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3 years ago

In triangle STU, u2 = s2 + t2.

Mathematics

1 answer:

sineoko [7]3 years ago

3 0

Answer:

(B) The measure of angle SUT is equal to 90 degrees

Step-by-step explanation:

It is given that in In triangle STU, $u^2=s^2+t^2$ .

We can see from the triangle and also the given condition that the triangle will be right triangle as it is satisfying the Pythagoras theorem.

$(Hyp)^2=(Base)^2+(Perpendicular)^2$

that is:

$u^{2}=s^{2}+t^{2}$

where u is the hypotenuse, s is the base and t is the perpendicular.

Thus, the angle SUT should be of 90°.

Thus, The measure of angle SUT is equal to 90 degrees.

Hence, option B is correct.

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Is 380 less than 248

Oliga [24]

Answer:

not unless its a negative integer

Step-by-step explanation:

380 is more than 248 but if 380 was put as a negative, it would be seenon the number line as less.

example: -380.......0........248

example 2: 0.....248.....380

8 0

3 years ago

Factorize: x² + 3x + 2

zvonat [6]

Step-by-step explanation:

x² + 3x + 2

x² + x + 2x + 2 because 1+2=3 and 2×1=2

By taking out common terms, we get

x(x+1)+2(x+1)

(x+1)(x+2)

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3 years ago

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riadik2000 [5.3K]

Let

$S_n = \displaystyle \sum_{k=0}^n r^k = 1 + r + r^2 + \cdots + r^n$

where we assume |r| < 1. Multiplying on both sides by r gives

$r S_n = \displaystyle \sum_{k=0}^n r^{k+1} = r + r^2 + r^3 + \cdots + r^{n+1}$

and subtracting this from $S_n$ gives

$(1 - r) S_n = 1 - r^{n+1} \implies S_n = \dfrac{1 - r^{n+1}}{1 - r}$

As n → ∞, the exponential term will converge to 0, and the partial sums $S_n$ will converge to

$\displaystyle \lim_{n\to\infty} S_n = \dfrac1{1-r}$

Now, we're given

$a + ar + ar^2 + \cdots = 15 \implies 1 + r + r^2 + \cdots = \dfrac{15}a$

$a^2 + a^2r^2 + a^2r^4 + \cdots = 150 \implies 1 + r^2 + r^4 + \cdots = \dfrac{150}{a^2}$

We must have |r| < 1 since both sums converge, so

$\dfrac{15}a = \dfrac1{1-r}$

$\dfrac{150}{a^2} = \dfrac1{1-r^2}$

Solving for r by substitution, we have

$\dfrac{15}a = \dfrac1{1-r} \implies a = 15(1-r)$

$\dfrac{150}{225(1-r)^2} = \dfrac1{1-r^2}$

Recalling the difference of squares identity, we have

$\dfrac2{3(1-r)^2} = \dfrac1{(1-r)(1+r)}$

We've already confirmed r ≠ 1, so we can simplify this to

$\dfrac2{3(1-r)} = \dfrac1{1+r} \implies \dfrac{1-r}{1+r} = \dfrac23 \implies r = \dfrac15$

It follows that

$\dfrac a{1-r} = \dfrac a{1-\frac15} = 15 \implies a = 12$

and so the sum we want is

$ar^3 + ar^4 + ar^6 + \cdots = 15 - a - ar - ar^2 = \boxed{\dfrac3{25}}$

which doesn't appear to be either of the given answer choices. Are you sure there isn't a typo somewhere?

7 0

2 years ago

What is the factor of <br><img src="https://tex.z-dn.net/?f=%20%7Bx%7D%5E%7B2%7D%20%20-%206x%20-%2027" id="TexFormula1" title="

Ksju [112]

Answer:

(x - 9)(x + 3)

Step-by-step explanation:

Given

x² - 6x - 27

Consider the factors of the constant term (- 27) which sum to give the coefficient of the x- term (- 6)

The factors are - 9 and + 3, since

- 9 × 3 = - 27 and - 9 + 3 = - 6, thus

x² - 6x - 27 = (x - 9)(x + 3)

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3 years ago

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lana [24]

Answer:

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3 years ago