Select the correct answer.
Based on these segment lengths, which group of segments cannot form a triangle?
OA 12,7,8
OB. 8, 7, 13
OC. 1,2,3
OD. 80,140, 70
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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:

Step-by-step explanation:

14 1⁄16

Answer:

the answer is attached herewith

ps.hv a great day ahead! :)

Answer:

b. 9s^2 t^2

Step-by-step explanation:

If (x*y) ^ 2 = multiple of x and y then power 2

(x * y) ^2 is equal to (x^2 * y^2)

That makes 9s^2t^2

Answer:

Option B. 9s^2t^2 is the correct answer.

Step-by-step explanation:

I just completed it.

Answer:

so we have 4/3 divided by 7/4

= 4/3 * 4/7

= 4*4/3*7

= 16/21

Step-by-step explanation:

answer from gauth math

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:

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:

45

Step-by-step explanation:

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:

6

Step-by-step explanation:

-3 | 6 - 2| + 1

-3 + 6 + 2 + 1

-3 + 9 = 6

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:

38416/10000

plss mark me brainliestt

Step-by-step explanation:

3.8416=38416/10000

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

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:

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:

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:

True

Step-by-step explanation:

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

The scaled factor is 3 for all of them.

Answer:

nzbzbzZnzbznzbzbzhzhsbsjs

Answer:

s,-8*2-`±_'682-¥´owhsi

Answer:

A

Step-by-step explanation:

Mark me brainliest.

Answer:

a.8100

b.3200

Step-by-step explanation:

hope you're okay

Answer:

D

Step-by-step explanation:

1 2/3÷4/5

=5/3×5/4

=25/12

2 and 1 over 12

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.

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

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]