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LuckyWell [14K]
3 years ago
9

What is the answer to 7.26 x 106 ft​

Mathematics
1 answer:
Diano4ka-milaya [45]3 years ago
7 0

Answer:

769.56

Step-by-step explanation:

its like doing 726 times 106 then just move the decimal. and i looked it up

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Rewrite the equation below so that it does not have fractions.
lidiya [134]

Answer:

\frac{15}{20} x-2=\frac{8}{20}

Step-by-step explanation:

\frac{3}{4} x-2=\frac{2}{5}

3/4 and 2/5 have a common denominator which is twenty so you can rewrite the equation with just knowing that

\frac{3}{4} =\frac{15}{20}

\frac{2}{5} =\frac{8}{20}

\frac{15}{20} x-2=\frac{8}{20}

You can then multiply both sides by twenty to solve this equation.

Hope this helps.

4 0
3 years ago
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What is 23d=109<br> pls answer
grandymaker [24]

Answer:

what kinda... what do they give yall these days but anyways,

Step-by-step explanation:

Divide each term by

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Exact Form:

d=109/23

Decimal Form:

d=4.73913043

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Mixed Number Form:

d=4 17/23

5 0
3 years ago
George ate 3/8 of a container of food.silvia ate 1/5 from the same container of food.what amount remaining in the contianer
MAVERICK [17]

Amount remaining would be: 1-(3/8+1/5) = 1-23/40 = 17/40

So, 17/40 is your answer.......

7 0
4 years ago
The spinner at the right is used to play a certain game. On your turn, you must spin the spinner twice. How many different combi
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There are over a thousand possibilities for it so the answer is inevitable
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3 years ago
Find the equation of the plane passing through the points
erica [24]

Extract the normal vectors from the given planes:

-2x+y+z+2=0\implies\vec n_1=(-2,1,1)

x+y-3z+1=0\implies\vec n_2=(1,1,-3)

(which are unique up to their signs, meaning either \vec n_1 or -\vec n_1 are valid choices for the normal vector)

The third plane must be perpendicular to both these given planes, which means it would be parallel to both \vec n_1 and \vec n_2, which in turn means its own normal vector \vec n_3 should be perpendicular to both \vec n_1 and \vec n_2.

Enter the cross product:

\vec n_3=\vec n_1\times\vec n_2=(-4,-5,-3)

or (4, 5, 3), which also works.

The given plane passes through (-1, 1, 4), so its equation is

(x+1,y-1,z-4)\cdot\vec n_3=0

Simplify:

(x+1,y-1,z-4)\cdot(4,5,3)=0

4(x+1)+5(y-1)+3(z-4)=0

\boxed{4x+5y+3z=13}

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