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3 years ago

What statement describes the decimal equivalent of 3/ 8 ?

Mathematics

1 answer:

HACTEHA [7]3 years ago

5 0

Answer:

0.375

Step-by-step explanation: All you need to do is divide 3 and 8

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Which equation represents the hyperbola in general form?

valina [46]

ANSWER

$9 {y}^{2} - 4 {x}^{2}- 36y + 16x - 16 = 0$

EXPLANATION

The standard equation of the hyperbola is

$\frac{ {(y - 2)}^{2} }{4} - \frac{ {(x - 2)}^{2} }{9} = 1$

We multiply through by 36 to obtain:

$9 {(y - 2)}^{2} - 4( {x - 2)}^{2} = 36$

We now expand to get,

$9( {y}^{2} - 4y + 4) - 4( {x}^{2} - 4x + 4) = 36$

Expand :

$9 {y}^{2} - 36y + 36 - 4 {x}^{2} + 16x - 16 = 36$

To get the general form, we equate everything to zero to get,

$9 {y}^{2} - 4 {x}^{2}- 36y + 16x - 16 = 0$

The correct choice is D.

7 0

3 years ago

Whats is the difference between a factor and multiple

Deffense [45]

Answer:

<h3> a factor is a number that divides another number without leaving any remainder. , A multiple is a number that, when divided a certain of times by another number, no remainder is left.</h3>

Step-by-step explanation:

6 0

2 years ago

Read 2 more answers

Use estimation to determine which of the following expressions is greater. Explain how you got your answer. 5 × (1.9 + 3.15) and

Sonja [21]

The first one is greater. Adding 3.2 and 0.2 is easy! It's 3.4! Make that 3. Make 6.9, 7. 7×3=21. For the first one, make 1.9, 2. Then Make 3.15, 3. 5×5=25. That is why the first one is greater.

5 0

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 <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

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

Find the base of a parallelogram with the area of 60 in.² and height of 4 inches use the formula for the area of the parallelogr

Alexxx [7]

A = bh
60 in.² = b(4)
60/4 = b
b = 15 inches

Hope this helped!

7 0

3 years ago