You are given that angle RST and angle TUR are right angles. What additional piece of info allows you to prove they are congruent

Answers

Answer 1

Answer:

Answer:

the additional info required is - the hypotenuse and one leg of one right-angled triangle are equal to the corresponding hypotenuse and leg of another right-angled triangle.

Step-by-step explanation:

As given , RST and angle TUR are right angles.

So the triangles are right angle triangles.

If the hypotenuse and one leg of one right-angled triangle are equal to the corresponding hypotenuse and leg of another right-angled triangle, the two triangles are congruent.

So, the additional info required is - the hypotenuse and one leg of one right-angled triangle are equal to the corresponding hypotenuse and leg of another right-angled triangle.

Answer 2

Answer:

Final answer:

To prove that angle RST and angle TUR are congruent, we need the additional information that the length of the sides RS and TU are equal.

Explanation:

In order to prove that angle RST and angle TUR are congruent, we need the additional piece of information that the length of the sides RS and TU are equal. This is because congruent right angles must have congruent sides.

Learn more about Congruent Angles here:

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If I had 60 units needed and units per case was 14 how many cases and additional units I got

Answers

The additional number of units required is 10 units.

Given that:

Total number of units needed = 60 units

Total number of units per case = 14

How to find the total number of cases required for the given question?

So, the total number of cases required=Number of units needed / number of units per case.

Number of cases required = 60/14 = 4.285 (this means that 5 cases are required as 4 cases won't be up to 60 units)

With 5 cases, we have exceeded the required units needed :

Additional units will be : (14×5) - 60

Additional units = 70 - 60 = 10 units

Therefore, the additional number of units required is 10 units.

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Answer:

4 with a remainder of 4

Step-by-step explanation:

60 divided by 14 is 4 with a remainder of 4

Write the complex number 4(cos 60 + i sin 60) in standard form 10. Use DeMoivre's Theorem to find (2+3i)6

Answers

Answer:

a) The standard form of $z = 4\cdot (\cos 60^(\circ)+i\cdot \sin 60^(\circ))$ is $z = 2 + i\cdot 2√(3)$ , b) $z = (2+i\cdot 3)^(6) = 1219.585 + i \cdot 1829.381$ .

Step-by-step explanation:

a) The standard form of the complex number is $z = a + i\cdot b$ , $\forall \,a,b \in \mathbb{R}$ . If we get that $z = 4\cdot (\cos 60^(\circ)+i\cdot \sin 60^(\circ))$ , whose standard form is obtained by algebraic means:

1) $z = 4\cdot (\cos 60^(\circ)+i\cdot \sin 60^(\circ))$ Given

2) $z = (4\cdot \cos 60^(\circ))+i\cdot (4\cdot \sin 60^(\circ))$ Distributive and Associative properties.

3) $z = 2 + i\cdot 2√(3)$ Multiplication/Result.

The standard form of $z = 4\cdot (\cos 60^(\circ)+i\cdot \sin 60^(\circ))$ is $z = 2 + i\cdot 2√(3)$ .

b) The De Moivre's Theorem states that:

$z = (a+i\cdot b)^(n)= r^(n)\cdot (\cos \theta + i\cdot \sin \theta)$

Where:

$r =\sqrt{a^(2)+b^(2)}$ and $\theta = \tan^(-1) \left((b)/(a)\right)$ .

If we know that $z = (2+i\cdot 3)^(6)$ , then:

$r = \sqrt{2^(2)+3^(2)}$

$r =√(13)$

$r \approx 3.606$

$\theta = \tan^(-1)\left((3)/(2) \right)$

$\theta \approx 56.310^(\circ)$

The resulting expression is:

$z = 3.606^(6)\cdot (\cos 56.310^(\circ)+i\cdot \sin 56.310^(\circ))$

$z = 1219.585+i\cdot 1829.381$

Therefore, $z = (2+i\cdot 3)^(6) = 1219.585 + i \cdot 1829.381$ .

Find the real value of x,y if

(1/(x^2+y^2))+(1/(x+yi))=1

Answers

It sounds like $x,y$ are supposed to be real numbers. If so, then we can do the following.

$\frac1{x^2+y^2}+\frac1{x+yi}=1$

Multiply the second term's numerator and denominator by the conjugate of the denominator:

$\frac1{x^2+y^2}+(x-yi)/((x+yi)(x-yi))=1$

$\frac1{x^2+y^2}+(x-yi)/(x^2+y^2)=1$

$(x+1-yi)/(x^2+y^2)=1$

Since the left hand side is equal to 1, this means it has no imaginary part, so that $y=0$ . Then the real parts of both sides of the equation give us

$(x+1)/(x^2)=1\implies x+1=x^2\implies x^2-x-1=0\implies x=\frac{1\pm\sqrt5}2$

PLSSS ANSWER I DONT HAVE LOTS OF TIMEEE!!?!?!?!?!?!??!??!??! Three families took a vacation to Phoenix. The Abbott Family spent $840 on four nights at their hotel and three airline tickets. The Baker Family spent $1,130 on nine nights at their hotel and two airline tickets. The Carson Family spent $2,080 on six nights at their hotel and five airline tickets. Two of the families paid the same price per night at their hotel and per airline ticket.Which two families would have had to pay the same price per night at their hotel and per airline ticket to have the largest possible nightly hotel cost?
A.
the Abbott and Baker Families

B.
the Abbott and Carson Families

C.
the Carson and Baker Families

D.
all three families would have to pay the same price

Answers

Answer:

Step-by-step explanation:

both of the familys have higher cost and have to stay longer

Final answer:

After solving systems of equations for each pair of families, the Abbott and Carson Families are the ones that fit the conditions allowing the largest possible nightly hotel cost.

Explanation:

Let's examine the vacation expenses for each family to determine which two families could have paid the same price per night at their hotel and per airline ticket for the largest possible nightly hotel cost. We must establish a system of equations to represent the total vacation expenses in terms of hotel cost per night (H) and airline ticket cost (A).

For the Abbott Family:
4H + 3A = $840

For the Baker Family:
9H + 2A = $1,130

For the Carson Family:
6H + 5A = $2,080

To determine which two families could have the same hotel and ticket costs, we should look for the two equations that can be true simultaneously for the highest values of H. For ease of calculation, we can eliminate one family at a time and solve the remaining pair of equations for H and A.

By trying different combinations, we can find that the Abbott and Carson Families fit the conditions allowing the largest possible nightly hotel cost. Here's the reasoning:

If the Abbott and Baker families paid the same rates, then solving the equations yields illogical negative values for the costs.
If the Carson and Baker families paid the same rates, the solution yields a lower hotel rate compared to the Abbott and Carson combination.
Using the Abbott and Carson families' equations, and solving for H and A yields feasible positive values, with the highest cost per night for the hotel.

Hence, option B, the Abbott and Carson Families, is the right choice for the largest possible nightly hotel cost.