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Marrrta [24]
3 years ago
8

Select the GCF of these numbers.

Mathematics
2 answers:
kiruha [24]3 years ago
8 0
ANSWER

{2}^{3}  \times 5



EXPLANATION


We want to find the greatest common factor of

{2}^{5}  \times 5 \times 11

and

{2}^{3}  \times  {5}^{2}  \times7

To find the greatest common factor, we need to identify the factors that are common to both numbers.

Then we pick the factors with the lowest degree and multiply them


This will give us,


{2}^{3}  \times 5

The correct answer is D.
Marysya12 [62]3 years ago
4 0
The Greatest Common Factor (GCF) of 25, 5, 11 and 23, 52, 7 is 1, same with 22,3 and 23, 5 and 13, 193, 232.
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A metal strip is being installed around a workbench that is 5 feet long and 3 feet wide. Find how much stripping is needed.
Ksivusya [100]
<span>The workbench represents a rectangle with two sides 5 feet long each and two sides 3 feet wide each. The perimeter of said rectangle is the sum of the four sides or 5 + 5 + 3 + 3 for a total of 16 feet. therefore the amount of metal tape required is 16 feet.</span>
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3 years ago
Kiran scored 36 baskets in 1 week. She scored 1/2 of the baskets on the weekend. She scored 1/4 of them on Tuesday, she scored 1
notka56 [123]

Answer:

6 baskets did Kiran score on Thursday

Step-by-step explanation:

As per the statement:

Kiran scored 36 baskets in 1 week.

On weekend she scored = \frac{1}{2} \cdot 36 = 18 baskets.

She scored 1/4 of them on Tuesday

⇒On Tuesday she scored = \frac{1}{4} \cdot 36 = 9 baskets

She scored 1/12 of them on Wednesday

⇒On Wednesday she scored = \frac{1}{12} \cdot 36 = 3 baskets.

Then;

She scored = On Weekend + On Tuesday + On Wednesday = 18+9+3 = 30 baskets.

Rest on Thursday she scored = 36 - 30 = 6 baskets.

Therefore, 6 baskets did Kiran score on Thursday

8 0
3 years ago
Arrange the hyperbolas in increasing order of the horizontal widths of their asymptote rectangles.
almond37 [142]

The  increasing order of the horizontal widths of their asymptote rectangles is dependent on the values gotten from  y = ± x.

<h3>What is a Hyperbola?</h3>

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The rectangular hyperbola has two asymptotes which are defined as y = ± x in this scenario.

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2 years ago
Can someone help me with this math homework please!
Korolek [52]

Answer:

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Step-by-step explanation:

4 0
2 years ago
Read 2 more answers
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morpeh [17]

Answer:

distance island dock to Dock A = 4.99 km

distance island dock to Dock K = 6.35 km

Step-by-step explanation:

Always make a scetch to visualize the situation.

You need to construct two triangle both with a streight angle, so you can use Pythagoras to calculate the unknown distances between the island dock L, and each of the other two docks A an K.

I chose to introduce an extra letter, the letter C. In total you have the letters A K L and the letter C.

The letter C has a streight angle of 90° between ACL and it has the same streight angle of 90° with KCL. It is crucial that you see that the distance of LC is exactly the same in triangle LAC and that LC has exactly the same distance in the other triangleLKC.

The distance between AK = 2.3 km.

I define the distance between K and point C as 2.3 + x, because the distance x is unknown.

KC = 2.3 + x

Further more, when you make a picture, you can see that the distance between A and point C = x.

From such a picture, it would show clearly, that K is further away in respect to L then point A. From the picture it would be clear that the angle of LKC is smaller then the angle of LAC, so LKC = 45° and LAC = 64°.

Because angle LKC = 45° and we choose C to have an angle of 90°, the TRIANGLE LKC must be a special triangle... In any triangle, the sum of the three angles together, must add up to 180° .

If that is true, then we have 45 + 90 + 45 (because that adds up to 180). Now that means triangle LKC must have two equal sides (because of the same angels of 45° ).

So we know the distance KC = LC and we already defined KC = 2.3 + x.

Now we know enough to solve the problem.

AK = 2.3 km

angle of LKC = 45°

angle of LAC = 64°

AC = x

KC = 2.3 + x

LC = KC

LC = 2.3 + x

Try to calculate the distance x by using tan. After that you can use Pythagoras to find the other distances.

tan(LKC) = ( LC ) / ( KC )

tan(LKC) = ( x+2.3 ) / ( x+2.3 )

That is not helpful. Let's try the other triangle...

tan(LAC) = LC / AC

tan(LAC) = ( x+2.3 ) / x

tan(64) = ( x+2.3 ) / x

Solve the equation which means you try to find the value for x.

x * tan(64) = ( x+2.3 )

tan(64) * x -x = 2.3

tan(64) * x - 1* x = 2.3

Try to get x outside of the braquets...

x* ( tan(64) - 1 ) = 2.3

x* (2.0503038415793 - 1 ) = 2.3

1.0503038415793 * x = 2.3

x = 2.3 / 1.0503038415793

x = 2.19

Now use Pythagoras a² + b² = c² in triangle LAC to find distance LA.

LA² = AC² + LC²

AC = x = 2.19

LC = 2.3 + x = 4.39

LA² = 2.19² + 4.39²

LA = SQRT( 4.79 + 20.16 )

LA = SQRT( 24.95 )

LA = 4.99 km

Now use Pythagoras a² + b² = c² in triangle LKC to find distance LK.

LK² = KC² + LC²

KC = 2.3 + x = 4.39

LC = 2.3 + x = 4.39

LK² = 4.39² + 4.39²

LK = SQRT( 20.16 + 20.16 )

LK = SQRT( 40.32 )

LK = 6.35 km

7 0
3 years ago
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