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Nuetrik [128]
4 years ago
12

Which of the following must be shown in order to prove that two irregular polygons are similar?

Mathematics
1 answer:
mafiozo [28]4 years ago
5 0
All corresponding angles are congruent.
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Find the volume of the figure.<br> 2.25 m<br> 4<br> m<br> 2.5 m<br> no
liberstina [14]

<u>Given</u>:

Given that the figure is a triangular prism.

The length of the prism is 4 m.

The base of the triangle is 2.5 m.

The height of the triangle is 2.25 m.

We need to determine the volume of the triangular prism.

<u>Volume of the triangular prism:</u>

The volume of the triangular prism can be determined using the formula,

V=\frac{1}{2}bhl

where b is the base of the triangle,

h is the height of the triangle and

l is the length of the prism.

Substituting b = 2.5, h = 2.25 and l = 4 in the above formula, we get;

V=\frac{1}{2}(2.5)(2.25)(4)

V=\frac{1}{2}(22.5)

V=11.25 \ m^3

Thus, the volume of the triangular prism is 11.25 m³

4 0
3 years ago
Consider the degree 5 polynomial function P(x)=x^5-4x^3+2x^2+3x-5. You do not need to factor this polynomial to answer the quest
olga2289 [7]

Answer:

5.

Step-by-step explanation:

According to the Linear Factorization Theorem, a polynomial function will have the same number of factors as its degree. We also know that these factors will have the form (x - c) where c is a complex number.

In this problem, and without factoring the polynomial, we observe that the <u>polynomial function is a 5 degree function so we can know that it will have 5 linear factors. </u>

<u />

3 0
3 years ago
Solve the following equation:
Rama09 [41]

Complete the square.

z^4 + z^2 - i\sqrt 3 = \left(z^2 + \dfrac12\right)^2 - \dfrac14 - i\sqrt3 = 0

\left(z^2 + \dfrac12\right)^2 = \dfrac{1 + 4\sqrt3\,i}4

Use de Moivre's theorem to compute the square roots of the right side.

w = \dfrac{1 + 4\sqrt3\,i}4 = \dfrac74 \exp\left(i \tan^{-1}(4\sqrt3)\right)

\implies w^{1/2} = \pm \dfrac{\sqrt7}2 \exp\left(\dfrac i2 \tan^{-1}(4\sqrt3)\right) = \pm \dfrac{2+\sqrt3\,i}2

Now, taking square roots on both sides, we have

z^2 + \dfrac12 = \pm w^{1/2}

z^2 = \dfrac{1+\sqrt3\,i}2 \text{ or } z^2 = -\dfrac{3+\sqrt3\,i}2

Use de Moivre's theorem again to take square roots on both sides.

w_1 = \dfrac{1+\sqrt3\,i}2 = \exp\left(i\dfrac\pi3\right)

\implies z = {w_1}^{1/2} = \pm \exp\left(i\dfrac\pi6\right) = \boxed{\pm \dfrac{\sqrt3 + i}2}

w_2 = -\dfrac{3+\sqrt3\,i}2 = \sqrt3 \, \exp\left(-i \dfrac{5\pi}6\right)

\implies z = {w_2}^{1/2} = \boxed{\pm \sqrt[4]{3} \, \exp\left(-i\dfrac{5\pi}{12}\right)}

3 0
2 years ago
I WILL GIVE BRAINLY
Alenkinab [10]

Answer:

5(6)+0.65y

Step-by-step explanation:

$5(6hrs)+65÷100=0.65

=5(6)+0.65y

6 0
3 years ago
1. which statement completes the syllogism?
Alex777 [14]
A) all squares are polygons. I think
5 0
3 years ago
Read 2 more answers
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