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2 years ago

Evaluate -32 + (2-6)(10).

Mathematics

2 answers:

vitfil [10]2 years ago

3 0

Answer:

-72

Step-by-step explanation:

-32 + (2-6)(10)

PEMDAS

Parenthases first

(-4)(10)

multiply together

(-40)

Then just add the -32 and since a negative plus a negative is always negative the answer is going to be negative

-32 + (-40)

= -72

hodyreva [135]2 years ago

3 0

Answer:

-49

Step-by-step explanation:

hope it helped :)

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If r=[x,y,z] and r0=[x0,y0,z0], describe the set of all points (x,y,z) such that Ir-r0I =1.

sdas [7]

Answer:

The points (x,y,z) that respond to Ir-r0I =1, are all that describes the form $(x-x_0)^2+(y-y_0)^2+(z-z_0)^2=1$ with:

-1+x₀<x<1+x₀

-1+y₀<y<1+y₀

-1+z₀<z<1+z₀

Step-by-step explanation:

All points required in this problem came from applying the definition of modulus of a vector:

Ir-r0I =1.

$|(x,y,z)-(x_{0},y_{0},z_{0})|=|(x-x_{0},y-y_{0},z-z_{0})|=\sqrt{(x-x_{0})^2+(y-y_{0})^2+(z-z_{0})^2}=1\\(x-x_{0})^2+(y-y_{0})^2+(z-z_{0})^2=1^2=1$

5 0

2 years ago

The graphs of the quadratic functions f(x) = 6 – 10x2 and g(x) = 8 – (x – 2)2 are provided below. Observe there are TWO lines si

natta225 [31]

Answer:

a) y = 7.74*x + 7.5

b) y = 1.148*x + 6.036

Step-by-step explanation:

Given:

f(x) = 6 - 10*x^2

g(x) = 8 - (x-2)^2

Find:

(a) The line simultaneously tangent to both graphs having the LARGEST slope has equation

(b) The other line simultaneously tangent to both graphs has equation,

Solution:

- Find the derivatives of the two functions given:

f'(x) = -20*x

g'(x) = -2*(x-2)

- Since, the derivative of both function depends on the x coordinate. We will choose a point x_o which is common for both the functions f(x) and g(x). Point: ( x_o , g(x_o)) Hence,

g'(x_o) = -2*(x_o -2)

- Now compute the gradient of a line tangent to both graphs at point (x_o , g(x_o) ) on g(x) graph and point ( x , f(x) ) on function f(x):

m = (g(x_o) - f(x)) / (x_o - x)

m = (8 - (x_o-2)^2 - 6 + 10*x^2) / (x_o - x)

m = (8 - (x_o^2 - 4*x_o + 4) - 6 + 10*x^2)/(x_o - x)

m = ( 8 - x_o^2 + 4*x_o -4 -6 +10*x^2) /(x_o - x)

m = ( -2 - x_o^2 + 4*x_o + 10*x^2) /(x_o - x)

- Now the gradient of the line computed from a point on each graph m must be equal to the derivatives computed earlier for each function:

m = f'(x) = g'(x_o)

- We will develop the first expression:

m = f'(x)

( -2 - x_o^2 + 4*x_o + 10*x^2) /(x_o - x) = -20*x

Eq 1. (-2 - x_o^2 + 4*x_o + 10*x^2) = -20*x*x_o + 20*x^2

And,

m = g'(x_o)

( -2 - x_o^2 + 4*x_o + 10*x^2) /(x_o - x) = -20*x

-2 - x_o^2 + 4*x_o + 10*x^2 = -2(x_o - 2)(x_o - x)

Eq 2 -2 - x_o^2 + 4*x_o+ 10*x^2 = -2(x_o^2 - x_o*(x + 2) + 2*x)

- Now subtract the two equations (Eq 1 - Eq 2):

-20*x*x_o + 20*x^2 + 2*x_o^2 - 2*x_o*(x + 2) + 4*x = 0

-22*x*x_o + 20*x^2 + 2*x_o^2 - 4*x_o + 4*x = 0

- Form factors: 20*x^2 - 20*x*x_o - 2*x*x_o + 2*x_o^2 - 4*x_o + 4*x = 0

20*x*(x - x_o) - 2*x_o*(x - x_o) + 4*(x - x_o) = 0

(x - x_o)(20*x - 2*x_o + 4) = 0

x = x_o , x_o = 10x + 2

- For x_o = 10x + 2 ,

(g(10*x + 2) - f(x))/(10*x + 2 - x) = -20*x

(8 - 100*x^2 - 6 + 10*x^2)/(9*x + 2) = -20*x

(-90*x^2 + 2) = -180*x^2 - 40*x

90*x^2 + 40*x + 2 = 0

- Solve the quadratic equation above:

x = -0.0574, -0.387

- Largest slope is at x = -0.387 where equation of line is:

y - 4.502 = -20*(-0.387)*(x + 0.387)

y = 7.74*x + 7.5

- Other tangent line:

y - 5.97 = 1.148*(x + 0.0574)

y = 1.148*x + 6.036

6 0

1 year ago

The following diagram shows the path of a planet around the Sun. Kepler discovered that _____.

Phantasy [73]

The right answer is:

area B = area C

We can solve this problem by using Kepler's laws of planetary motion. There are three Kepler's laws. In this exercise, we need to use the second law. According to this law, a line segment joining a planet and the sun sweeps out equals areas during equals intervals of time. So, a certain planet sweeps out an area B from the point P3 to P4 in an interval of time t. On the other hand, for the same interval of time t, the planet sweeps out an area C from point P4 to P5, that is equal to the previous area according to second kepler's law.

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2 years ago

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Let S(x) denote "x has a good attitude," where the domain is the set of people in the class. Express the negation of the proposi

user100 [1]

Answer:

See below

Step-by-step explanation:

Remember, we have two quantifiers, the existential quantifier ∃, and the universal quantifier ∀. The existential ∃ translates to English as "for some" or "there exists", whereas ∀ means "for all" or "every". We will also use the negation operator ¬.

First, let's write the proposition using quantifiers. "There is someone in this class who does not have a good attitude" translates to "(∃x)(¬S(x))". ∃x means that there exists a person in this class x. ¬S(x) means that x, the person that exists because of the quantifier, does not have a good attitude.

The negation is "¬(∃x)(¬S(x))" or equivalently "(∀x)(S(x))". To negate a proposition using quantifiers, change the quantifier (existential to universal and viceversa) and negate the predicate (in this case we negated ¬S(x)).

In English, "(∀x)(S(x))" means "Every person in this class has a good attitude".

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1 year ago

Which statements can be represented by this equation? Select three options.

mars1129 [50]

Answer:

Options A,C and E.

Step-by-step explanation:

The given equation is

$2n-9.2=n+\dfrac{4}{5}$

Let n be the unknown number, then

Twice a number minus nine and two-tenths is the same as the number plus four-fifths.

$2n-9.2=n+\dfrac{4}{5}$

So, option A is correct.

Nine point two decreased by double a number is the same as the number added to four-fifths.

$9.2-2n=n+\dfrac{4}{5}$

Nine point two less than twice a number is equal to the number added to four-fifths.

$2n-9.2=n+\dfrac{4}{5}$

So, option C is correct.

Two times the difference of a number and nine and two-tenths is equal to the number plus four-fifths.

$2(n-9.2)=n+\dfrac{4}{5}$

Double a number decreased by nine and two-tenths is the same as the sum of four-fifths and the number.

$2n-9.2=n+\dfrac{4}{5}$

So, option E is correct.

Therefore, the correct options are A,C and E.

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2 years ago

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