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3 years ago

Evaluate f(x) = 2x at x = 5. A) 32 B) 10 C) 64 D) 25

Mathematics

1 answer:

maks197457 [2]3 years ago

7 0

Answer:

A is the answer 32

Step-by-step explanation:

use a digital calculator or a scientific

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Consider the vectors a =3i +j −k, b =i +j +4k, c=i +3j +k, d =−i −3j +k, g =−3i −j +k. Which pairs (if any) of these vectors are

11111nata11111 [884]

Answer:

a and b are perpendicular to each other, as are b and d, b and g

Step-by-step explanation:

To check whether two vectors are perpendicular to each other, we need the angle between these vectors to be 90 degrees.

We can find the angle between to vectors a and b from the following relation:

The cosine of the angle $\theta$ between two vectors is equal to the dot product of this vectors divided by the product of vector magnitude.

$cos(\theta) = \frac{a.b}{|a||b|}$

cos(90) = 0, so when the dot product between vectors a and b is 0, it means that these vectors are perpendicular to each other.

Now, for your exercise, let's compute the dot product between these vectors.

-----------

a.b = (3,1,-1).(1,1,4) = 3+1-4 = 0

So a and b are perpendicular to each other.

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a.c = (3,1,-1).(1,3,1) = 3+3-1 = 5

a and c are not perpendicular to each other.

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a.d = (3,1,-1).(-1,-3,1) = -3-3-1 = -7

So not perpendicular

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a.g = (3,1,-1).(-3,-1,1) = -9-1-1 = -11

Not

-----------------

b.c = (1,1,4).(1,3,1) = 1+3+4 = 8

Not

------------------

b.d = (1,1,4).(-1,-3,1) = -1 -3 +4 = 0

b.d = 0, so b and d are perpendicular to each other

--------------------

b.g = (1,1,4).(-3,-1,1) = -3-1+4 = 0

b.g = 0, perpendicular

---------------------

c.d = (1,3,1).(-1,-3,1) = -1-9+1 = -9

-----------------------

c.g = (1,3,1).(-3,-1,1) = -3-3+1 = -5

------------------------

d.g = (-1,-3,1).(-3,-1,1) = 3+3+1 = 7

7 0

3 years ago

Make groups of ten. Write the tens and ones. 1) 90 Ones

Alisiya [41]

90 ones would make 9 tens. It makes 9 tens because theres a 0 after 90. The 0 represents another place value. You can write this as 9 groups of ten

7 0

3 years ago

the following table shows the number of hours some high schools students in two towns spend riding the bus

11Alexandr11 [23.1K]

There is no table (just a comment)

8 0

3 years ago

Which are the solutions of x^2 = –11x + 4?

strojnjashka [21]

This is a quadratic equation, i.e. an equation involving a polynomial of degree 2. To solve them, you must rearrange them first, so that all terms are on the same side, so we get

$x^2 + 11x - 4 = 0$

i.e. now we're looking for the roots of the polynomial. To find them, we can use the following formula:

$x_{1,2} = \frac{-b\pm\sqrt{b^2-4ac}}{2}$

where $x_{1,2}$ is a compact way to indicate both solutions $x_1$ and $x_2$ , while $a,b,c$ are the coefficients of the quadratic equation, i.e. we consider the polynomial $ax^2+bx+c$ .

So, in your case, we have $a=1,\ \ b=11,\ \ c=-4$

Plug those values into the formula to get

$x_{1,2} = \frac{-11\pm\sqrt{121+16}}{2} = \frac{-11\pm\sqrt{137}}{2}$

So, the two solutions are

$x_1 = \frac{-11+\sqrt{137}}{2}$

$x_2 = \frac{-11-\sqrt{137}}{2}$