the numeral 16 is a _____ of 4 because it is the product of a given number sbd any whole number

Answer:

the graph crosses the y-axis at -5 and x-axis at -6

Answer:

a = - 2, remainder = 21

Step-by-step explanation:

The Remainder theorem states that if f(x) is divided by (x - a) the remainder is f(a)

Since f(x) is divisible by (x - a) then remainder is zero , then

f(a) = 2a³ - 7a² + 7a² + 16 = 0 , that is

2a³ + 16 = 0 ( subtract 16 from both sides )

2a³ = - 16 ( divide both sides by 2 )

a³ = - 8 ( take the cube root of both sides )

a = [tex]\sqrt[3]{-8}[/tex] = - 2

Then

f(x) = 2x³ - 7x² - 14x + 16

Evaluate f(- [tex]\frac{1}{2}[/tex] ) for remainder on division by (2x + 1)

f(- [tex]\frac{1}{2}[/tex] ) = 2(- [tex]\frac{1}{2}[/tex] )³ - 7(- [tex]\frac{1}{2}[/tex] )² - 14(- [tex]\frac{1}{2}[/tex] ) + 16

        = 2(- [tex]\frac{1}{8}[/tex] ) - 7([tex]\frac{1}{4}[/tex] ) + 7 + 16

        = - [tex]\frac{1}{4}[/tex] - [tex]\frac{7}{4}[/tex] + 23

        = - [tex]\frac{8}{4}[/tex] + 23

       = - 2 + 23

       = 21

Answer:

is 5

Step-by-step explanation:

Answer:

5

Step-by-step explanation:

So first when dividing fractions you need to invert and multiply. Thus the fraction becomes:

1/4 * 20/1.

When simplified it becomes 20/4 which is 5.

HOPE THIS HELPED

This is simply done by Method of Completing the Square.

3x - x²

Add and subtract half the coefficient of x and square it.

( This is done so there'd be no alterations to the quadractic Expression)

So To start

We'd like x² to be positive...(So factor out a negative).

–(x² - 3x )

Now In this case

You see there's a Negative Outside the the bracket

Instead of adding and Subtracting squared values of half the coefficient of X... We'd Add Twice and do not subtract.

Reason: If you add outside the bracket and subtract the other inside the bracket... This will be wrong because there's a negative patiently waiting outside the bracket to Interact with the negative you subtracted to make it Positive.

See what I mean.

Let's say you added 2² and subtracted 2² in this problem

2² – ( x² - 3x - 2²)

If you decide to open the bracket

You'll have

2² – x² + 3x + 2²

NOW THIS IS WRONG BECAUSE WE ALTERED THIS EXPRESSION. WHERE'D 2² + 2² COME FROM?

THIS IS WHY YOU'LL ADD THE SQUARED COEFFICIENT OF X TWICE IN CASES LIKE THESE.

SO GOING BACK TO THE ORIGINAL QUESTION.

– (x² - 3x )

Adding the half the coefficient of x twice and squaring them...

Coefficient of x = 3

Half of 3 = 3/2

Squaring it gives (3/2)²

NOW PROCEEDING

(3/2)² – [ x² - 3x + (3/2)²]

If you open this bracket... (3/2)² will cancel out with —(3/2)²

Meaning that we haven't altered the expression in any way

Moving On...

Applying basic factorizing principle

9/4 – ( x - 3/2)².

Answer = 9/4 – ( x - 3/2)² Which is in the Form m – ( x - n )²

Therefore m = 9/4 and n = 3/2.

Hope This Helps

Answer:

C

Step-by-step explanation:

If you add up the length of two sides, the sum must be greater than the third side of the triangle.

If you add 1 and 2, it equals 3

But that means it will be equal to the length of the third side, 3

It will be impossible to make a triangle with those lengths of sides no matter how the sides or angles are set.

It's sort of difficult to explain this without any visual

You can look up "triangle inequality" to find out more about this

Answer:

A

Step-by-step explanation:

Mark me brainliest.

Answer:

45

Step-by-step explanation:

Answer:

a.8100

b.3200

Step-by-step explanation:

hope you're okay

Answer:

nzbzbzZnzbznzbzbzhzhsbsjs

Answer:

k = 2

Step-by-step explanation:

2 x 2 = 4

3 x 2 = 6

so (2,3) multiplied by 2 will be (4,6)

Answer:

x ≠ -1/3

Step-by-step explanation:

(f/g)(x) = f(x)/g(x) = ∛2x/(3x+1)

since division by zero is undefined,

Answer:

Step-by-step explanation:

x = - 1/2

Answer: No, a triangle is not possible in this case

Why not? Because the two sides 6 and 6 add to 6+6 = 12, which does not exceed the remaining side.

Refer to the triangle inequality theorem. This theorem says that for a triangle to be possible, we need these three facts to be true

where a,b,c are the three sides of the triangle. In short: adding any two sides must be larger than the third side.

I recommend cutting out slips of paper to try making a triangle with sides 12,6, and 6. You'll find its not possible. The two smaller sides combine to perfectly line up with the larger side. The three sides form a line and not a triangle.

Step-by-step explanation:

f(10)=10

f(-2)=-2

f(a)=a

f(a+b)=a+b

g(10)=5×10-12=3

g(-2)=5×(-2)-12=-10-12=-22

g(a)=5×a-12=5a-12

g(a+b)=5×(a+b)-12=5a+5b-12

h(10)=(10)^2 +4(10)-7=100+40-7=133

h(-2)=(-2)^2 +4(-2)-7=4-8-7=-11

h(a)=a^2 +4a -7

h(a+b)=(a+b)^2 +4a+4b-7=a^2+2ab+b^2+4b-7

Answer:

s,-8*2-`±_'682-¥´owhsi

Answer:

w = 7

Step-by-step explanation:

L = 22

L =  2*w + 8

Since the Ls are the same thing, you can equate them

2w + 8 = 22                 Subtract 8 from both sides

2w = 22 - 8                  Combine

2w = 14                         divide by 2

w = 14/2

w = 7

Answer:

the answer is attached herewith

ps.hv a great day ahead! :)

Answer:

6

Step-by-step explanation:

-3 | 6 - 2| + 1

-3 + 6 + 2 + 1

-3 + 9 = 6

Answer:

[tex]ax + bx = 10 \\ \\ x(a + b) = 10 \\ \\ x = \frac{10}{a + b} [/tex]

I hope I helped you^_^

Answer:

38416/10000

plss mark me brainliestt

Step-by-step explanation:

3.8416=38416/10000

Answer:

D

Step-by-step explanation:

1 2/3÷4/5

=5/3×5/4

=25/12

2 and 1 over 12

Answer:

True

Step-by-step explanation:

The sum of any rational number and any irrational number will always be an irrational number

Answer:

It is 9 months

Step-by-step explanation:

[tex] = 20\% \times 3 \frac{3}{4} \\ \\ = \frac{20}{100} \times \frac{15}{4} \\ \\ = \frac{300}{400} \\ \\ = \frac{3}{4} \: \: { \sf{years}}[/tex]

convert them to months by multiplying by 12:

[tex]{ \sf{ = \frac{3}{4} \times 12}} \\ \\ = { \sf{9 \: months}}[/tex]

Answer:

12 -16. 19

4. 8. 22

3. -9. -4

Step-by-step explanation:

13-1 = 12

-10-6= -16

12 - -7 = 12+7=19

6-2 = 4

Keep subtracting

Answer:

Step-by-step explanation:

14 1⁄16

Answer:

the probability would be 3/8

Step-by-step explanation:

Since there are 3 red and 5 green, there is a total of 8 balls in the bag making it a chance of 3/8 to get a red ball. You're Welcome :) <3

Probability of picking a red ball is  [tex]\frac{3}{8}[/tex]

Probability of getting head coin is  [tex]\frac{1}{2}[/tex]

"Probability is the possibility or chance of an event to occur." Ratio of number of favorable outcomes and total number of favorable outcomes is called probability.

We have,

Red balls = 3

Green balls =5

Formula for probability

 [tex]\frac{Number of Favorable Outcomes}{Total Number of Favorable Outcomes}[/tex]

Total number of outcomes n(S) = 8

Number of favorable outcomes for getting a red ball n(E) = 3

Probability of picking a red ball

= [tex]\frac{Number of Favorable Outcomes}{Total number of Favorable Outcomes}[/tex]

= [tex]\frac{3}{8}[/tex]

We have a fair coin,

Total number of outcomes n(S) = 2

Number of favorable outcomes for getting head n(E) = 1

Probability of getting head coin

= [tex]\frac{Number of Favorable Outcomes}{Total Number of Favorable Outcomes}[/tex]

= [tex]\frac{1}{2}[/tex]

Hence, Probability of picking a red ball is  [tex]\frac{3}{8}[/tex]

Probability of getting head coin is  [tex]\frac{1}{2}[/tex].

Learn more about probability here

https://brainly.com/question/15052059

#SPJ2

Answer:

keyboard and mouse are two examples of input device

Answer:

keyboard and mouse

Step-by-step explanation:

If you like my answer than please mark me brainliest

It looks like the limit you want to compute is

[tex]\displaystyle L = \lim_{x\to0}\left(\frac x{\tan(x)}\right)^{\cot^2(x)}[/tex]

Rewrite the limand with an exponential and logarithm:

[tex]\left(\dfrac{x}{\tan(x)}\right)^{\cot^2(x)} = \exp\left(\cot^2(x) \ln\left(\dfrac{x}{\tan(x)}\right)\right) = \exp\left(\dfrac{\ln\left(\dfrac{x}{\tan(x)}\right)}{\tan^2(x)}\right)[/tex]

Now, since the exponential function is continuous at 0, we can write

[tex]\displaystyle L = \lim_{x\to0} \exp\left(\dfrac{\ln\left(\dfrac{x}{\tan(x)}\right)}{\tan^2(x)}\right) = \exp\left(\lim_{x\to0}\dfrac{\ln\left(\dfrac{x}{\tan(x)}\right)}{\tan^2(x)}\right)[/tex]

Let M denote the remaining limit.

We have [tex]\dfrac x{\tan(x)}\to1[/tex] as [tex]x\to0[/tex], so [tex]\ln\left(\dfrac x{\tan(x)}\right)\to0[/tex] and [tex]\tan^2(x)\to0[/tex]. Apply L'Hopital's rule:

[tex]\displaystyle M = \lim_{x\to0}\dfrac{\ln\left(\dfrac{x}{\tan(x)}\right)}{\tan^2(x)} \\\\ M = \lim_{x\to0}\dfrac{\dfrac{\tan(x)-x\sec^2(x)}{\tan^2(x)}\times\dfrac{\tan(x)}{x}}{2\tan(x)\sec^2(x)}[/tex]

Simplify and rewrite this in terms of sin and cos :

[tex]\displaystyle M = \lim_{x\to0} \dfrac{\dfrac{\tan(x)-x\sec^2(x)}{\tan^2(x)}\times\dfrac{\tan(x)}{x}}{2\tan(x)\sec^2(x)} \\\\ M= \lim_{x\to0}\dfrac{\sin(x)\cos^3(x) - x\cos^2(x)}{2x\sin^2(x)}[/tex]

As [tex]x\to0[/tex], we get another 0/0 indeterminate form. Apply L'Hopital's rule again:

[tex]\displaystyle M = \lim_{x\to0} \frac{\sin(x)\cos^3(x) - x\cos^2(x)}{2x\sin^2(x)} \\\\ M = \lim_{x\to0} \frac{\cos^4(x) - 3\sin^2(x)\cos^2(x) - \cos^2(x) + 2x\cos(x)\sin(x)}{2\sin^2(x)+4x\sin(x)\cos(x)}[/tex]

Recall the double angle identity for sin:

sin(2x) = 2 sin(x) cos(x)

Also, in the numerator we have

cos⁴(x) - cos²(x) = cos²(x) (cos²(x) - 1) = - cos²(x) sin²(x) = -1/4 sin²(2x)

So we can simplify M as

[tex]\displaystyle M = \lim_{x\to0} \frac{x\sin(2x) - \sin^2(2x)}{2\sin^2(x)+2x\sin(2x)}[/tex]

This again yields 0/0. Apply L'Hopital's rule again:

[tex]\displaystyle M = \lim_{x\to0} \frac{\sin(2x)+2x\cos(2x)-4\sin(2x)\cos(2x)}{2\sin(2x)+4x\cos(2x)+4\sin(x)\cos(x)} \\\\ M = \lim_{x\to0} \frac{\sin(2x) + 2x\cos(2x) - 2\sin(4x)}{4\sin(2x)+4x\cos(2x)}[/tex]

Once again, this gives 0/0. Apply L'Hopital's rule one last time:

[tex]\displaystyle M = \lim_{x\to0}\frac{2\cos(2x)+2\cos(2x)-4x\sin(2x)-8\cos(4x)}{8\cos(2x)+4\cos(2x)-8x\sin(2x)} \\\\ M = \lim_{x\to0} \frac{4\cos(2x)-4x\sin(2x)-8\cos(4x)}{12\cos(2x)-8x\sin(2x)}[/tex]

Now as [tex]x\to0[/tex], the terms containing x and sin(nx) all go to 0, and we're left with

[tex]M = \dfrac{4-8}{12} = -\dfrac13[/tex]

Then the original limit is

[tex]L = \exp(M) = e^{-1/3} = \boxed{\dfrac1{\sqrt[3]{e}}}[/tex]