All categories

3 years ago

The corresponding arcs of two congruent chords are equal. A)True B)False

Mathematics

1 answer:

kramer3 years ago

7 0

Answer:

it is B

Step-by-step explanation:

You might be interested in

Betty has 741 coins in her piggy bank, worth a total of 10,725 cents. The piggy bank contains ONLY quarters and dimes. How many

rjkz [21]

This is the 2nd time I've posted this answer... not sure why it didn't go through the first time. Let me know if you have questions. :)

6 0

4 years ago

na2s + hcl _____ nacl + h2s Which of these would represent the coefficients needed to balance this equation? A) 1, 1, 1, 1 B) 1,

USPshnik [31]

Answer:

B) 1, 2, 2, 1

Step-by-step explanation:

In given equation:

Left side: Right Side:

2 Na 1 Na

1 S 1 S

1 H 2 H

1 Cl 1 Cl

To equal the number of Na and H atoms on both sides, HCl and NaCl will have a coefficient of 2 to balance the equation while Na₂S and H₂S will have a coefficient of 1.

1 Na₂S + 2 HCl ⇒ 2 NaCl + 1 H₂S

5 0

3 years ago

What is 0.96 rounded to the neatest tenth.

Alona [7]

Answer:

1.0

Step-by-step explanation:

6 is greater than 5 therefore it rounds 9 up to the next number which is 10, 1.0 or just 1.

4 0

3 years ago

Read 2 more answers

Please help this is a 30 60 90 triangle

natka813 [3]

Answer:

See below

Step-by-step explanation:

$x = \frac{1}{2} \times \sqrt{6} \\ \\ x = \frac{ \sqrt{6} }{2} \\ \\ y = \frac{ \sqrt{3} }{2} \times \sqrt{6} \\ \\ y = \frac{ \sqrt{3 \times 6} }{2} \\ \\ y = \frac{ \sqrt{18} }{2} \\ \\ y = \frac{ 3\sqrt{2} }{2}$

8 0

3 years ago

Read 2 more answers

If 3x^2 + y^2 = 7 then evaluate d^2y/dx^2 when x = 1 and y = 2. Round your answer to 2 decimal places. Use the hyphen symbol, -,

S_A_V [24]

Taking $y=y(x)$ and differentiating both sides with respect to $x$ yields

$\dfrac{\mathrm d}{\mathrm dx}\bigg[3x^2+y^2\bigg]=\dfrac{\mathrm d}{\mathrm dx}\bigg[7\bigg]\implies 6x+2y\dfrac{\mathrm dy}{\mathrm dx}=0$

Solving for the first derivative, we have

$\dfrac{\mathrm dy}{\mathrm dx}=-\dfrac{3x}y$

Differentiating again gives

$\dfrac{\mathrm d}{\mathrm dx}\bigg[6x+2y\dfrac{\mathrm dy}{\mathrm dx}\bigg]=\dfrac{\mathrm d}{\mathrm dx}\bigg[0\bigg]\implies 6+2\left(\dfrac{\mathrm dy}{\mathrm dx}\right)^2+2y\dfrac{\mathrm d^2y}{\mathrm dx^2}=0$

Solving for the second derivative, we have

$\dfrac{\mathrm d^2y}{\mathrm dx^2}=-\dfrac{3+\left(\frac{\mathrm dy}{\mathrm dx}\right)^2}y=-\dfrac{3+\frac{9x^2}{y^2}}y=-\dfrac{3y^2+9x^2}{y^3}$

Now, when $x=1$ and $y=2$ , we have

$\dfrac{\mathrm d^2y}{\mathrm dx^2}\bigg|_{x=1,y=2}=-\dfrac{3\cdot2^2+9\cdot1^2}{2^3}=\dfrac{21}8\approx2.63$

3 0

3 years ago