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3 years ago

If Babe Ruth average 46 home runs per year, how many home runs did he hit in 12 years?

Mathematics

2 answers:

lara [203]3 years ago

5 0

12x46 = 552 home runs

lilavasa [31]3 years ago

3 0

552 home runs or more if he hit above average and less if he hit below average

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The cafeteria manager took a survey of the third grade students and found that 83 of 127 students said they did not want hot dog

Natalka [10]

The answer is 588 students , because you must make proportions.

Shown work below
You're welcome ;)

6 0

3 years ago

Read 2 more answers

Sewing patterns recommended buying extra fabric in case of mistakes. Darlene bought 7 yards of fabric, plus 20% extra fabric. Ch

Debora [2.8K]

Answer:

Darlene bought more fabric

Darlene bought 3.4 yards more

Step-by-step explanation:

Darlene = 7 yards + 20% extra

= 7 yards + (20% of 7 yards)

= 7 + (20% × 7)

= 7 + (0.2 × 7)

= 7 + 1.4

= 8.4 yards

Chris = 4 yards + 25% of 4 yards

= 4 yards + (25% of 4 yards)

= 4 + (25% × 4)

= 4 + (0.25 × 4)

= 4 + 1

= 5 yards

Darlene bought more fabric

How much more?

Darlene - Chris

= 8.4 yards - 5 yards

= 3.4 yards

Darlene bought 3.4 yards more

6 0

3 years ago

Explain how you can use base 10 blocks to find one. 54+2.37

jasenka [17]

So you mean add 54+2.37? Well, I do know that the answer to 53+2.37= 2.90. I hope I helped, even if I only helped by an inch.

6 0

3 years ago

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]

8 0

3 years ago

Please awnser asap Factor out the trinomial 10x^2+3x-1?

g100num [7]

Answer- (5x-1)(2x+1)

let me know if you need to see work

8 0

3 years ago