The tempature at 5 P.M. is 20 degrees f. The temperature at 10 P.M. is -5 degrees f. How many degrees did the temperature fall?

Answer:

[tex]-\sqrt{3} /2[/tex]

Step-by-step explanation:

300 degrees is in the fourth quadrant (it's between 270 and 360); sine is negative in the fourth quadrant.

Given we're in the fourth quadrant, the reference angle is 360 - 300 = 60 degrees

sin(60°) = [tex]\sqrt{3} /2[/tex]

And since sine is negative, this value turns negative:

sin(300°) = [tex]-\sqrt{3} /2[/tex]

Answer:

3

Step-by-step explanation:

[tex]log(3x^{3}) - log(x^{2}) = log(\frac{3x^{3}}{x^{2}})\\log(27) - log(x) = log(\frac{27}{x} )\\[/tex]

therefore,

[tex]\frac{3x^{3} }{x^{2} } = \frac{27}{x} \\3x=\frac{27}{x} \\3x^{2} =27\\x= +3\\x=-3[/tex]

however, since logarithms cannot have negative arguments, x can only be +3

i.e. log(-3) is impossible, and will return MATH ERROR on a calculator.

B: is correctly in my opinion

Answer:

60

Step-by-step explanation:

its the answer

Answer:

the space (usually measured in degrees) between two intersecting lines or surfaces at or close to the point where they meet.

Step-by-step explanation:

Answer:

-3.64

Step-by-step explanation:

Step-by-step explanation:

1. 6(5)^2-1/(5)^2 = 149/25

2. 6(-5)^2-1/25 = -150/25

3. 6x^2-2/x^2

Answer:

[tex]1) \huge\boxed{ \sf f(5) = 5 \frac{24}{25} }[/tex]

[tex]2) \huge\boxed{ \sf f(-5) = 5\frac{24}{25} }[/tex]

[tex]3) \huge\boxed{\sf f(-x) = \frac{6x^2-1}{x^2} }[/tex]

Step-by-step explanation:

[tex]\displaystyle f(x) = \frac{6x^2-1}{x^2}[/tex]

For f(5):

Put x = 5

[tex]\displaystyle f(5) = \frac{6(5)^2-1}{(5)^2} \\\\f(5) = \frac{6(25)-1}{25} \\\\f(5) = \frac{150-1}{25} \\\\f(5) = \frac{149}{25} \\\\f(5) = 5 \frac{24}{25}[/tex]

For f(-5):

Put x = -5

[tex]\displaystyle f(-5) = \frac{6(-5)^2-1}{(-5)^2} \\\\f(-5) = \frac{6(25)-1}{25} \\\\f(-5) = \frac{150-1}{25} \\\\f(-5) = \frac{149}{25} \\\\f(-5) = 5\frac{24}{25}[/tex]

For f(-x):

Put x = -x

[tex]\displaystyle f(-x) =\frac{6(-x)^2-1}{(-x)^2} \\\\f(-x) = \frac{6x^2-1}{x^2} \\\\\rule[225]{225}{2}[/tex]

Hope this helped!

Answer:

x is 2 and -8

Step-by-step explanation:

[tex] {x}^{2} + 6x - 16 = 0[/tex]

general equation

[tex] {ax}^{2} + (sum)x + product = 0[/tex]

sum is 6, product is -16

for completing squares,

first divide the sum by 2:

[tex] = \frac{6}{2} = 3[/tex]

add the square of the result on (x² + 6x) and subtract it from the product:

[tex] ( {x}^{2} + 6x + {3}^{2} ) - 16 - {3}^{2} = 0 \\ {(x + 3)}^{2} - 25 = 0 \\ {(x + 3)}^{2} = 25[/tex]

take square root:

[tex] \sqrt{ {(x + 3)}^{2} } = \sqrt{25 } \\ x + 3 = ±5 \\ x = ±5 - 3[/tex]

x is either: 5-3 or -5-3

[tex]x = 2 \: \: and \: - 8[/tex]

Answer:

Step-by-step explanation:

Area of ΔABC is:

We have:

Find CD using Pythagorean:

Find DB using the following identity, coming from ratios of corresponding sides of similar triangles:

Find AB:

Find the area of ΔABC:

Step-by-step explanation:

[tex]b(1) = 4[/tex]

Since this is an arithmetic sequence, notice that if you subtract b(1) (i.e., 4) from b(2)(i.e., 22), you get 18. Likewise, if you subtract b(2) from b(3) you also get 18. Therefore,

[tex]b(n) = b(n-1) + 18[/tex]

9514 1404 393

Answer:

  a = ±2

Step-by-step explanation:

For similar figures, the same scale factor applies to both x and y.

  12 = 3a²

  4 = a²

  a = ±√4 = ±2

Answer:

(5,2)

(6,-6)

Step-by-step explanation:

Differentiate both sides with respect to x and solve for the derivative dy/dx :

[tex]\dfrac{\mathrm d}{\mathrm dx}\left[x^2y^2+xy\right] = \dfrac{\mathrm d}{\mathrm dx}[2] \\\\ \dfrac{\mathrm d}{\mathrm dx}\left[x^2\right]y^2 + x^2\dfrac{\mathrm d}{\mathrm dx}\left[y^2\right] + \dfrac{\mathrm d}{\mathrm dx}\left[x\right]y + x\dfrac{\mathrm dy}{\mathrm dx} = 0 \\\\ 2xy^2 + x^2(2y)\dfrac{\mathrm dy}{\mathrm dx} + y + x\dfrac{\mathrm dy}{\mathrm dx} = 0 \\\\ (2x^2y+x)\dfrac{\mathrm dy}{\mathrm dx} = -2xy^2-y \\\\ \dfrac{\mathrm dy}{\mathrm dx} = -\dfrac{2xy^2+y}{2x^2y+x}[/tex]

This gives the slope of the tangent to the curve at the point (x, y).

If the slope of some tangent line is -1, then

[tex]-\dfrac{2xy^2+y}{2x^2y+x} = -1 \\\\ \dfrac{2xy^2+y}{2x^2y+x} = 1 \\\\ 2xy^2+y = 2x^2y+x \\\\ 2xy^2-2x^2y + y - x = 0 \\\\ 2xy(y-x)+y-x = 0 \\\\ (2xy+1)(y-x) = 0[/tex]

Then either

[tex]2xy+1 = 0\text{ or }y-x=0 \\\\ \implies y=-\dfrac1{2x} \text{ or }y=x[/tex]

In the first case, we'd have

[tex]x^2\left(-\dfrac1{2x}\right)^2+x\left(-\dfrac1{2x}\right) = \dfrac14-\dfrac12 = -\dfrac14\neq2[/tex]

so this case is junk.

In the second case,

[tex]x^2\times x^2+x\times x=x^4+x^2=2 \\\\ x^4+x^2-2 = (x^2-1)(x^2+2)=0[/tex]

which means either

[tex]x^2-1 = 0 \text{ or }x^2+2 = 0 \\\\ x^2 = 1 \text{ or }x^2 = - 2[/tex]

The second case here leads to non-real solutions, so we ignore it. The other case leads to [tex]x=\pm1[/tex].

Find the y-coordinates of the points with x = ±1 :

[tex]x=1 \implies y^2+y=2 \implies y=-2 \text{ or }y=1 \\\\ x=-1\implies y^2-y=2\implies y=-1\text{ or }y=2[/tex]

so the points of interest are (1, -2), (1, 1), (-1, -1), and (-1, 2).

9514 1404 393

Answer:

  24 in square by 6 in deep

Step-by-step explanation:

Let x represent the side of the square cut from each corner. Then the dimensions of the base of the box are 36-2x in each direction. The total volume of the box is ...

  V = LWH = (36 -2x)(36 -2x)x = x(4x² -144x +1296)

The volume will be a maximum where dV/dx = 0.

  dV/dx = 12x^2 -288x +1296 = 0

  x² -24x +108 = 0 . . . . divide by 12

  (x -6)(x -18) = 0 . . . . . factor

  x = 6 or 18 . . . . . . x = 18 gives a minimum volume; we want x = 6

Then the dimensions are 36 -2(6) = 24 inches square by 6 inches deep.

Answer:

You need to bring the function to the standard form.

The term with highest degree exponent is the leading term and its coefficient is the leading coefficient.

Let it be axⁿ.

Depending on the n and a, the end behavior of the function will change.

This is an odd function and:

This is an odd function and:

This is an even function and:

This is an even function and:

Answer:

Yeah your right

Step-by-step explanation:

Answer:

D

Step-by-step explanation:

When all the members in a domain has only but one member in the do main then function has been satisfied.

considering a situation of,

4 1

6 1

8 2

the domain has only one member in the Co domain hence which makes it a function

Answer:

Pretty sure you'd be in hell by then 0_0

Step-by-step explanation:

think of it-2,200 ft and your at 0. Your in hell!

Answer:

Uhm, I see there's a pattern, but there are only 2 green boxes, whole there are 4 orange boxes in every figure.

Answer:

B

Step-by-step explanation:

i did this a couple years ago should be right if im correct

Answer:

Perimeter = 38.8m

Step-by-step explanation:

If you like my answer than please mark me brainliest thanks

Answer:

39cm

Step-by-step explanation:

When you find the upper and lower bounds of values with decimals, you will decrease or increase the value by increments of 0.05. Since we are just trying to find the upper bound we will add 0.05 to the values we are given.

15.6 + 0.05 = 15.65cm

3.8 + 0.05 = 3.85cm

Now that we have those values, we can find the perimeter using the formula [ 2(L + W) ]

= 2(15.65 + 3.85)

= 2(19.5)

= 39cm

Best of Luck!

9514 1404 393

Answer:

  58 km/h

Step-by-step explanation:

Let s represent James's speed. Then s-24 is Kim's speed. The relationship between time, speed, and distance is ...

  d = st

Since the two were traveling in opposite directions, their distance apart is the sum of the distances they drove. James drove for 1.6 more hours than the 1.8 hours Kim drove, so their total distance is ...

  (1.6 +1.8)s +1.8(s -24) = 258.4

  5.2s -43.2 = 258.4 . . . . . . . . . . . . collect terms

  5.2s = 301.6 . . . . . . . . . . . add 43.2

  s = 58 . . . . . . . . . . . divide by 5.2

James drove at 58 km/h.

Treating the data as a Venn set, it is found that:

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

I am going to say that:

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

9 are good in all of the three courses.

This means that: [tex]A \cap B \cap C = 9[/tex]

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

13 are good in both mathematics and psychology

This means that:

[tex](A \cap C) + (A \cap B \cap C) = 13[/tex]

[tex](A \cap C) + 9 = 13[/tex]

[tex](A \cap C) = 4[/tex]

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

15 are good in both mathematics and in English

This means that:

[tex](A \cap B) + (A \cap B \cap C) = 15[/tex]

[tex](A \cap B) + 9 = 15[/tex]

[tex](A \cap B) = 6[/tex]

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

16 are good in both English and psychology

This means that:

[tex](B \cap C) + (A \cap B \cap C) = 16[/tex]

[tex](B \cap C) + 9 = 16[/tex]

[tex](B \cap C) = 7[/tex]

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

20 are good in psychology

This means that:

[tex]c + (A \cap C) + (B \cap C) +  (A \cap B \cap C) = 20[/tex]

[tex]c + 4 + 7 + 9 = 20[/tex]

[tex]c = 0[/tex]

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

45 are good in mathematics

This means that:

[tex]a + (A \cap B) + (A \cap C) +  (A \cap B \cap C) = 45[/tex]

[tex]a + 6 + 4 + 9 = 45[/tex]

[tex]a = 26[/tex]

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

Question a:

[tex]a = 26[/tex], which means that 26 students are good in mathematics only.

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

Question b:

At least one is:

[tex]a + (A \cap B) + (A \cap C) + (B \cap C) +  (A \cap B \cap C) = 26 + 6 + 4 + 7 + 9 = 52[/tex]

Thus, 80 - 52 = 28

28 students are not good in any of the three courses.

A similar problem is given at: https://brainly.com/question/22003843

Answer:

Polynomial Equation                          Degree                                  Example

Linear Equations                                     1                                    -3x + 1 = 4x + 5

Quadratic Equations                               2                                   x^2 – 6x + 9 = 0

Cubic Equations                                      3                              x^3 – 2x^2 + 3x = -5

Quartic Equations                                   4                                      x^4 – 2x^2 = -4

Step-by-step explanation:

Linear Equations

Linear equations are polynomial equations that have a degree of 1.

ax + b = 0

Solving for solutions for this type of equation will require us to isolate the unknown variable on one side of the equation. Master your craft in solving linear equations here.

Quadratic Equations

Quadratic equations are polynomial equations with a degree of 2.

ax2 + bx + c = 0

There are different ways we can solve quadratic equations – it mostly depends on the form of the quadratic expression on the right-hand side.

We can factor quadratic expressions and apply the zero-property.

Applying special algebraic properties such as the difference of two squares, perfect square trinomial properties, and completing the square.

Lastly, we can also use the quadratic formula to find the zeroes of quadratic equations.

Polynomial Equations (with a degree of 3 or higher)

Here’s the exciting part: what if we need to find the zeros of the solutions of a polynomial equation with degrees that are 3 or higher?

Some cubic and quartic equations can be factored by grouping and be reduced to equations with a smaller degree. There are times, however, that finding the actual factors can be challenging.

Answer:

[tex]f(x)^{-1}=-x+5[/tex]

Step-by-step explanation:

It looks like the function is f(x)=-x+5.

To find the inverse, first replace f(x) with y.

y=-x+5

Switch the y and x.

x=-y+5.

Solve for y.

x-5=-y

-x+5=y

[tex]f(x)^{-1}=-x+5[/tex]

I hope this helps!

pls ❤ and mark brainliest pls

Answer:

-60 , -19 , 6.28 , 14

Step-by-step explanation:

Order the number in ascending order, or from least to greatest. Note that the bigger the negative number, the smaller the number. The bigger the positive number, the bigger the number:

-60 , -19 , 6.28 , 14

~

Note: For 6.28, the decimal point results in 6 being in the ones place, 2 being in the tenths place, 8 being in the hundredths place. By the rules regarding the usage of decimals, place values after the decimal point will be significantly less then the place values before the decimal point, and would be used to determine a more exact value.

Answer:

a = [tex]\frac{2A}{p}[/tex]

Step-by-step explanation:

Given

A = [tex]\frac{1}{2}[/tex] ap ( multiply both sides by 2 to clear the fraction )

2A = ap ( isolate a by dividing both sides by p )

[tex]\frac{2A}{p}[/tex] = a

Answer:

plane EBG

Step-by-step explanation:

plane C is also named plane EBG

Answer:

[tex]\\ \sf\longmapsto Plane EBG[/tex]

We can't name it EBF as it is a axis and coplanar points

Also nEF is not a satisfactory name .

So the correct option is C

Perimeter is the distance around the rectangle and the formula is 2 x length + 2 x height.

Perimeter = 2 x45 + 2x20

Perimeter = 90 + 40

Perimeter = 130 yards

Answer:

Length = 45 yard

Height = 20 yards

Perimeter

We know that

[tex] \: perimeter = 2(l + b)[/tex]

» Perimeter = 2(45 + 20)

» Perimeter = 2(65)

» Perimeter = 130 yards

Answer:

A

Step-by-step explanation:

Because lot of people love it