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2 years ago

Find m∠U. Write your answer as an integer or as a decimal rounded to the nearest tenth.

Mathematics

1 answer:

Marysya12 [62]2 years ago

4 0

Answer:

Answer:

m∠U.=50.2

Step-by-step explanation:

Answer:

50.2

Step-by-step explanation:

$ut = \sqrt{ {6}^{2} + {5}^{2} } \\ \sqrt{36 + 25} \\ \sqrt{61 } \\ using \: cosine \: rule \\ cos \: u = \frac{ {5}^{2} + 61 - 36 }{2 \times 5 \times \sqrt{61} } \\ \\ = \frac{25 + 61 - 36}{78} \\ \\ = \frac{50}{78 } \\ \cos \: u = 0.64 \\ m \:u = 50.2$

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Select all of the following expressions that are equivalent to 0.25.

lora16 [44]

1. (0.5)2=1

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3. |-3 + 2.75| = 0.25

4. -(-0.25) = 0.25

5. 2x - 1.75x = 0.25x

So your answers are 3 and 4.

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2 years ago

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Step-by-step explanation:

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2 years ago

Read 2 more answers

The six numbers in a set are 137, 128, 250, 136, 129, and 138. Which measures of the data increase when the outlier is included?

julia-pushkina [17]

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3 years ago

Find the length of the curve y = integral from 1 to x of sqrt(t^3-1)

Arlecino [84]

$y=\displaystyle\int_1^x\sqrt{t^3-1}\,\mathrm dt$

By the fundamental theorem of calculus,

$\dfrac{\mathrm dy}{\mathrm dx}=\dfrac{\mathrm d}{\mathrm dx}\displaystyle\int_1^x\sqrt{t^3-1}\,\mathrm dt=\sqrt{x^3-1}$

Now the arc length over an arbitrary interval $(a,b)$ is

$\displaystyle\int_a^b\sqrt{1+\left(\frac{\mathrm dy}{\mathrm dx}\right)^2}\,\mathrm dx=\int_a^b\sqrt{1+x^3-1}\,\mathrm dx=\int_a^bx^{3/2}\,\mathrm dx$

But before we compute the integral, first we need to make sure the integrand exists over it. $x^{3/2}$ is undefined if $x$ , so we assume $a\ge0$ and for convenience that $a$ . Then

$\displaystyle\int_a^bx^{3/2}\,\mathrm dx=\frac25x^{5/2}\bigg|_{x=a}^{x=b}=\frac25\left(b^{5/2}-a^{5/2}\right)$

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3 years ago

What are 3 consecutive integers whose product is -120

iren2701 [21]

Consecutive integers are 1 apart
x,x+1,x+2
(x)(x+1)(x+2)=-120
x^3+3x^2+3x=-120
add 120 to both sides
x^3+3x^2+3x+120=0
factor
(x+6)(x^2-3x+20)=0
set each to zero

x+6=0
x=-6

x^2-3x+20=0
will yeild non-real result, discard
x=-6
x+1=-5
x+2=-4
the numbers are -4,-5,-6

use trial and error and logic
factor 120
120=2*2*2*3*5
how can we rearange these numbers in (x)(y)(z) format such that they multiply to 120?
obviously, the 5 has to stay since 2*5=10 which is out of range
so 2*2*2*3 has to arrange to get 3,4 or 4, 6 or 6,7
obviously, 7 cannot happen since it is prime
3 and 4 results in in 12, but 2*2*2*3=24
therfor answer is 4 and 6
they are all negative since negaive cancel except 1

the numbers are -4,-5,-6

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2 years ago