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3 years ago

11

Which of the p-values satisfy the following inequality?

1 answer:

Likurg_2 [28]3 years ago

3 0

Answer:

All of the answers

Step-by-step explanation:

P=0, p=1, p=2

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How to find x for this equation : 3x-3(x-1)=3

Romashka [77]

The answer for that is infinitely many solutions

5 0

3 years ago

How do you solve this?

Naddika [18.5K]

$\dfrac{3+4i}{2+i}=\dfrac{(3+4i)(2-i)}{(2+i)(2-i)^{(*)}}=\dfrac{3\cdot2-3\cdot i+4i\cdot2-4i\cdot i}{2^2-i^2^{(**)}}\\\\=\dfrac{6-3i+8i+4}{4-(-1)}}=\dfrac{10+5i}{4+1}=\dfrac{10+5i}{5}=\dfrac{10}{5}+\dfrac{5i}{5}=\boxed{2+i}\to \boxed{a=2}\\\\used:\\(*)\ (a+b)(a-b)=a^2-b^2\\(**)\ i^2=-1$

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4 years ago

In the following diagram, parallelogram ABCD contains triangle CED. Line CE and line ED intersect at point E. The measures of ∠C

nirvana33 [79]

Answer:

∠x = 59°

Step-by-step explanation:

∠ECD = 52° because they are alternate angles and will be equal

∠x = 180° - 52° - 69° = 59° because the sum of the interior angles of a triangle = 180°

5 0

3 years ago

Help me Idk how to do this one

Musya8 [376]

Answer:

Hmm im not sure

Step-by-step explanation:

6 0

3 years ago

Please explain 12 and13and please please please hurry!

pantera1 [17]

Answer:

12. a = 2

13. $x = \dfrac{4}{3}$

Step-by-step explanation:

12.

6a(a + 3) = 2a(3a + 5) + 16

First, distribute 6a on the left side, and distribute 2a on the right side.

6a² + 18a = 6a² + 10a + 16

Subtract 6a² from both sides.

18a = 10a + 16

Subtract 10a from both sides.

8a = 16

Divide both sides by 8.

a = 2

13.

$\dfrac{1}{3}x + \dfrac{1}{2}x = \dfrac{1}{3}(x + 2)$

The LCD of 2 and 3 is 6. We multiply both sides of the equation by 6 to get rid of all denominators.

$6 \times \dfrac{1}{3}x + 6 \times \dfrac{1}{2}x = 6 \times \dfrac{1}{3}(x + 2)$

$\dfrac{6}{3}x + \dfrac{6}{2}x = \dfrac{6}{3}(x + 2)$

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

Combine like terms on the left side. Distribute the 2 on the right side.

$5x = 2x + 4$

Subtract 2x from both sides.

$3x = 4$

Divide both sides by 3.

$x = \dfrac{4}{3}$

6 0

2 years ago