755,082 round to nearest ten thousand

Pls help me with this page I used all of my points

Travka [436]

Alright so the first one involves in shading.

You can see that Jack and Renee have two bars with rectangles inside the bar.

Jack took 3/4 hours or in words, three out of four. So in Jack's bar, shade 3 boxes and leave one unshaded.

Renee took 2/3 or two out of three. Shade 2 boxes in Renee's bar and see the difference.

As you can see, Jack has more shade length than Renee and that could only mean one thing:

Jack took longer to fix the tire than Renee.

The symbol that represents this is 3/4 > 2/3

The second activity has the same case.

Water: 3/5 means that you have to shade three boxes and leave the other two unshaded.

Sports Drink: 4/6 means that you have to shade four boxes and leave the other two unshaded.

As you can see again, there is more sports drink than the water since the shade length is longer in the sports drink than on the water.

This meant that she poured in more sports drink than water.

The third activity is just as simple, you just need to visually make up your own sections. Do the shading then see which one is bigger then just input the sign (hint: the first one is < and the second one is >)

cheers

7 0

2 years ago

Q2.The two triangles shown below are congruent by ?

dalvyx [7]

Answer:

cccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccc

Step-by-step explanation:

7 0

3 years ago

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Rachael is 5 ft 6 in tall . She notices that her shadow is 4 ft long and the shadow of a nearby water tower is 100 ft long . Wha

Nimfa-mama [501]

5.5/4=x/100
4x=550
x=137.5

the water tower is 137 ft. 6 in. tall.

3 0

3 years ago

Dy/dx = 2xy^2 and y(-1) = 2 find y(2)

Anarel [89]

If you're using the app, try seeing this answer through your browser: brainly.com/question/2887301

—————

Solve the initial value problem:

dy
——— = 2xy², y = 2, when x = – 1.
dx

Separate the variables in the equation above:

$\mathsf{\dfrac{dy}{y^2}=2x\,dx}\\\\
\mathsf{y^{-2}\,dy=2x\,dx}$

Integrate both sides:

$\mathsf{\displaystyle\int\!y^{-2}\,dy=\int\!2x\,dx}\\\\\\
\mathsf{\dfrac{y^{-2+1}}{-2+1}=2\cdot \dfrac{x^{1+1}}{1+1}+C_1}\\\\\\
\mathsf{\dfrac{y^{-1}}{-1}=\diagup\hspace{-7}2\cdot \dfrac{x^2}{\diagup\hspace{-7}2}+C_1}\\\\\\
\mathsf{-\,\dfrac{1}{y}=x^2+C_1}$

$\mathsf{\dfrac{1}{y}=-(x^2+C_1)}$

Take the reciprocal of both sides, and then you have

$\mathsf{y=-\,\dfrac{1}{x^2+C_1}\qquad\qquad where~C_1~is~a~constant\qquad (i)}$

In order to find the value of C₁ , just plug in the equation above those known values for x and y, then solve it for C₁:

y = 2, when x = – 1. So,

$\mathsf{2=-\,\dfrac{1}{1^2+C_1}}\\\\\\
\mathsf{2=-\,\dfrac{1}{1+C_1}}\\\\\\
\mathsf{-\,\dfrac{1}{2}=1+C_1}\\\\\\
\mathsf{-\,\dfrac{1}{2}-1=C_1}\\\\\\
\mathsf{-\,\dfrac{1}{2}-\dfrac{2}{2}=C_1}$

$\mathsf{C_1=-\,\dfrac{3}{2}}$

Substitute that for C₁ into (i), and you have

$\mathsf{y=-\,\dfrac{1}{x^2-\frac{3}{2}}}\\\\\\
\mathsf{y=-\,\dfrac{1}{x^2-\frac{3}{2}}\cdot \dfrac{2}{2}}\\\\\\
\mathsf{y=-\,\dfrac{2}{2x^2-3}}$

So y(– 2) is

$\mathsf{y\big|_{x=-2}=-\,\dfrac{2}{2\cdot (-2)^2-3}}\\\\\\
\mathsf{y\big|_{x=-2}=-\,\dfrac{2}{2\cdot 4-3}}\\\\\\
\mathsf{y\big|_{x=-2}=-\,\dfrac{2}{8-3}}\\\\\\
\mathsf{y\big|_{x=-2}=-\,\dfrac{2}{5}}\quad\longleftarrow\quad\textsf{this is the answer.}$

I hope this helps. =)

Tags: <em>ordinary differential equation ode integration separable variables initial value problem differential integral calculus</em>

7 0

3 years ago

Answer for<br> -4 (-x + 2) = 2 (3x - 7)

Lana71 [14]

Step-by-step explanation:

-4 (-x + 2) = 2 (3x - 7)

2x-4=3x-7

x=3

5 0

3 years ago

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