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3 years ago

12

Round 751446 to the nearest hundred thousand

1 answer:

exis [7]3 years ago

7 0

800000 becuase the 5 we can round that up

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Please answer correctly !!!!!!!!!!!!! Will mark Brianliest !!!!!!!!!!!!!!!

kramer

Answer:

MN = 4

m∠MLN = 42°

Step-by-step explanation:

If ΔMLN is congruent to ΔYZX, then

ML ↔ YZ MN ↔ YX LN ↔ ZX ∠M ↔ ∠Y ∠L ↔ ∠Z ∠N ↔ ∠X

and all those correspondences are congruent when the triangles are congruent

5 0

2 years ago

Show that the following rectangular form x^2+y^2=2xy can be expressed as 1 = sin(2) in polar form.

lara [203]

Answer:

Step-by-step explanation:

substitute x = r*cos(θ), y = r*sin(θ) ==> r²(cos²(θ) + sin²(θ)) = 2r²cos(θ)sin(θ). Cancel the r² on both sides. On the left, use pythagorean identity cos²(θ) + sin²(θ) = 1. On the right apply double angle identity sin(2θ) = 2cos(θ)sin(θ).

This yields 1=sin(2θ). (I assume you meant to type sin(2θ) on the right hand side of the equation).

7 0

2 years ago

What is the answer to 64+(-15)+(-5)

MissTica

Answer:

44

Step-by-step explanation:

8 0

3 years ago

Read 2 more answers

-8(1+8n)-8(6-4n)=-24

Naily [24]

-8(1 + 8n) - 8(6 - 4n) = -24

-8 - 64n - 48 + 32n = -24

-32n - 56 = -24

-32n = 32

n = -1

3 0

3 years ago

For the following linear system, put the augmented coefficient matrix into reduced row-echelon form.

Anni [7]

Answer:

The reduced row-echelon form of the linear system is $\left[\begin{array}{cccc}1&0&-5&0\\0&1&3&0\\0&0&0&1\end{array}\right]$

Step-by-step explanation:

We will solve the original system of linear equations by performing a sequence of the following elementary row operations on the augmented matrix:

Interchange two rows
Multiply one row by a nonzero number
Add a multiple of one row to a different row

To find the reduced row-echelon form of this augmented matrix

$\left[\begin{array}{cccc}2&3&-1&14\\1&2&1&4\\5&9&2&7\end{array}\right]$

You need to follow these steps:

Divide row 1 by 2 $\left(R_1=\frac{R_1}{2}\right)$

$\left[\begin{array}{cccc}1&3/2&-1/2&7\\1&2&1&4\\5&9&2&7\end{array}\right]$

Subtract row 1 from row 2 $\left(R_2=R_2-R_1\right)$

$\left[\begin{array}{cccc}1&3/2&-1/2&7\\0&1/2&3/2&-3\\5&9&2&7\end{array}\right]$

Subtract row 1 multiplied by 5 from row 3 $\left(R_3=R_3-\left(5\right)R_1\right)$

$\left[\begin{array}{cccc}1&3/2&-1/2&7\\0&1/2&3/2&-3\\0&3/9&9/2&-28\end{array}\right]$

Subtract row 2 multiplied by 3 from row 1 $\left(R_1=R_1-\left(3\right)R_2\right)$

$\left[\begin{array}{cccc}1&0&-5&16\\0&1/2&3/2&-3\\0&3/9&9/2&-28\end{array}\right]$

Subtract row 2 multiplied by 3 from row 3 $\left(R_3=R_3-\left(3\right)R_2\right)$

$\left[\begin{array}{cccc}1&0&-5&16\\0&1/2&3/2&-3\\0&0&0&-19\end{array}\right]$

Multiply row 2 by 2 $\left(R_2=\left(2\right)R_2\right)$

$\left[\begin{array}{cccc}1&0&-5&16\\0&2&3&-6\\0&0&0&-19\end{array}\right]$

Divide row 3 by −19 $\left(R_3=\frac{R_3}{-19}\right)$

$\left[\begin{array}{cccc}1&0&-5&16\\0&2&3&-6\\0&0&0&1\end{array}\right]$

Subtract row 3 multiplied by 16 from row 1 $\left(R_1=R_1-\left(16\right)R_3\right)$

$\left[\begin{array}{cccc}1&0&-5&0\\0&1&3&-6\\0&0&0&1\end{array}\right]$

Add row 3 multiplied by 6 to row 2 $\left(R_2=R_2+\left(6\right)R_3\right)$

$\left[\begin{array}{cccc}1&0&-5&0\\0&1&3&0\\0&0&0&1\end{array}\right]$

8 0

3 years ago