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3 years ago

J^2 over 2j times 2j over 3g

Mathematics

1 answer:

Serhud [2]3 years ago

5 0

Answer:

Step-by-step explanation:

given $\frac{j^2}{2j}$ × $\frac{2j}{3g}$

the 2j on the denominator of the first fraction cancels with the 2j on the numerator of the second fraction, leaving the expression in simplified form as

= $\frac{j^2}{3g}$ → D

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How do i solve 2(x+4)=5x-2

IceJOKER [234]

=2(5x2+20x-1)
The 2 by the x has to be on top of the x

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3 years ago

Julia is going to the store to buy candies. Small candies cost $4 and extra-large candies cost $12.She needs to purchase at leas

BartSMP [9]

Answer:

Small candies $=9$

Extra large candies $=12$

Step-by-step explanation:

Let small candies $=x$

Extra large candies $=y$

the number of candies is at least $20$ .

$x+y\geq20$

Cost of $1$ small candy $=\$4$

Cost of $1$ extra large candy $=\$12$

but she has only $\$180$ to spend

$4x+12y\leq180$

Solve for

$x+y=20.......(1)\\4x+12y=180.....(2)\\eqn(2)-eqn(1)\times4\\8y=100\\y=\frac{100}{8} \\y=\frac{25}{8} \\from\ eqn(1)\\x+\frac{25}{2}=20\\ x=20-\frac{25}{2} \\x=\frac{15}{2}$

Since number of candies should be integer.

let $x=7,y=13$

total spend $4\times7+12\times13=184$ which is more than $\$180$ , so this combination is not possible.

$let\ x=8,y=12\\8\times4+12\times12=176$

She has $\$4$ more so she can buy $1$ more small candy.

Hence small candy $=9$

extra large candy $=12$

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3 years ago

Which is the graph of f(x) = x2 - 2X + 3?

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3 years ago

Read 2 more answers

Question and choices are in the photo please explain the answer

Natasha_Volkova [10]

Answer:

$\text{A) }68\:\mathrm{cm}$

Step-by-step explanation:

The perimeter of a polygon is equal to the sum of all the sides of the polygon. Quadrilateral PTOS consists of sides TP, SP, TO, and SO.

Since TO and SO are both radii of the circle, they must be equal. Thus, since TO is given as 10 cm, SO will also be 10 cm.

To find TP and SP, we can use the Pythagorean Theorem. Since they are tangents, they intersect the circle at a $90^{\circ}$ , creating right triangles $\triangle TOP$ and $\triangle SOP$ .