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3 years ago

10

The mass of rileys math book is 4,458 grams. What is the mass of 4 math books in kilogram

2 answers:

Bumek [7]3 years ago

8 0

Your answer would be 17.832

zzz [600]3 years ago

3 0

17.832 Kilograms. 4,458 x 4 and 1 gram = .001 of a kilo

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Convert 11/13 and 17/20 to fractions with same denominator

alexandr402 [8]

Answer:

220/260 and 221/260

Step-by-step explanation:

multiply 11/13 by 20/20 and 17/20 by 13/13

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4 years ago

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Find the vertex for the parabola given by the function ƒ(x) = −3x^2 − 6x.

telo118 [61]

Answer:

B) (−1, 3)

Step-by-step explanation:

The standard form of a quadratic function is

y = ax² + bx + c

The vertex form of a parabola is

y = a(x - h)² + k

where (h, k) is the vertex of the parabola.

h = -b/(2a) and k = f(h)

In your equation, ƒ(x) = −3x² − 6x

a = -3; b = -6; c = 0

Calculate h

h = -(-6)/2(-3)]

h = 6/(-6)

h = -1

Calculate k

k = -3(-1)² -6(-1)

k = -3 + 6

k = 3

So, h = -1, k = 3, a = -3

The vertex form of the equation is f(x) = -3(x + 1)² + 3.

The vertex is at (-1, 3).

The figure below shows the graph of ƒ(x) = −3x² − 6x with the vertex

at (-1, 3).

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3 years ago

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Tems11 [23]

Step-by-step explanation:

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3 years ago

Solve each equation by using the square root property. 7x^2–17=39.

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2.83 is the answer for this

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The expressions A, B, C, D, and E are left-hand sides of trigonometric identities. The expressions 1, 2, 3, 4, and 5 are right-h

Semenov [28]

Answer:

$A.\ \tan(x) \to 2.\ \sin(x) \sec(x)$

$B.\ \cos(x) \to 5. \sec(x) - \sec(x)\sin^2(x)$

$C.\ \sec(x)csc(x) \to 3. \tan(x) + \cot(x)$

$D. \frac{1 - (cos(x))^2}{cos(x)} \to 1. \sin(x) \tan(x)$

$E.\ 2\sec(x) \to\ 4.\ \frac{\cos(x)}{1 - \sin(x)} +\frac{1-\sin(x)}{\cos(x)}$

Step-by-step explanation:

Given

$A.\ \tan(x)$

$B.\ \cos(x)$

$C.\ \sec(x)csc(x)$

$D.\ \frac{1 - (cos(x))^2}{cos(x)}$

$E.\ 2\sec(x)$

Required

Match the above with the appropriate identity from

$1.\ \sin(x) \tan(x)$

$2.\ \sin(x) \sec(x)$

$3.\ \tan(x) + \cot(x)$

$4.\ \frac{cos(x)}{1 - sin(x)} + \frac{1 - \sin(x)}{cos(x)}$

$5.\ \sec(x) - \sec(x)(\sin(x))^2$

Solving (A):

$A.\ \tan(x)$

In trigonometry,

$\frac{sin(x)}{\cos(x)} = \tan(x)$

So, we have:

$\tan(x) = \frac{\sin(x)}{\cos(x)}$

Split

$\tan(x) = \sin(x) * \frac{1}{\cos(x)}$

In trigonometry

$\frac{1}{\cos(x)} =sec(x)$

So, we have:

$\tan(x) = \sin(x) * \sec(x)$

$\tan(x) = \sin(x) \sec(x)$ --- proved

Solving (b):

$B.\ \cos(x)$

Multiply by $\frac{\cos(x)}{\cos(x)}$ --- an equivalent of 1

So, we have:

$\cos(x) = \cos(x) * \frac{\cos(x)}{\cos(x)}$

$\cos(x) = \frac{\cos^2(x)}{\cos(x)}$

In trigonometry:

$\cos^2(x) = 1 - \sin^2(x)$

So, we have:

$\cos(x) = \frac{1 - \sin^2(x)}{\cos(x)}$

Split

$\cos(x) = \frac{1}{\cos(x)} - \frac{\sin^2(x)}{\cos(x)}$

Rewrite as:

$\cos(x) = \frac{1}{\cos(x)} - \frac{1}{\cos(x)}*\sin^2(x)$

Express $\frac{1}{\cos(x)}\ as\ \sec(x)$

$\cos(x) = \sec(x) - \sec(x) * \sin^2(x)$

$\cos(x) = \sec(x) - \sec(x)\sin^2(x)$ --- proved

Solving (C):

$C.\ \sec(x)csc(x)$

In trigonometry

$\sec(x)= \frac{1}{\cos(x)}$

and

$\csc(x)= \frac{1}{\sin(x)}$

So, we have:

$\sec(x)csc(x) = \frac{1}{\cos(x)}*\frac{1}{\sin(x)}$

Multiply by $\frac{\cos(x)}{\cos(x)}$ --- an equivalent of 1

$\sec(x)csc(x) = \frac{1}{\cos(x)}*\frac{1}{\sin(x)} * \frac{\cos(x)}{\cos(x)}$

$\sec(x)csc(x) = \frac{1}{\cos^2(x)}*\frac{\cos(x)}{\sin(x)}$

Express $\frac{1}{\cos^2(x)}\ as\ \sec^2(x)$ and $\frac{\cos(x)}{\sin(x)}\ as\ \frac{1}{\tan(x)}$

$\sec(x)csc(x) = \sec^2(x)*\frac{1}{\tan(x)}$

$\sec(x)csc(x) = \frac{\sec^2(x)}{\tan(x)}$

In trigonometry:

$tan^2(x) + 1 =\sec^2(x)$

So, we have:

$\sec(x)csc(x) = \frac{\tan^2(x) + 1}{\tan(x)}$

Split

$\sec(x)csc(x) = \frac{\tan^2(x)}{\tan(x)} + \frac{1}{\tan(x)}$

Simplify

$\sec(x)csc(x) = \tan(x) + \cot(x)$ proved

Solving (D)

$D.\ \frac{1 - (cos(x))^2}{cos(x)}$

Open bracket

$\frac{1 - (cos(x))^2}{cos(x)} = \frac{1 - cos^2(x)}{cos(x)}$

$1 - \cos^2(x) = \sin^2(x)$

So, we have:

$\frac{1 - (cos(x))^2}{cos(x)} = \frac{sin^2(x)}{cos(x)}$

Split

$\frac{1 - (cos(x))^2}{cos(x)} = \sin(x) * \frac{sin(x)}{cos(x)}$

$\frac{sin(x)}{\cos(x)} = \tan(x)$

So, we have:

$\frac{1 - (cos(x))^2}{cos(x)} = \sin(x) * \tan(x)$

$\frac{1 - (cos(x))^2}{cos(x)} = \sin(x) \tan(x)$ --- proved

Solving (E):

$E.\ 2\sec(x)$

In trigonometry

$\sec(x)= \frac{1}{\cos(x)}$

So, we have:

$2\sec(x) = 2 * \frac{1}{\cos(x)}$

$2\sec(x) = \frac{2}{\cos(x)}$

Multiply by $\frac{1 - \sin(x)}{1 - \sin(x)}$ --- an equivalent of 1

$2\sec(x) = \frac{2}{\cos(x)} * \frac{1 - \sin(x)}{1 - \sin(x)}$

$2\sec(x) = \frac{2(1 - \sin(x))}{(1 - \sin(x))\cos(x)}$

Open bracket

$2\sec(x) = \frac{2 - 2\sin(x)}{(1 - \sin(x))\cos(x)}$

Express 2 as 1 + 1

$2\sec(x) = \frac{1+1 - 2\sin(x)}{(1 - \sin(x))\cos(x)}$

Express 1 as $\sin^2(x) + \cos^2(x)$

$2\sec(x) = \frac{\sin^2(x) + \cos^2(x)+1 - 2\sin(x)}{(1 - \sin(x))\cos(x)}$

Rewrite as:

$2\sec(x) = \frac{\cos^2(x)+1 - 2\sin(x)+\sin^2(x)}{(1 - \sin(x))\cos(x)}$

Expand

$2\sec(x) = \frac{\cos^2(x)+1 - \sin(x)- \sin(x)+\sin^2(x)}{(1 - \sin(x))\cos(x)}$

Factorize

$2\sec(x) = \frac{\cos^2(x)+1(1 - \sin(x))- \sin(x)(1-\sin(x))}{(1 - \sin(x))\cos(x)}$

Factor out 1 - sin(x)

$2\sec(x) = \frac{\cos^2(x)+(1- \sin(x))(1-\sin(x))}{(1 - \sin(x))\cos(x)}$

Express as squares

$2\sec(x) = \frac{\cos^2(x)+(1-\sin(x))^2}{(1 - \sin(x))\cos(x)}$

Split

$2\sec(x) = \frac{\cos^2(x)}{(1 - \sin(x))\cos(x)} +\frac{(1-\sin(x))^2}{(1 - \sin(x))\cos(x)}$

Cancel out like factors

$2\sec(x) = \frac{\cos(x)}{1 - \sin(x)} +\frac{1-\sin(x)}{\cos(x)}$ --- proved

3 0

3 years ago