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3 years ago

876.34 rounded to the nearest thousandth

Mathematics

2 answers:

kogti [31]3 years ago

3 0

Answer:

There is no thousands place. It only goes up to a hundred

Step-by-step explanation:

You need another number after 4 to be able to round thousands

just olya [345]3 years ago

3 0

Answer:

876.340

Step-by-step explanation:

The thousandth is in the third place from the right of the decimal point. There's no figure on there, to the nearest thousandth, just add zero after the 4 hundredth

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Use long division to find the quotient of 990 and 45.

solmaris [256]

Answer:

Step-by-step explanation:

because if you divide 45 first it will be wrong so you must always first will the biggest number in your problem

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3 years ago

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Find the volume of the rectangular prism. Express your answer as a simplified mixed number.

OverLord2011 [107]

Answer:

Volume = = 138⁶/32 cubic meters

Step-by-step explanation:

Given:

Length = 8⅜ m

Width = 2 m

Height = 8¼ m

Required:

Volume as a mixed number

Solution:

Volume of rectangular prism = L*W*H

Plug in the values

V = 8⅜ × 2 × 8¼

Change to improper fraction

V = 67/8 × 2 × 33/4

= (67 × 2 × 33)/(8 × 4)

= 4,422/32

= 138⁶/32 m³

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3 years ago

Plz help with me math questions

Ludmilka [50]

Question five

D is correct

10x-35

Question six

These are true

4(4y+7)

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8(2y+4) - 4

2y+7(2y+4)

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3 years ago

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Find the largest interval which includes x = 0 for which the given initial-value problem has a unique solution. (Enter your answ

Ivan

Answer:

$(-\infty,3)$

Step-by-step explanation:

We are given that

$(x-3)y''+4y=x$

$y''+\frac{4}{x-3}y=\frac{x}{x-3}$

y(0)=0

y'(0)=1

By comparing with

$y''+p(x)y'+q(x)y=g(x)$

We get

$p(x)=\frac{4}{x-3}$

$g(x)=\frac{x}{x-3}$

q(x)=0

p(x),q(x) and g(x) are continuous for all real values of x except 3.

Interval on which p(x),q(x) and g(x) are continuous

$(-\infty,3)$ and (3, $\infty)$

By unique existence theorem

Largest interval which contains 0= $(-\infty,3)$

Hence, the larges interval on which includes x=0 for which given initial value problem has unique solution= $(-\infty,3)$

7 0

3 years ago

Question 4

astraxan [27]

Answer:

No real roots

Step-by-step explanation:

The given equations are:

$3x-1+2y=0$

$3x^2-y^2+4=0$

We make y the subject in the first equation to get

$y=\frac{1-3x}{2}$

We substitute into the second expression to get:

$3x^2-(\frac{1-3x}{2})^2+4=0$

We expand to get:

$3x^2-\frac{(1-6x+9x^2)}{4}+4=0$

Multiply through by 4 to get:

$12x^2-1+6x-9x^2+16=0$

$3x^2+6x+15=0$

The discriminant is given by $D=b^2-4ac$

$D=3^2-4*12*15$

$D=-711$

Since the discriminant is less than zero, the two curves never intersects.

Therefore the system has no real roots

6 0

4 years ago