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3 years ago

Round 42395 to the nearest ten thousand

Mathematics

1 answer:

RUDIKE [14]3 years ago

3 0

Answer:

40000

Step-by-step explanation:

4(2)395

if it smaller than 5 round down

if its lager than 5 round up

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Graph Y= 1/3x {If someone could help me on this I would really appreciate it}

gizmo_the_mogwai [7]

Y=1/3x while m=0................

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7 0

3 years ago

Read 2 more answers

2x − y = −4→10x − 5y = −20 3x 5y = 59→3x 5y = 59 13x = 39 which equation can replace 3x 5y = 59 in the original system and still

malfutka [58]

13x = 39 can replace 3x + 5y = 59 in the original system and still produce the same solution.

8 0

3 years ago

X-5y-z=-20 -2x-y-6z=25 -x-4y-z=-15 Solve by elimination

Tju [1.3M]

Answer:

x = 0

y = 5

z = -5.

Step-by-step explanation:

X-5y-z=-20

-2x-y-6z=25

-x-4y-z=-15

Add equation 1 and 3:

-9y - 2z = -35 .......... A

Multiply the first equation by 2:

2x - 10y - 2z = -40

Now add this to equation 2:

-11y -8z = -15.............B

Now multiply equation A by -4:

36y + 8x = 140.........C

Add equations B and C:

25y = 125

y = 5.

Substitute for y in equation A:

-9*5 - 2z = -35

2z = -45 + 35 = - 10

z = -5.

Now find x by substituting for y and z in the first equation:

x - 5(5) - (-5) = -20

x - 20 = -20

x = 0.

6 0

3 years ago

A student factors a6 - 64 to (a2 - 4)(a4 + 4a2 + 16).Which statement about (a2 − 4)(a4 + 4a2 + 16) is correct?

IrinaK [193]

Answer:

Option (b) is correct.

The expression is equivalent, but the term is not completely factored.

Step-by-step explanation:

Given : a student factors to

We have to choose the correct statement about from the given options.

Given is factored to

Consider

Using algebraic identity,

comparing and b = 4, we have,

Thus, the factorization is equivalent but we can simplify it further also, as

Using algebraic identity,

Thus,

Can be written as

Thus, the expression is equivalent, but the term is not completely factored.

Option (b) is correct.

4 0

2 years ago

.. Which of the following are the coordinates of the vertices of the following square with sides of length a?

atroni [7]

Option A: $O(0,0), S(0,a), T(a,a), W(a,0)$

Option D: $O(0,0), S(a,0), T(a,a), W(0,a)$

Step-by-step explanation:

Option A: $O(0,0), S(0,a), T(a,a), W(a,0)$

To find the sides of a square, let us use the distance formula,

$d=\sqrt{\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}}$

Now, we shall find the length of the square,

$\begin{array}{l}{\text { Length } O S=\sqrt{(0-0)^{2}+(a-0)^{2}}=\sqrt{a^{2}}=a} \\{\text { Length } S T=\sqrt{(a-0)^{2}+(a-a)^{2}}=\sqrt{a^{2}}=a} \\{\text { Length } T W=\sqrt{(a-a)^{2}+(0-a)^{2}}=\sqrt{a^{2}}=a} \\{\text { Length } O W=\sqrt{(a-0)^{2}+(0-0)^{2}}=\sqrt{a^{2}}=a}\end{array}$

Thus, the square with vertices $O(0,0), S(0,a), T(a,a), W(a,0)$ has sides of length a.

Option B: $O(0,0), S(0,a), T(2a,2a), W(a,0)$

Now, we shall find the length of the square,

$\begin{aligned}&\text { Length } O S=\sqrt{(0-0)^{2}+(a-0)^{2}}=\sqrt{a^{2}}=a\\&\text {Length } S T=\sqrt{(2 a-0)^{2}+(2 a-a)^{2}}=\sqrt{5 a^{2}}=a \sqrt{5}\\&\text {Length } T W=\sqrt{(a-2 a)^{2}+(0-2 a)^{2}}=\sqrt{2 a^{2}}=a \sqrt{2}\\&\text {Length } O W=\sqrt{(a-0)^{2}+(0-0)^{2}}=\sqrt{a^{2}}=a\end{aligned}$

This is not a square because the lengths are not equal.

Option C: $O(0,0), S(0,2a), T(2a,2a), W(2a,0)$

Now, we shall find the length of the square,

$\begin{array}{l}{\text { Length OS }=\sqrt{(0-0)^{2}+(2 a-0)^{2}}=\sqrt{4 a^{2}}=2 a} \\{\text { Length } S T=\sqrt{(2 a-0)^{2}+(2 a-2 a)^{2}}=\sqrt{4 a^{2}}=2 a} \\{\text { Length } T W=\sqrt{(2 a-2 a)^{2}+(0-2 a)^{2}}=\sqrt{4 a^{2}}=2 a} \\{\text { Length } O W=\sqrt{(2 a-0)^{2}+(0-0)^{2}}=\sqrt{4 a^{2}}=2 a}\end{array}$

Thus, the square with vertices $O(0,0), S(0,2a), T(2a,2a), W(2a,0)$ has sides of length 2a.

Option D: $O(0,0), S(a,0), T(a,a), W(0,a)$

Now, we shall find the length of the square,

$\begin{aligned}&\text { Length OS }=\sqrt{(a-0)^{2}+(0-0)^{2}}=\sqrt{a^{2}}=a\\&\text { Length } S T=\sqrt{(a-a)^{2}+(a-0)^{2}}=\sqrt{a^{2}}=a\\&\text { Length } T W=\sqrt{(0-a)^{2}+(a-a)^{2}}=\sqrt{a^{2}}=a\\&\text { Length } O W=\sqrt{(0-0)^{2}+(a-0)^{2}}=\sqrt{a^{2}}=a\end{aligned}$

Thus, the square with vertices $O(0,0), S(a,0), T(a,a), W(0,a)$ has sides of length a.

Thus, the correct answers are option a and option d.

8 0

3 years ago