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3 years ago

(3x^2 - 2x)(2x^2 + 3x - 1) Polynomials

Mathematics

1 answer:

dedylja [7]3 years ago

3 0

Answer:

Step-by-step explanation:

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Pls find x, has to do with trig and answer has to be in radical form

MrMuchimi

I don’t have a calculator with me use sine 20divided by Sin(60) will give you x

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3 years ago

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Which statement is true of triangles FGH and JKL?

Mars2501 [29]

Answer:

They are similar because their corresponding angles are congruent and their corresponding sides are proportional in length.

Step-by-step explanation:

From the statements,

∠G≅∠K

∠F≅∠J

∠H≅∠L

The sides of triangle FGH are proportional to those of triangle JKL

FG/JK =GH/KL=FH/JL =1/2

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Find the arc length of the partial circle.

nalin [4]

Answer:

1/2 pi

Step-by-step explanation:

This is 1/4 of a circle

Find the circumference and multiply by 1/4

1/4 C

1/4 ( 2*pi*r) where r is the radius

The radius is 1

1/2 pi*1

1/2 pi

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3 years ago

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Jacky read at the rate of 15 pages per hour. Johnny read at the rate of 1/3 of a pages per minute. Who was reading faster? By ho

o-na [289]

We can compare by converting Johnny's reading time into the same unit of Jacky's, hours.

There are 60 minutes in an hour, so use this:

$\frac{1}{3} * \frac{60}{1} = 20$

15 < 20, so Johnny read faster. This is 5 pages faster per hour than Jacky.

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3 years ago

Triangle ABC has AB=5, BC=7, and AC=9. D is on AC with BD=5. Find the length of DC

e-lub [12.9K]

Answer:

$DC=\frac{8}{3}\ units$

Step-by-step explanation:

The picture of the question in the attached figure

we know that

The triangle ABD is an isosceles triangle

because

AB=BD

The segment BM is a perpendicular bisector segment AD

In the right triangle ABM

Applying the Pythagorean Theorem

$BM^2=AB^2-AM^2$

we have

$AB=5\ units\\AM=x\ units$

substitute

$BM^2=5^2-x^2$

$BM^2=25-x^2$ -----> equation A

In the right triangle BMC

Applying the Pythagorean Theorem

$BM^2=BC^2-MC^2$

we have

$BC=7\ units\\MC=AC-AM=(9-x)\ units$

substitute

$BM^2=7^2-(9-x)^2$

$BM^2=49-(81-18x+x^2)$

$BM^2=49-81+18x-x^2$

$BM^2=-x^2+18x-32$ ----> equation B

equate equation A and equation B

$-x^2+18x-32=25-x^2$

solve for x

$18x=25+32\\18x=57\\\\x=\frac{57}{18}$

Simplify

$x=\frac{19}{6}$

Find the length of DC

$DC=AC-2x$

substitute the given values

$DC=9-2(\frac{19}{6})$

$DC=9-\frac{19}{3}\\\\DC=\frac{8}{3}\ units$

8 0

3 years ago