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3 years ago

Isiah determined that 5a2 is the GCF of the polynomial a^3 – 25a^2b^5 – 35b^4. Is he correct?

Mathematics

2 answers:

Charra [1.4K]3 years ago

8 0

Answer:

here is the sample response

Step-by-step explanation:

Shkiper50 [21]3 years ago

6 0

Isiah determined that 5a2 is the GCF of the polynomial a^3 – 25a^2b^5 – 35b^4-<u>No,Isiah is not correct if he states that that 5a2 is the GCF of the polynomial</u>

Step-by-step explanation:

Firstly ,we divide each term by the GCF and if the term does not divide evenly, then it is not valid

so,

5a^2 divide by a^3

we see that we cannot divide 5a^2 by a^3 because firstly the power is higher and then there is no 5 number.Thus it is not the GCF

The GCF of the coefficients is 1, and there are no common variables among all three terms of the polynomial.

Also we find that 5b^4 is a factor of -25a^2b^5 and -35b^4, but not a3. Additionally, a2 is a factor of a3 and -25a^2b^5, but not -35b^4.

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The total surface area of this cuboid is 112cm squared find the value of x

Liono4ka [1.6K]

Answer:

The height of cuboid is 3 cm

Step-by-step explanation:

Total surface area of cuboid = $2lb+2bh+2hl$

Length of cuboid = 10 cm

Width of cuboid = 2 cm

Let the height be x

Total surface area of cuboid = $2(10)(2)+2(2)(x)+2(x)(10)$

Total surface area of this cuboid is 112 cm squared

$2(10)(2)+2(2)(x)+2(x)(10)=112$

$40+4x+20x=112$

$40+24x=112$

$24x=112-40$

$24x=72$

$x=\frac{72}{24}$

$x=3$

Hence the height of cuboid is 3 cm

5 0

3 years ago

The average of 16 consecutive positive integers is 30.5. What is the average of the first 8 integers of this set?

GREYUIT [131]

consecutive integers are 1 apart

16 of them can be represented as

x, x+1, x+2, x+3, x+4, x+5, x+6, x+7, x+8, x+9, x+10, x+11, x+12, x+13, x+14, x+15

the average is $\frac{x+x+1+x+2+x+3+x+4+x+5+x+6+x+7+x+8+x+9+x+10 x+11+x+12+x+13+x+14+x+15}{16}=30.5$

there might be some trick to get the average of the first 8 integers only

the average of the first 8 integers would be $\frac{x+x+1+x+2+x+3+x+4+x+5+x+6+x+7}{8}=?$

how can we get $\frac{x+x+1+x+2+x+3+x+4+x+5+x+6+x+7}{8}=?$ from $\frac{x+x+1+x+2+x+3+x+4+x+5+x+6+x+7+x+8+x+9+x+10 x+11+x+12+x+13+x+14+x+15}{16}=30.5$

first, let's match the denomenators (make them both 1 for ease)

multiply the first eqution by 8 to get

x+x+1+x+2+x+3+x+4+x+5+x+6+x+7=8?

multiply the 2nd equation by 16 to get

x+x+1+x+2+x+3+x+4+x+5+x+6+x+7+x+8+x+9+x+10 x+11+x+12+x+13+x+14+x+15=488

notice that we can try and force the 1st equation into the 2nd equation by adding some numbers

we can already do this:

x+x+1+x+2+x+3+x+4+x+5+x+6+x+7+x+8+x+9+x+10 x+11+x+12+x+13+x+14+x+15=488

[x+x+1+x+2+x+3+x+4+x+5+x+6+x+7]+x+8+x+9+x+10 x+11+x+12+x+13+x+14+x+15=488

[8?]+x+8+x+9+x+10 x+11+x+12+x+13+x+14+x+15=488

we can try to force another 8? into it

[8?]+x+7+1+x+7+2+x+7+3+x+7+4+x+7+5+x+7+6+x+7+7+x+7+8=488

[8?]+(x)+7+(1+x)+7+(2+x)+7+(3+x)+7+(4+x)+7+(5+x)+7+(6+x)+7+(7+x)+7+8=488

[8?]+(x)+7+(x+1)+7+(x+2)+7+(x+3)+7+(x+4)+7+(x+5)+7+(x+6)+7+(x+7)+7+8=488

group

[8?]+[x+x+1+x+2+x+3+x+4+x+5+x+6+x+7]+7+7+7+7+7+7+7+7+8=488

[8?]+[8?]+8(7)+8=488

16?+64=488

minus 64 from both sides

16?=424

divide both sides by 16

?=26.5

the average of the first 8 integers is 26.5

I'm not sure if there is a simpler way to do it or not

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3 years ago

What are the next three terms in the sequence 2, -4, 8, -16, 32, ...?

Advocard [28]

Answer:

Hey there!

The next three terms are: -64, 128, -256.

Let me know if this helps :)

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4 years ago

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Consider the quadratic function:

adoni [48]

Answer:

Step-by-step explanation:

hello :

f(x) = x² – 8x – 9 = (x²-8x+16)-16-9

f(x) = (x-4)²-25 ..... vertex form

the vertex is : (4 ; -25)

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4 years ago

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What is the density of an object with a mass of 20 g and volume of 10 cm3?

Anton [14]

Answer:

2 g/cm^3

Step-by-step explanation:

D=m/V

=20g/10cm^3

=2g/cm^3

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3 years ago