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3 years ago

What does the standard form of a linear equation require?

Mathematics

1 answer:

Maru [420]3 years ago

8 0

Answer:

the standard form of a linear equation required is B must be positive

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The diagram shows a 3 cm x 5 cm x 4 cm cuboid

Rainbow [258]

Answer:

a) The length of segment AC is approximately 5.83 centimeters.

b) The angle ACD is approximately 34.5º.

Step-by-step explanation:

a) Since $AB \perp BC$ , the length of segment $AC$ is determined by Pythagorean Theorem, that is:

$AC = \sqrt{(5\,cm)^{2}+(3\,cm)^{2}}$

$AC \approx 5.831\,cm$

The length of segment AC is approximately 5.831 centimeters.

b) Since $AB \perp BC \perp AD$ , the length of segment $AD$ is determined by this Pythagorean identity:

$AD = \sqrt{(3\,cm)^{2}+(5\,cm)^{2}+(4\,cm)^{2}}$

$AD \approx 7.071\,cm$

The angle ACD is determined by the following trigonometric expression:

$\cos C = \frac{AC}{CD}$

$\cos C = \frac{5.831\,cm}{7.071\,cm}$

$\cos C = 0.825$

$C = \cos^{-1} 0.825$

$C \approx 34.448^{\circ}$

The angle ACD is approximately 34.448º.

4 0

2 years ago

If you deposit $50 in checks in your account with $100 cash amount, what is your total deposit?

Free_Kalibri [48]

What do you mean? You deposited $50 didn’t you? HUuH

3 0

2 years ago

Hope u can do it HELP MATH

Nostrana [21]

Answer:

the right answer would be 13

5 0

3 years ago

Read 2 more answers

What precent is 12 outta 40

Lubov Fominskaja [6]

Answer: 30%

Step-by-step explanation:

12/40 × 100

= 30%

5 0

3 years ago

Read 2 more answers

Does the function satisfy the hypotheses of the Mean Value Theorem on the given interval? f(x) = 3x2 − 2x + 1, [0, 2] Yes, it do

expeople1 [14]

Answer:

Yes , function is continuous in [0,2] and is differentiable (0,2) since polynomial function are continuous and differentiable

Step-by-step explanation:

We are given the Function

f(x) = $3x^2 -2x +1$

The two basic hypothesis of the mean valued theorem are

The function should be continuous in [0,2]

The function should be differentiable in (1,2)

upon checking the condition stated above on the given function

f(x) is continuous in the interval [0,2] as the functions is quadratic and we can conclude that from its graph

also the f(x) is differentiable in (0,2)

f'(x) = 6x - 2

Now the function satisfies both the conditions

so applying MVT

6x-2 = f(2) - f(0) / 2-0

6x-2 = 9 - 1 /2

6x-2 = 4

6x=6

x=1

so this is the tangent line for this given function.

4 0

2 years ago