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2 years ago

Need help asap i’ll vote brain liest

Mathematics

2 answers:

djverab [1.8K]2 years ago

7 0

Answer:

y = 3x + 1

Step-by-step explanation:

it appears the line is crossing the y-axis at 1 so this eliminates the first and third option

the last option is incorrect because '3x²' would not create a straight line

this leaves y = 3x + 1 which means the line crosses the y-axis at one and the slope is 3

Novay_Z [31]2 years ago

6 0

Answer:

y=3x+1

Step-by-step explanation:

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Convert 3/8 to its decimal form and round to the nearest thousandths

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Answer:

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Step-by-step explanation:

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3 years ago

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Given that vector p=i+2j-2k and q=2i-j+2k, find two vectors m and m satisfying the condition;

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Answer:

m = j +k
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Step-by-step explanation:

There are <em>an infinite number of possibilities</em>. Any vector whose dot-product with p is zero will be perpendicular to p.

Let m = 0i +1j +ak. Then we require ...

m·p = 0 = 0×1 +1×2 +a(-2) ⇒ 0 = 2 -2a ⇒ a = 1

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2 years ago

A rectangular prism has a volume of 729ft^3. The length width and height are the same. What is the length of each side

viktelen [127]

Answer:

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Step-by-step explanation:

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2 years ago

Find all solutions of each equation on the interval 0 ≤ x < 2π.

Korvikt [17]

Answer:

$x = 0$ or $x = \pi$ .

Step-by-step explanation:

How are tangents and secants related to sines and cosines?

$\displaystyle \tan{x} = \frac{\sin{x}}{\cos{x}}$ .

$\displaystyle \sec{x} = \frac{1}{\cos{x}}$ .

Sticking to either cosine or sine might help simplify the calculation. By the Pythagorean Theorem, $\sin^{2}{x} = 1 - \cos^{2}{x}$ . Therefore, for the square of tangents,

$\displaystyle \tan^{2}{x} = \frac{\sin^{2}{x}}{\cos^{2}{x}} = \frac{1 - \cos^{2}{x}}{\cos^{2}{x}}$ .

This equation will thus become:

$\displaystyle \frac{1 - \cos^{2}{x}}{\cos^{2}{x}} \cdot \frac{1}{\cos^{2}{x}} + \frac{2}{\cos^{2}{x}} - \frac{1 - \cos^{2}{x}}{\cos^{2}{x}} = 2$ .

To simplify the calculations, replace all $\cos^{2}{x}$ with another variable. For example, let $u = \cos^{2}{x}$ . Keep in mind that $0 \le \cos^{2}{x} \le 1 \implies 0 \le u \le 1$ .