3=\dfrac32\pm\dfrac{\sqrt{23}}2$

So, there are two solutions here:

$(x,y)=\left(-\dfrac32+\dfrac{\sqrt{23}}2,\dfrac32+\dfrac{\sqrt{23}}2\right)$
$(x,y)=\left(-\dfrac32-\dfrac{\sqrt{23}}2,\dfrac32-\dfrac{\sqrt{23}}2\right)$

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3 years ago

8 to 2 power, -3 x 23 is

Nana76 [90]

23 x 8 is equal to 184. 23+23+23+23+23+23+23+23. 184 ÷ 8 = 123

5 0

4 years ago

Read 2 more answers