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3 years ago

How can postulates and theromes relating to similar and congruent triangles be used to write a proof

1 answer:

7 0

As per this theorem the two triangles are congruent if two angles and a side not between these two angles of one triangle are congruent to two corresponding angles and the corresponding side not between the angles of the other triangle.

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Simplify the expression ASAP 8p3

Reika [66]

Answer:

336

Step-by-step explanation:

8P3

=8!(8-3)!

=8!5!

=8x7x6

=336

5 0

3 years ago

The mass of an ant is 1/10 of the load, which it can carry in one try. What is the mass of the ant, if in one time it can carry

Viefleur [7K]

If you call $m$ the mass of the ant and $l$ the load, we have the equation

$m = \cfrac{l}{10}$

In fact, the mass of the ant is one tenth of the load, which is exactly what this equation states.

Since we are given the load, we simply need to plug its value in the equation to deduce the mass of the ant:

$m = \cfrac{\frac{7}{250}}{10} = \cfrac{7}{250}\cdot\cfrac{1}{10} = \cfrac{7}{2500}$

8 0

3 years ago

Identify one way to rewrite the expression using the Distributive Property. 2(9 + 3) A. (2 × 9) − (2 × 3) B. (2 × 9) + (9 × 3) C

Dmitry_Shevchenko [17]

The answer is D.

Because you want to find the rewrite way of distributive right? 2(9+3)

Therefore it’s (2x9) + (2x3)

3 0

3 years ago

Please solve this, will rate 5 stars and mark as STAR!

Nina [5.8K]

Answer:

$\boxed{5 \cdot \sqrt{2} \cdot \sqrt[6]{5} }$

Step-by-step explanation:

$\sqrt[3]{250} \cdot \sqrt{\sqrt[3]{10} }$

$\sqrt{\sqrt[3]{10} } \implies (10^\frac{1}{3} )^\frac{1}{2} =10^\frac{1}{6} =\sqrt[6]{10}$

$\therefore \sqrt{\sqrt[3]{10} }=\sqrt[6]{10}$

$\text{Solving }\sqrt[3]{250} \cdot \sqrt{\sqrt[3]{10} }$

$250=2 \cdot 5^3$

$\sqrt[3]{250}=\sqrt[3]{2\cdot 5^3}=5 \sqrt[3]{2}$

Once

$\sqrt[6]{2} \cdot \sqrt[6]{5} = \sqrt[6]{10}$

We have

$5 \sqrt[3]{2} \cdot \sqrt[6]{2} \cdot \sqrt[6]{5}$

We can proceed considering the common base of exponentials

$\sqrt[3]{2} \cdot \sqrt[6]{2} = 2^{\frac{1}{3}} \cdot 2^{\frac{1}{6} } = 2^{\frac{3}{6} } = 2^{\frac{1}{2} }=\sqrt{2}$

Therefore,

$5 \sqrt[3]{2} \cdot \sqrt[6]{2} \cdot \sqrt[6]{5} = 5 \cdot \sqrt{2} \cdot \sqrt[6]{5}$

7 0

3 years ago

Can anyone help me on this problem? Please it doesn’t make sense to me...

Sav [38]

Answer:

Step-by-step explanation:

Cost per item is found by dividing the cost by the number of items. If the woman bought n items for $120, the cost of each item is $120/n. If the woman bought 24 more items, n+24, at the same price, then the cost per item is $120/(n+24). The problem statement tells us this last cost is $16 less than the first cost:

120/(n+24) = (120/n) -16

Multiplying by n(n+24) gives ...

120n = 120(n+24) -16(n)(n+24)

0 = 120·24 -16n^2 -16·24n . . . . . . subtract 120n and collect terms

n^2 +24n -180 = 0 . . . . . . . . . . . . . divide by -16 to make the numbers smaller

(n +30)(n -6) = 0 . . . . . . . . . . . . . . factor the quadratic

The solutions to this are the values of n that make the factors zero: n = -30, n = 6. The negative value of n has no meaning in this context, so n=6 is the solution to the equation.

The woman bought 6 items.

_____

Check

When the woman bought 6 items for $120, she paid $120/6 = $20 for each of them. If she bought 6+24 = 30 items for the same money, she would pay $120/30 = $4 for each item. That amount, $4, is $16 less than the $20 she paid for each item.

3 0

3 years ago