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3 years ago

Age If you add Natalie's age and Fred's age, the result is 43. If you add Fred's age to 4 times

1 answer:

8 0

Let’s take fred age as x and Natalie’s age as y.
So,
A/Q
X + Y = 43. Let it be eqution 1
and
X + 4Y = 79. Let it be equation 2

Combining equation 1 and 2, we get

( X + Y) + 3Y = 79
43 + 3Y = 79
3Y = 36
Y = 12

Now, evaluating the value of Y in equation 1 we get,

X + 12 = 43
X = 31

Therefore, Fred is 31years old and Natalie is 12 years old.

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a crew is made up of 20 men; the rest are women. 66 2/3% of the crew are men. How many people are in the crew? Show the work plz

liberstina [14]

20 men

66 2/3% men

66 2/3% of 30= 20

30+20= 50

Hope this helps!

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3 years ago

One should continue to monitor his or her goals and adjust his or her plan for achieving them when it is necessary A:true B:fals

kupik [55]

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3 years ago

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What is the direct variation between x and y when x=10 and y=250

ioda

You can use equations to describe the relationship between the two numbers.

x+240=y

y-240=x

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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]

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3 years ago

True or false, A function is a relation if and only if the x-values do not repeat in a given set

salantis [7]

Its false because it can't repeat in a set.

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3 years ago

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