Ask question

Ask question

All categories

3 years ago

9

Is y=3x linear or nonlinear

1 answer:

BlackZzzverrR [31]3 years ago

7 0

Linear that is it ugh i needed to make it 20 characters ok

You might be interested in

Write 6 17/20 as a decimal

djverab [1.8K]

Answer:

6.85

Step-by-step explanation:

6 17/20

Convert to improper fraction.

137/20

Divide.

= 6.85

3 0

3 years ago

Read 2 more answers

What is the difference between the temperature -3 and 1

lisov135 [29]

That 1 is greater than 1 because 1 is positive and -3 is negative

8 0

3 years ago

Read 2 more answers

A sphere has a diameter of 4 x 10^-3 mm. What is the approximate volume of the sphere? Use 3.14 for pi.

kolezko [41]

Answer:

$3.349 \times {10}^{ - 10} \: {mm}^{3}$

Step-by-step explanation:

$d = 4 \times {10}^{ - 3} \: mm \\ r = 2 \times {10}^{ - 3} \: mm \\ V_{sphere} = \frac{4}{3} \pi {r}^{3} \\ = \frac{4}{3} \times 3.14 \times {(2 \times {10}^{ - 3} )}^{3} \\ = \frac{4}{3} \times 3.14 \times 8\times {10}^{ - 9} \\ = \frac{100.48}{3} \times {10}^{ - 9} \\ = 33.4933333\times {10}^{ - 9} \\ = 33.49 \times {10}^{ - 9} \\ = 3.349 \times {10}^{ - 10} \: {mm}^{3}$

8 0

3 years ago

Area unit conversion with whole numbers value 5 yards long and 4 wide = what in square feet?

Law Incorporation [45]

Poopjjjvgjg tj bejvevkq rvrj vrvirgivd

4 0

3 years ago

[tex]cos {}^{4} α+sin {}^{4} α= \frac{1}{4} (3+cos4α)<br>Prove:<br>

asambeis [7]

Given:

$\cos^4 \alpha+\sin^4\alpha=\dfrac{1}{4}(3+\cos 4 \alpha)$

To prove:

The given statement.

Proof:

We have,

$\cos^4 \alpha+\sin^4\alpha=\dfrac{1}{4}(3+\cos 4 \alpha)$

$LHS=\cos^4 \alpha+\sin^4\alpha$

$LHS=(\cos^2 \alpha)^2+(\sin^2 \alpha)^2$

$LHS=(\cos^2 \alpha+\sin^2\alpha)^2-2\sin ^2\alpha\cos^2 \alpha$ $[\because a^2+b^2=(a+b)^2-2ab]$

$LHS=(1)^2-2(1-\cos^2 \alpha)\cos^2 \alpha$ $[\because \cos^2 \alpha+\sin^2\alpha=1]$

$LHS=1-2\cos^2 \alpha+2\cos^4 \alpha$

Now,

$RHS=\dfrac{1}{4}(3+\cos 4 \alpha)$

$RHS=\dfrac{1}{4}[3+(2\cos^2 2\alpha-1)]$ $[\because \cos 2\theta=2\cos^2\theta -1]$

$RHS=\dfrac{1}{4}[2+2\cos^2 2\alpha]$

$RHS=\dfrac{1}{4}[2+2(2\cos^2 \alpha-1)^2]$ $[\because \cos 2\theta=2\cos^2\theta -1]$

$RHS=\dfrac{1}{4}[2+2(4\cos^4 \alpha-4\cos \alpha+1)]$ $[\because (a-b)^2=a^2-2ab+b^2]$

$RHS=\dfrac{1}{4}[2+8\cos^4 \alpha-8\cos \alpha+2]$

$RHS=\dfrac{1}{4}[4+8\cos^4 \alpha-8\cos \alpha]$

$RHS=1+2\cos^4 \alpha-2\cos \alpha$

$RHS=1-2\cos^2 \alpha+2\cos^4 \alpha$

$LHS=RHS$

Hence proved.

8 0

3 years ago