Find the greatest number you can add to 4315 without having to regroup

Answers

Answer 1

Answer: 4315
+5684
9999 So 5684 is the greatest number you can write.

Answer 2

Answer:

Final answer:

The greatest number that can be added to 4315 without having to regroup is 5644. When added together, the sum doesn't result in a carryover to the next place value.

Explanation:

In Mathematics, particularly arithmetic, regrouping is the process of making groups of tens when adding or subtracting two numbers. When the question asks for the greatest number that can be added to 4315 without having to regroup, we are essentially looking for a number that when added to 4315 doesn't result in a carryover to the next place value.

Considering the number 4315, you would notice that the units digit is 5, the tens digit is 1, the hundreds digit is 3, and the thousands digit is 4. This means for each position we can add a number up to 9 without needing to regroup. Hence, the greatest number you can add to 4315 without needing to regroup would be 5644 (which gives 9 in the units place, 1+5=6 in the tens place, 3+4=7 in the hundreds place, and 4+4=8 in the thousands place).

Learn more about Arithmetic here:

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The sum of three consecutive intergers is 174. What are the intergers?

Answers

I have no idea what intergers are.

If you were talking about integers (whole numbers) instead,
then you could make 174 with 57, 58, and 59 .

Let the middle integer = x.
That means we have (x-1) + x + (x+1) = 174
3x = 174
x = 58
x-1 = 57
x+1 = 59

an exponential function is expressed in the form y=a*b/\x. the relation represents a growth when ________ and a decay when ________.

Answers

That function grows when 'b' is more than 1 and 'x' is positive, or when b <1 and x<0. It decays when b> 1 and x <0, or when b<1 and x> 0.

Answer 1: b>1

Answer 2: 0b<b<1

An exponential function is expressed in the form y=axb^x. The relation represents a growth when "b>1" and a decay when "0<b<1".

What is the greatest possible error if Bruce measured a buckle as 3.2 cm using the ruler?

Answers

Answer:

The greatest possible error for the measurement 3.2 cm is 0.05 cm.

Step-by-step explanation:

We have given that,

Bruce measured a buckle as 3.2 cm using the ruler

We have to find,

What is the greatest possible error?

Solution is given by,

The greatest possible error is defined as the error is half of the measuring unit.

Bruce measured a buckle as 3.2 cm using the ruler.

The measuring unit is nearest tenth of a cm or 0.1 cm.

The greatest possible error is half of 0.1.

i.e. $GPE=(0.1)/(2)$

$GPE=0.05}$

Therefore, The greatest possible error for the measurement 3.2 cm is 0.05 cm.

If A, B,C are the angles of a triangle then prove: (the following in picture)Please help me to prove this.

Answers

Answer: see proof below

Step-by-step explanation:

Given: A + B + C = π → A + B = π - C

→ C = π - (A + B)

Use Sum to Product Identity: cos A + cos B = 2 cos [(A + B)/2] · cos [(A - B)/2]

Use Product to Sum Identity: 2 sin A · sin B = cos [(A + B)/2] - cos [(A - B)/2]

Use the Double Angle Identity: cos 2A = 1 - 2 sin² A

Use the Cofunction Identity: cos (π/2 - A) = sin A

Proof LHS → RHS:

LHS: cos A + cos B + cos C

= (cos A + cos B) + cos C

$\text{Sum to Product:}\qquad 2\cos \bigg((A+B)/(2)\bigg)\cdot \cos \bigg((A-B)/(2)\bigg)+\cos C$

$\text{Given:}\qquad 2\cos \bigg((\pi -C)/(2)\bigg)\cdot \cos \bigg((A-B)/(2)\bigg)+\cos C\n\n\n.\qquad \qquad =2\cos \bigg((\pi)/(2) -(C)/(2)\bigg)\cdot \cos \bigg((A-B)/(2)\bigg)+\cos C$

$\text{Cofunction:}\qquad 2\sin \bigg((C)/(2)\bigg)\cdot \cos \bigg((A-B)/(2)\bigg)+\cos C$

$\text{Double Angle:}\qquad 2\sin \bigg((C)/(2)\bigg)\cdot \cos \bigg((A-B)/(2)\bigg)+\cos\bigg(2\cdot (C)/(2)\bigg)\n\n\n.\qquad \qquad \qquad =2\sin \bigg((C)/(2)\bigg)\cdot \cos \bigg((A-B)/(2)\bigg)+1-2\sin^2 \bigg((C)/(2)\bigg)\n\n\n.\qquad \qquad \qquad =1+2\sin \bigg((C)/(2)\bigg)\cdot \cos \bigg((A-B)/(2)\bigg)-2\sin^2\bigg((C)/(2)\bigg)$

$\text{Factor:}\qquad 1+2\sin \bigg((C)/(2)\bigg)\bigg[\cos \bigg((A-B)/(2)\bigg)-\sin\bigg((C)/(2)\bigg)\bigg]$

$\text{Given:}\qquad 1+2\sin \bigg((C)/(2)\bigg)\bigg[\cos \bigg((A-B)/(2)\bigg)-\sin\bigg((\pi-(A+B))/(2)\bigg)\bigg]\n\n\n.\qquad \qquad 1+2\sin \bigg((C)/(2)\bigg)\bigg[\cos \bigg((A-B)/(2)\bigg)-\sin\bigg((\pi)/(2)-(A+B)/(2)\bigg)\bigg]$

$\text{Cofunction:}\qquad 1+2\sin \bigg((C)/(2)\bigg)\bigg[\cos \bigg((A-B)/(2)\bigg)-\cos\bigg((A+B)/(2)\bigg)\bigg]$

$\text{Product to Sum:}\qquad 1+2\sin \bigg((C)/(2)\bigg)\bigg[2\sin \bigg((A)/(2)\bigg)\cdot \sin\bigg((B)/(2)\bigg)\bigg]\n\n\n.\qquad \qquad \qquad \qquad =1+4\sin \bigg((C)/(2)\bigg)\bigg[\sin \bigg((A)/(2)\bigg)\cdot \sin\bigg((B)/(2)\bigg)\bigg]\n\n\n.\qquad \qquad \qquad \qquad =1+4\sin \bigg((A)/(2)\bigg)\sin \bigg((B)/(2)\bigg) \sin\bigg((C)/(2)\bigg)$

$\text{LHS = RHS:}\ 1+4\sin \bigg((A)/(2)\bigg)\sin \bigg((B)/(2)\bigg) \sin\bigg((C)/(2)\bigg)=1+4\sin \bigg((A)/(2)\bigg)\sin \bigg((B)/(2)\bigg) \sin\bigg((C)/(2)\bigg)\quad \checkmark$

The proof for this is simple. Let's say that A + B + C = π. From here on we require several trigonometric identities that must be applied.

$\cos \left(A\right)+\cos \left(B\right)+\cos \left(C\right) \n= 2 * cos((A + B) / 2) * cos((A - B) / 2) + \cos C \n= 2 * cos((\pi /2) - (C/2)) * cos((A - B) / 2) +\cos C \n= 2 * sin(C/2) * cos((A - B) / 2) + (1 - 2 * sin^2 (C/2)) \n= 1 + 2 sin (C/2) * cos((A - B) / 2) - sin (C/2) \n= 1 + 2 sin (C/2) * cos((A - B) / 2) - sin((\pi /2) - (A + B)/2 ))\n= 1 + 2 sin (C/2) * cos((A - B) / 2) - cos((A + B)/ 2)\n= 1 + 2 sin (C/2) * 2 sin (A/2) * sin(B/2) \n= 1 + 4 sin(A/2) sin(B/2) sin(C/2)$

Hope that helps!

A sequence is defined by the recursive formula f(n + 1) = 1.5f(n). Which sequence could be generated using the formula? –12, –18, –27, ... –20, 30, –45, ... –18, –16.5, –15, ... –16, –17.5, –19, ...

Answers

Formula f(n+1) = 1.5 f(n) => f(n+1) / f(n) = 1.5. That means, that you have to search for a sequence where the ratio of two consecutive terms is 1.5. The first sequence is the only one that meets that: - 18 / - 12 = 1.5 and - 27 / - 18 = 1.5. So, the answer is the first option: -12, -18, -27, ...

Answer: –12, –18, –27, ... [ A.K.A: (A.) ]

Simplify each expressiona. 5n2 − 3n2
b. 8 + 3b − 2c − 4 + 6b
(these are the answer choices)
A. a. −2n2
b. 9b − 2c + 4

B. a. −8n2
b. 9b − 2c + 4

C. a. 2n2
b. 9b − 2c + 4

D. a. 8n2
b. 9b − 2c + 4

Answers

Alright now combine all like terms remember no matter what you can allways add like terms ok now it would be 5-3 you should get two now add the n and the squared 2 and 2n2 should be your answer witch is ANow for the other equation 8+-4 =4 3b +6b = 9b now drag the 2c down and you get4+9b-2c I hope this helps you

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