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1 year ago

How many like terms are in the expression:5a^2+6b+a^2-3b-2+4c.

Mathematics

1 answer:

Goryan [66]1 year ago

7 0

Solution

Part 3

For this case we have 6 different terms

Part 4

We can do the following:

$5a^2+a^2+6b-3b+4c-2$

Part 5

And we can simplify on this way:

$6a^2+3b+4c-2$

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Help out please thank you

Helga [31]

Answer:

x = 26

Step-by-step explanation:

The three angles form a straight line, so they will sum to 180

4x - 8 + 34 + 76- x = 180

Combine like terms

3x+ 102 = 180

Subtract 102 from each side

3x+102-102 = 180-102

3x= 78

Divide each side by 3

3x/3 = 78/3

x = 26

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3 years ago

Find the volume of the square pyramid below

Snowcat [4.5K]

Answer:

Step-by-step explanation:

The volume (V) of a pyramid is

V = $\frac{1}{3}$ area of base × perpendicular height (h)

area of square base = 4² = 16 and h = 4, hence

V = $\frac{1}{3}$ × 16 × 4 = $\frac{64}{3}$ = 21 $\frac{1}{3}$ ft³

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2 years ago

Math: Slope help please

pishuonlain [190]

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2 years ago

Read 2 more answers

8 x 6 = (4 x 6) (__x 6) = 24 + __ =

guajiro [1.7K]

Answer:

8x6= (4x6) + (4x6)

= 24+24

= 48

Step-by-step explanation:

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3 years ago

Read 2 more answers

Find T5(x) : Taylor polynomial of degree 5 of the function f(x)=cos(x) at a=0 . (You need to enter function.) T5(x)= Find all va

Burka [1]

Answer:

$\bf cos(x)\approx1-\displaystyle\frac{x^2}{2}+\displaystyle\frac{x^4}{4!}=\\\\=1-\displaystyle\frac{x^2}{2}+\displaystyle\frac{x^4}{24}$

The polynomial is an approximation with an error less than or equals to 0.002652 for x in the interval

[-1.113826815, 1.113826815]

Step-by-step explanation:

According to Taylor's theorem

$\bf f(x)=f(0)+f'(0)x+f''(0)\displaystyle\frac{x^2}{2}+f^{(3)}(0)\displaystyle\frac{x^3}{3!}+f^{(4)}(0)\displaystyle\frac{x^4}{4!}+f^{(5)}(0)\displaystyle\frac{x^5}{5!}+R_6(x)$

with

$\bf R_6(x)=f^{(6)}(c)\displaystyle\frac{x^6}{6!}$

for some c in the interval (-x, x)

In the particular case f

f(x)=cos(x)

we have

$\bf f'(x)=-sin(x)\\f''(x)=-cos(x)\\f^{(3)}(x)=sin(x)\\f^{(4)}(x)=cos(x)\\f^{(5)}(x)=-sin(x)\\f^{(6)}(x)=-cos(x)$

therefore

$\bf f'(x)=-sin(0)=0\\f''(0)=-cos(0)=-1\\f^{(3)}(0)=sin(0)=0\\f^{(4)}(0)=cos(0)=1\\f^{(5)}(0)=-sin(0)=0$

and the polynomial approximation of T5(x) of cos(x) would be

$\bf cos(x)\approx1-\displaystyle\frac{x^2}{2}+\displaystyle\frac{x^4}{4!}=\\\\=1-\displaystyle\frac{x^2}{2}+\displaystyle\frac{x^4}{24}$

In order to find all the values of x for which this approximation is within 0.002652 of the right answer, we notice that

$\bf R_6(x)=-cos(c)\displaystyle\frac{x^6}{6!}$

for some c in (-x,x). So

$\bf |R_6(x)|\leq|\displaystyle\frac{x^6}{6!}|=\displaystyle\frac{|x|^6}{6!}$

and we must find the values of x for which

$\bf \displaystyle\frac{|x|^6}{6!}\leq0.002652$

Working this inequality out, we find

$\bf \displaystyle\frac{|x|^6}{6!}\leq0.002652\Rightarrow |x|^6\leq1.90944\Rightarrow\\\\\Rightarrow |x|\leq\sqrt[6]{1.90944}\Rightarrow |x|\leq1.113826815$

Therefore the polynomial is an approximation with an error less than or equals to 0.002652 for x in the interval

[-1.113826815, 1.113826815]

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2 years ago