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ipn [44]
2 years ago
14

Use the balanced scale to find the conversion factor that can be used to convert the number of blocks to the weight of the block

s in pounds.
Balanced scale has 3 one-pound weights on 1 side; 2 question mark weights on other side.



A. 1.5 pounds per block



B. 1.5 blocks per pound



C. 2 pounds per block



D. 3 blocks per 2 pounds
Mathematics
2 answers:
Sedbober [7]2 years ago
8 0

Answer:

A. 1.5 pounds per block

Step-by-step explanation:

The conversion factor shows 2 blocks on the right and three 1 pound blocks on the left. It takes two 1 pound blocks to make 2 and then you get your 3rd block, making it 3 (.5 more than 2.5) needing another block to add a .5

nignag [31]2 years ago
5 0

Answer:

Step-by-step explanation:

1.5

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bija089 [108]

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Step-by-step explanation:

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2 years ago
A student wishes to increase an amount by 10% and then by 30%. What is the single multiplier that can be used?
liraira [26]

Step-by-step explanation:

let an amount be a

increase an amount by 10% and then by 30%

= a*1.1 *1.3 = a*1.43

7 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 <em>0.002652</em> 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

<em>f(x)=cos(x) </em>

<em> </em>

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
Nathan buys 8 bottles of iced tea at the grocery store. He has coupons for $0.35 off the regular price of each bottle. After usi
solniwko [45]

Answer:

$1.25

Step-by-step explanation:

0.35 x 8 = 2.80

7.20 + 2.80 = 10

10 divided by 8 = 1.25

Thank you :)

3 0
2 years ago
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