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2 years ago

9

Brian is wrapping a present. If his box is in the shape of a cube with a side length of 7 inches, what is the minimum amount of

paper Brian would need to cover the present?

1 answer:

chubhunter [2.5K]2 years ago

5 0

Surface area is 6s^2, so 6(7^2) is 294. The minimum amount would be 294 inches^2 of wrapper paper.

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Kindly solve this question 6 too !!

Shkiper50 [21]

Answer:

$x=1$

Step-by-step explanation:

$~~~~~x = \dfrac{1}{2 - \dfrac 1{2-\dfrac{1}{2-x}}}\\\\\\\implies x = \dfrac{1}{2 - \dfrac 1{\dfrac{4-2x-1}{2-x}}}\\\\\\\implies x= \dfrac{1}{2 - \dfrac{2-x}{3-2x}}\\\\\\\implies x = \dfrac{1}{\dfrac{6-4x-2+x}{3-2x}}\\\\\\\implies x = \dfrac{3-2x}{4-3x}\\\\\\\implies 4x -3x^2 = 3-2x\\\\\\\implies 3x^2 -4x +3-2x = 0\\\\\\\implies 3x^2 -6x +3 = 0\\\\\\\implies 3(x^2 -2x +1) =0\\\\\\\implies x^2 -2x +1 = 0\\\\\\\implies (x-1)^2 = 0\\\\\\\implies x -1 = 0\\\\\\\implies x = 1$

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2 years ago

Do these numbers make a function why or why not?

Paladinen [302]

Answer:

No, because it has repeating domains (x-values) which is -2

Step-by-step explanation:

A function should not have repeating x-values.

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2 years ago

Josie buys a scoop of dried cranberries and a scoop of dried bananas to make trail mix. The

iris [78.8K]

Answer:

4.65

Step-by-step explanation:

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2 years ago

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I need to figure out if I'm doing this correctly

kogti [31]

To prove that triangles TRS and SUT are congruent we can follow these statements:

1.- SR is perpendicular to RT: Given

2.-TU is perpendicular to US: Given

3.-Angle STR is congruent with angle TSU: Given.

4.-Reflexive property over ST: ST is congruent with itself (ST = ST)

From here, we can see that both triangles TRS and SUT have one angle of 90 degrees, another angle that they both have, and also they share one side (ST) ,then:

5.- By the ASA postulate (angle side angle), triangles TRS and SUT are congruent

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1 year ago

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 $\triangle SOP$ .

The Pythagorean Theorem states that the following is true for any right triangle:

$a^2+b^2=c^2$ , where $c$ is the hypotenuse, or the longest side, of the triangle

Thus, we have:

$10^2+TP^2=26^2,\\TP^2=26^2-10^2,\\TP^2=\sqrt{576},\\TP=24$

Since both TP and SP are tangents of the circle and extend to the same point P, they will be equal.

What we know:

$TP=SP=24$
$TO=SO=10$

Thus, the perimeter of the quadrilateral PTOS is equal to $24+24+10+10=\boxed{\text{A) }68\:\mathrm{cm}}$

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3 years ago

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