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2 years ago

If y varies inversely as x and y=3 when x=8 what is x when y=4

Mathematics

1 answer:

I am Lyosha [343]2 years ago

5 0

Answer:

x = 6

Step-by-step explanation:

y varies inversely as x

y = k × 1/x. {where k is a constant}

when y = 3, x = 8

Therefore 3 = k/8 and k is 24

y = 24/x

When y = 4,

4 = 24/x

4x = 24

(dividing both sides by 4)

x = 6

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If x= 1/2-√3 , prove that x³-2x²-7x+5=3

prohojiy [21]

$x=\frac{1}{2}-\sqrt3\\\\x^3-2x^2-7x+5=3\\\\(\frac{1}{2}-\sqrt3)^3-2(\frac{1}{2}-\sqrt3)^2-7(\frac{1}{2}-\sqrt3)+5\\\\=(\frac{1}{2})^3-3\cdot(\frac{1}{2})^2\cdto\sqrt3+3\cdot\frac{1}{2}\cdot(\sqrt3)^2-(\sqrt3)^3-...\\\\...-2[(\frac{1}{2})^2-2\cdot\frac{1}{2}\cdot\sqrt3+(\sqrt3)^2]-\frac{7}{2}+7\sqrt3+5$

$=\frac{1}{8}-\frac{3\sqrt3}{4}+\frac{9}{2}-3\sqrt3-2(\frac{1}{4}-\sqrt3+3)-\frac{7}{2}+7\sqrt3+5\\\\=\frac{1}{8}+\frac{9}{2}-\frac{1}{2}-6-\frac{7}{2}+5-\frac{3}{4}\sqrt3-3\sqrt3+2\sqrt3+7\sqrt3\\\\=\frac{1}{8}+\frac{1}{2}-1+5\frac{1}{4}\sqrt3=\frac{1}{8}-\frac{1}{2}+5\frac{1}{4}\sqrt3=\frac{1}{8}-\frac{4}{8}+5\frac{1}{4}\sqrt3\\\\=-\frac{3}{8}+5\frac{1}{4}\sqrt3\neq3$

6 0

2 years ago

Determine cosine 23 degrees:

dexar [7]

$cos23^o\approx0,9205$

7 0

3 years ago

Please answer correctly !!!!!!!!!!!!!!!!!!!!! Will mark brainliest !!!!!!!!!!!!!!!!!!!!

Sergeeva-Olga [200]

Answer:

q = 14

General Formulas and Concepts:

Order of Operations: BPEMDAS
Equality Properties
Complementary Angles: Angles that add up to 90°

Step-by-step explanation:

Step 1: Set up equation

The 2 angles must add up to 90°.

(4q - 5)° + 39° = 90°

Step 2: Solve for q

Combine like terms: 4q + 34 = 90
Subtract 34 on both sides: 4q = 56
Divide both sides by 4: q = 14

6 0

3 years ago

Evaluate the expression. 621+5) - 3

Gennadij [26K]

Answer:

I think its 623

8 0

3 years ago

May somebody help me please .

amm1812

Okay, I'm pretty sure the answers are:

11) B

12) C

13) B

14) A

3 0

3 years ago