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4 years ago

Find the missing dimension of the triangle. DETAILED EXPLANATION.

Mathematics

2 answers:

Lisa [10]4 years ago

6 0

Answer:

$h=2\frac{2}{3}in.$

Step-by-step explanation:

Formula

$A=\frac{hb}{2}$

Equation

$h=2\frac{A}{b}$

$h=2(\frac{2}{1.5})$

$h=2\frac{2}{3}$

Check

$2\frac{2}{3}$ · $1\frac{1}{2}=4$

4 ÷ 2 = 2 in.²

sineoko [7]4 years ago

5 0

Answer:

1.75 in

Step-by-step explanation:

kyo na bhala dyan sa sgot

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Where is 1-6 at ????

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3 years ago

Multiply: 4/17 x − 3 /4 need help

beks73 [17]

$\bf \cfrac{4}{17}\times -\cfrac{3}{4}\implies \cfrac{4}{17}\times \cfrac{-3}{4}\implies \cfrac{4\times -3}{17\times 4}\implies \cfrac{-12}{68}\implies \cfrac{-3}{17}$

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4 years ago

The first card selected from a standard 52-card deck was a king. if it is not returned to the deck, what is the probability that

Sunny_sXe [5.5K]

4 kings total out of 52 cards

1 king was picked so 3 are left, 52 -1 = 51 cards left

so you have a 3/51 reduced to a 1/17 probability of picking another king

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3 years ago

Liem is 6 feet 2 inches, Eli is 5 feet 9 inches, Faith is 6 feet, and Simon is 5 feet 4 inches. In yards, what is the total of t

Oksi-84 [34.3K]

<span>First step is find total height of all the figures expressed in feet: Liem: 6 ft 2 inch = 6,167 ft Eli: 5 ft 9 inch = 5,75 ft Faith 6 ft Simon: 5 ft 4 inch = 5,34 ft To calculate the total height we need to know that 12 inches are 1 foot 6â€™2â€ť+5â€™9â€ť+6+5â€™4â€ť= 22 ft 15 inch = 23 ft 3 inch Another way is by adding decimal figurel: 6,167 + 5,75 + 6 + 5,34 = 23,25 feet Then we know that 3 feet are 1 yard, so then total height is: 23,25 / 3 = 7,75 yards</span>

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3 years ago

A student claims that 2i is the only imaginary root of a polynomial equation that has real coefficients. Explain the student's m

____ [38]

Answer:

The Fundamental Theorem of Algebra assures that any polynomial f(x)=0 whose degree is n ≥1 has at least one Real or Imaginary root. So by the Theorem we have infinitely solutions, including imaginary roots ≠ 2i

Step-by-step explanation:

1) This claim is mistaken.

2) The Fundamental Theorem of Algebra assures that any polynomial f(x)=0 whose degree is n ≥1 has at least one Real or Imaginary root. So by the Theorem we have infinitely solutions, including imaginary roots ≠ 2i with real coefficients.

$a_{0}x^{n}+a_{1}x^{2}+....a_{1}x+a_{0}$

For example:

3) Every time a polynomial equation, like a quadratic equation which is an univariate polynomial one, has its discriminant following this rule:

$\Delta < 0\\b^{2}-4*a*c$

We'll have <em>n </em>different complex roots, not necessarily 2i.

For example:

Taking 3 polynomial equations with real coefficients, with

$\Delta < 0$

$-4x^2-x-2=0 \Rightarrow S=\left \{ x'=-\frac{1}{8}-i\frac{\sqrt{31}}{8},\:x''=-\frac{1}{8}+i\frac{\sqrt{31}}{8} \right \}\\-x^2-x-8=0 \Rightarrow S=\left\{\quad x'=-\frac{1}{2}-i\frac{\sqrt{31}}{2},\:x''=-\frac{1}{2}+i\frac{\sqrt{31}}{2} \right \}\\x^2-x+30=0\Rightarrow S=\left \{ x'=\frac{1}{2}+i\frac{\sqrt{119}}{2},\:x''=\frac{1}{2}-i\frac{\sqrt{119}}{2} \right \}\\(...)$

2.2) For other Polynomial equations with real coefficients we can see other complex roots ≠ 2i. In this one we have also -2i

$x^5\:-\:x^4\:+\:x^3\:-\:x^2\:-\:12x\:+\:12=0 \Rightarrow S=\left \{ x_{1}=1,\:x_{2}=-\sqrt{3},\:x_{3}=\sqrt{3},\:x_{4}=2i,\:x_{5}=-2i \right \}\\$

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3 years ago